7.

2 DEFINITIONS
distribution and u and
Quantities appearing in distributions,
such as p in the binomial
in the normal distribution are called parameters.
a
No. 1), May 2012]
Estimate JNTU (H) Nov. 2009, (A) 201OS, Nov. 2011 (Set
An estimate is a statement made to findunknown population parameter.
an

Estimator JNTU (H) Nov. 2009, (A) 2010S, Nov. 2011 (Set No. 1), May 2012]
The rule to determine unknown population parameter is called an
procedure or an

estimator. For instance, sample mean is an estimator of population mean because sample
mean is a method of determining the population mean. Remember that an estimator must bea
statistic and it must depend only on the sample and not on the parameter to the estimated. So
an estimator is a statistic which for all practical purposes, can be used in place of unknowwn
parameter of the population.
Aparameter can have one or two or many estimators.
Types of Estimation
Basically, there kinds of estimates to determine the stati_tic of the
are two
population
parameters namely, (a) Point Estimation and (b) Interval Estimation.
Statistical Inference
 (1) Hypothesis testing -
to test some hypothesis about parent population from which
the sample is drawn.
(2) Estimation -
the statistics obtained from the sample as estimate of the unknown
to use
parameter of the population from which the sample is drawn.
An important problem of Statistical Inference is the estimation of population parameters
(i.e., population mean, population S.D, etc.) from the corresponding sample statistics (i.e.,
sample mean, sample S.D. etc)
Point Estimation and Interval Estimation JNTU(K) Nov. 2011 (Set No. 2)]
A point estimate of a parameter is a number (point on the real axis) which is computed
from a given sample and serves as an approximation of the unknown value of the
parameter. An
interval estimate is an interval obtained from a sample.
If an estimate of the population parameter is given by a single value, then the estimate is
called a Point Estimation of the parameter. But if an estimate of a population parameter is
given by two different values between which the parameter may be considered to lie, then the
estimate is called an interval estimation of the parameter.
Example : If the height of a student is measured as 162 cms, then the measurement
gives a point estimation. But if the hcight is given as (163 +3.5) cms, then the height lies
between 159.5 cms and 166.5 cms and the mea_urement gives an interval estimation.

The sample mean is a point estimate of population mean 4, sample variance s2 is a

point estimate of population variance os.

Definition: A point estimate of a parameter 0 is a single numerical value, which is
computed from a given sample and serves as an approximation of the unknown exact value of
the parameter.
Definition: A point estimator is a statistic for estimating the population parameter 6

and will be denoted by (read as theta hat).

Properties of Estimation: An estimator is not expected to estimate the population
parameter without error. An estimator should be close to the true value ofunknown parameter.
Unbiased and Biased Estimates
A statistic is said to be an unbiased estimator of the corresponding parameter if the
mean of the sampling distribution of the statistic is equal to the corresponding population
parameter. Otherwise the statistic is called a biased estimator of the corresponding
parameter. The values of statistics in the above two cases are called unbiased and biased
estimates respectively.
 Unbiased Estimator: Let 6 be an estimator of e. The statistic e is said to be an
unbiased estimator, or its value an unbiased estimate, if, and only if the mean or expected value

of
of 66 is
is equal
equal to 6.
0. This is equivalent to say that the mean of the probability distribution

af (or the mean of the sampling distribution of 6) is equal to 6. An estimator possessing this
property is said to be unbiased.
Definition. Unbiased estimator: A statistic or point estimator e is said to be an unbiased
estimator of the parameter 6 if E (0) = 0.
In other words, if E (statistic) = parameter, then statistic is said to be an unbiased
estimator of the parameter.

Example 1: Show that S is an unbiased estimator of the parameter o

Solution: Let us write

-) = [-)-7-*
i=l i=l

2-H-2(-H)
i=l
-H) +n (F-p?
i=1

- -)-nF-u* (1)
i=l

Now E(S ) =
E i=
n-1

n-1
E-u-n E(F-p)
Li=l
using (1)

o n-1 i=1
 However, ,= afori =
1, 2,.,n and o =

ES)n-1
Although S2 is an unbiased estimator of o,S, on the other hand, is a biased estimator of
with the bias becoming insignificant for
large samples. This example illustrates why we
divide by (n-1) rather than n when the variance is
estimated.
 7.3 INTERVAL ESTIMATION JNTU (A) Nov. 2011 (Set No. 3,4)]
Point estimates rarely coincide with quantities they are intended to estimate. So instead
ofpoint estimation where the quantity to be estimated is replaced by a single value a better way
of estimation is interval estimation, which determines an interval in which the parameter lies.
Interval Estimate: Even the most efficient unbiased estimator cannot estimate the
population parameter exactly. It is true that our accuracy increases with Jarge samples. But
there is still no reason why
should expect a point.estimate from
we a given sample to be
exactly equal to the population parameter, it is supposed to estimate.
Therefore in many situations it is preferable to determine an interval within which we
would expect to find the value of the parameter. Such an interval is called an interval estimate.
Thus interval estimate is an interval (confidence interval) obtained from
an a sample.
Interval Estimation: An interval estimate of a population parameter e is an interval of
the form
6 <0 <6,, where ê, and ê, depend on the value ofthe statistic ê for a particular
im
sample and also on the sampling distribution of 0

Since different samples will generally yield different values of ê and, therefore, different

e
values and 6,. These end points of the interval are values of coresponding random variables
6 and
 The formulae for confidence limits of
some well-known statistic for large random
samples are given below.
I Confldence limits for
Population Mean f
() 95% confidence limits are xt1.96 (S.E. of 7)
(i) 99% confidence limits are 7+2.58 (S.E. of x)
(ii) 99.73% confidence limits are 7t3 (S.E. of F)
(iv) 90% confidence limits are
Ft1.64 (S.E. of )
II. Confidence limits for
Population ProportionP
() 95% confidence limits are pt1.96(S.E. of p)
(it) 99% confidence limits are pt2.58 (S.E. of p)
(iii) 99.73% confidence limits are
pt3(S.E. of p)
(iv) 90% confidence limits are pt1.64 (S. E. ofp)
LConfidence limits for the difference - of two Population Means H and ^
() 95% confidence limits are (-7)t1.96 (S.E. of
(7-))
(it) 99% confidence limits are -2)£2.58 (S.E. of
(-))
(ii) 99.73% confidence limits are (G-F) t3 (S.E. of ( -T))
(iv) 90% confidence limits are
(-E)t1.64(S.E. of ( -,))
IV Confidence limits for the difference - P of two population
proportions
() 95% confidence limits are (p P2)£1.96 (S.E. of (p -P))
-
 99%
(i) 99% confidence limits are (pi-P,)t2.58 (S.E. of (P - P2))

( :h 9.73% confidence
i )99.73% limits are (P P2)t3 (S.E. of (P- P2))

(iv) 90% confidence limits are (p -Pa)t1.64 (S.E. of (p- Pa)

Note. I1f
f forstatistic, P(-3<z<3) ie., confidence limits cover 99.73% of the area
a
statistic
underthe standard normal curve, then we
standard say that the confidence limits for the
the
smst sure limits without mentioning the degree of confidence.
arealmostsure.
S.E. if population parameters are unknown, corresponding sample
In calculating
to find an approximate value of S. E. For example, if the population
istics are used
tistics are
of o .
sD. o is given, the S.E. of ï is calculated by using sample S.D. s in place
not
5 DETERMINATION OF PROPER SAMPLE SIZE
in businesS
Determination of proper sample size is important for testing of hypothesis the
nor too large. If the size of
roblems. The size of sample should neither be too small
hand, if the size
sarmple is too small, then it may not give a valid conclusion. On the other
is too large, then there be loss of time and money without getting the
ofthe sample may
required results.

1. Sample size for estimating Population Mean

and
Let be the mean of a random sample drawn from a population having
mean

I be approximately a normal
S.D. o. Let the sampling distribution of the sample mean
distribution with mean and S. D. o. If E be the permissible sampling error, then

E--.
The confidence interval for the population mean is 7tz (S.E. of 7) = FtE
where z = confidence coefficient or z- value (which is 1.96 at 5% level of significance)

and E=z (S.E. of 7) =z,'n' being the sample size and o= population S.D.
n

20
Thus E or or vn=

n= , which gives therequired sample size for estimating population mean.

E
2. Samnlo eizc fo. Eosfmotina Ponulation Proportion
 312 Computer Oriented Statistical Meh
Methods
Substituting for t, we write Pan S/ =-a
Multiplying each term in the inequality by s/Vn, and then subtracting x from each tem
and multiplying by
tem
-

1, we obtain

PF-a/m)<p<F+(aia (S/Vm)=1-a
Confidence interval for 4, o unknown: JNTU (A) Nov. 2011 (Set No.31
If ands are the mean and standard deviation of a random sample from a nommal
al
population with unknown variance o , a(1-a)100% confidence interval for is

-auas/wn) <p <++lan(s/vn)

where a/2 is the t- value with v= n -1 degrees offreedom, leaving an area of a/2 to the
right.
The maximum error of estimate for small samples is given by
E= a/a-s/Wn, where n =
Sample size, s =
S.D ofsample
7.6 BAYESIAN ESTIMATION JNTU (A), (H) Dec 2009,(A) Nov. 2011 (Set No.4)
(This Topic is for Student Reference only)
The new concept introduced in Bayesian methods is personal or
subjective probability.
Also, parameters are considered as random variable in Bayesian method. To estimate the mean
ofa population, is treated as a random variable whose distribution is indicative of the "strong
feclings" or assumption ofa person about the possible value of 4. Let po and oo be the mean
and standard deviation
of such a subjective "prior distribution".
Bayesian Estimation
Combining the prior feelings about the possible values of with direct sample evidence,
the "posterior" distribution of p in Bayesian estimation is
approximated by normal distribution
with

and o ns+o
no+o
where n sample size, ï =
sample mean and S is the standard deviation of sample.
Use S o.
Here and o are known as the mean and standard deviation
of the posterior distribution.
In the computation of 4, and oj , a is assumed to be known, when 2 is unknown, whichs
generally the case, is replaced by sample variance S" provided n> 30 (large sample).
Bayesian interval for :
(1- a) 100% Bayesian interval for H is given by
-Za/2 0j <<k1+Za/2 +O1
 SOLVED EXAMPLES
Example 1I &In a study of an automobile insurance a random sample of 80 body repair
Example
had a mean
had a
of Rs..
m e a n of Rs. 472.36 and the S.D of Rs. 62.35. If 7 is used as a point estimate
to
cOsts
average repair costs, with what confidence we can assert that the maximum error
the true averag

doesn't exceed Rs. 10.

JNTU Dec May 2010S (Set No. 2)|
2004S, (K)
Solution: Size ofa random sample, n = 80
The mean of random sample, f = Rs,. 472.36

Standard deviation, a= Rs. 62.35

Maximum error of estimate, E = Rs. 10

10/80 89.4427
We have E max a/2. = 1.4345
Vn 62.35 62.35
a/2 1.43
The area when z = 1.43 from tables is 0.4236.

= 0.4236a=0.8472
Confidence = (1 - a) 100% = 84.72 %

Hence we are 84.72% confidence that the maximum error is Rs. 10.
Example 2: What is the size of the smallest sample required to estimate an unknown
proportion to within a maximum error of 0.06 with atleast 95% confidence.
JNTU Dec. 2004 S, April 2006, (K) Nov. 2009,(A) Nov. 2010, (H) Sept. 2017
Solution: We are given
The maximum error, E =0.06
Confidence limit = 95%

ie. (1-a)]100=95

1-a= 0.95a =0.05a/2=0.025

Fa/2 =1.96
Here P is not given. So we take P= Thus
Q=
Hence n = Po)

When P is unknown, sample size n 1/2

E
=

11.96
4 L0.06
266.78 267
Example 3: If we can assert with 95% that the maximum error is 0.05 and P 0.2, =

find the size of

the
sample. JNTU Dec 2005, (H) Nov. 2010 (Set No. 1)]
Solution: Given P =0.2, E = 0.05
We have Q=0.8 and zal2 = 1.96 (tor 95%)
 We know that maximum error, E= *a/2

0.2 x 0.8
0.05 1.96
x0.8 (1.96)*
Sample size, n = 0.2
x
-246
246
(0.05)
Example4: Assuming that o = 20.0, how large a random sample be taken to ao
sample mean will not differ from the true mean by more thanassert
with probability 0.95 that the
points ? 3
JNTU Apr. 2005, Dec 2005
Solution: Given maximum errorE =3.0 and o = 20.0
We have Za/2 = 1.96

We know that, n

(1.96x20
- -170.74

.n 171

Example 5: It is desired to
estimate the mean number of hours of continuous use until
acertain computer will first
require repairs. If it can be assumed that o 48 hours, how large
=

a
sample needed so that one will be able to assert with 90% confidence that the
be
is off by at most 10 hours. sample mean
JNTU Dec 2004, April 2006, JNTU (K) Nov. 2009, (H) Nov.2010 (Set No.3)]
Solution: It is given that
Maximum error, E = 10 hours
48 hours
and a/2 = 1.645 (for 90%)

: -)- (1.645 x 48
10
= 62.3
= =
62
Hence sample size = 62

Example 6: What is the maximum error one can expect to make with probability u.>
when using the mean of a random sample of size n = 64 to estimate the mean of population witn

2.56. JNTU 2004S, (K) Nov. 2009, (H) May 2011 (Set No. 3), Sept. 2017
Solution: Here n = 64. The probability = 0.90

o 2.56 0 = v2.56 = 1.6

Confidence limit = 90%%

Fa/2 1.645
 315
sifmalion

1.6
e maximum error E= a/ 1.645 x =0.329.

What can you

rample 7:Arandom sample ofsize 100 has a standard deviation of 5.
the maximum error with 95%.confidence. JNTU (A)Apr. 2012 (H) Sept. 20171
ahout 1

iar) If n=100,g =5, Iind the maximum error with 95% confidence limits.
JNTU (H) May 2017

Solution: Given S. D. d =5, n=100

1.96
for95% confidence
=

We know that,
maximum error, E =
Za/2 .

E =
(1.96). V100 0.98

to
Example 8: The efficiency expert of a computer company tested 40 engineers
a mean or
imate the average time it takes to assemble a certain computer component, getting
12.73 minutes and S.D of 2.06 minutes.
actual average time required to
(a) If i =12.73 is used as a point estimate of the
perform the task, determine the maximum eror with 99% confidence.
h) Construct 98% confidence intervals for the true average time it takes to do the job.

(c) With what confidence can we assert that the sample mean does not differ from the
true mean by more than 30 seconds. JNTU (K) Nov. 2009 (Set No. 1)]
(OR)
To estimate the average time it takes to assemble a certain computer component, the
at an electronic firm timed 40 technicians in the performance of the task,
industrial engineer
geting a mean of 12.73 min. and a S. D. of 2.06 min.

(i) What say with 99% confidence about the

can we
maximum error if F =12.73 is

ed as a point estimate of the actual average time required to do the job?

u) Use the given data to construct 98% confidence interval.
ii) With what confidence we can assert that the sample mean does not differ from the
Iuc mean by more than 30 sec. JNTU (A)Apr. 2012 (Set No. 1)1
For 99%, Za/2 = 2.575
Solution: Here = 12.73, S= 2.06, n = 40

a) Maximum error of estimate E= za/2

(2.06)
E=
(2.575). =0.8387
40
)For 98% confidence, E= (2.35) .

(2.06)0,758915
40
98% confidence interval limits are
 Computer Oriented Statistical Methods
316

=+tE=12.73+ t0.7589
*ta/
ie., confidence interval is (11.97, 13.4889)
30 a/2
2.06
(c) minutes = minutes E = a/2 40
60

Fana/2 1.5350.
correspondingto a/2
1.5350 is 0.4370.
From normal distribution table, the area

0.8740
Then the area between z-a/2 to a/n is
2 (0.4370) =

Thus we have 87.4% confidence.

population are 11.795 and 14.054
standard deviation of a
Example 9: The mean and maximum error if
the
can one assert with 95% confidence about
respectively. What
interval for the true mean.
I =11.795 and n 50. And also construct 95% confidence
JNTU (A)Apr.2012(Set No.4)
(OR)
ofa population are 11.795 and 14.054 respectively,
The mean and the standard deviation
If n 50, find 95% confidence interval for the mean.

UNTU 2006, 2008S, 2008 (Set No. 3), (K)

May 2013 (Set No.2)

Mean of population,
=
11.795
Solution: Here
14.054
S.D of population, o =
= 11.795
n sample size = 50

Maximum error= za/2

a/z for 95% confidence =

1.96

Max. eror, E = a/2 =1.96, 4.054) (14.054) 3.8955

V50
Confidence interval = ( - a/2 t za/2
Vn
= (11.795 -3.8955, 11.795 +3.8955)
= (7.8994, 15.6905)
an
A sample of 10 cam shafts intended for use ingasoline engines has
Example 10: Assuming the data may
be
standard deviation of 0.044 inch.
average eccentricity of1.02 and a confidence interval for the
treated a random sample from a normal population, determine a 95%
shaft ? UNTU 2004, (K) May 2010S (Set
No. 4)
actual mean eccentricity of the cam

Solution: We know that confidence interval is x

t E where E = a/2
vn

Given n= 10, za/2 1.96, o =0.044, r = 1.02

 317
E s t i m a t i o n

0.044
Confidence interval is 1.02t (1.96) 1.02 0.027

Confidence interval = (0.993, 1.047)

29.23 aro
Example Il:A random sample of size 81 was taken whose variance is
(Set No. 2)
is 32, construct 98% confidence interval. JNTU Dec 2005
mean
mea

Solution: Given
s a m p l e mean= 3 2 , n = 81,

o 20.25 a = 4.5 and a/n = 2.33 (for 98%)

We know that 98% confidence interval is 7-u / 2 t Fain T

Now aa -(2.33)=2.33x0.5=1.165
V81
Confidence interval = (32-1.165,32+1.165) = (30.835,33.165)

the average time it takes a mechanic

Example 12: A research worker wants to determine the
to rotate the tyres of a car
and he wants to be able to assert with 95%. Confidence that
that
mean of his sample is off by atmost 0.5
minutes. If he can presume from past experience
how large a sample will have to take? JNTU 1999
1.6 minutes,
Solution: We have Max. error, E =
0.5
n = sample size = ?

o standard deviation of the population =

1.6 minutes

a/2 1.96

Hence n [1.96x0.51.6= 40
40 mechanics will have to perform the task.
Example 13: It is desired to estimate the mean time ofcontinuous use until an answering
machine will first require service. I f it can be assumed thato = 60 days, how large a sample

is needed so that one will be able to assert with 90% confidence that the sample mean is ofT by
10 JNTU (H) Nov. 2012
at most days.
Solution: Wehave
Maximum error = 10 days = E, a = 60 days and z,/2 = 1.645

n E
1.645 x60 -97
Example 14: The dean of a college wants to use the mean of a random sample to
estimate the average amount of time students take to get from one class to the next and she
wants to be able to assert with 99% confidence that the error is at most 0.25 minute. If it can
be presumed from experience that a = 1.40 minutes. How large a sample will she have

to take? 1JNTU 2000,(H) May 20131

318 Computer Oriented Statistical
Methor
Solution:We are given
Maximum error, E = 0.25 minutes

standard deviation 1.40 minutes

=

Fa/2.575

Sample size, n 2.575x1.4=208

025
=

Evample 15:A random sample of size 100 is taken from a population witho=c
S..
Given that the sample mean is = 21.6. Construct a 95% confidence interval for the populati.
mean .
ion
JNTU 2001
Solution: Given = sample mean = 21.6,

a/21.96, n = sample size = 100, a=5.1

Confidence interval =
(7 -

Fa/2.G/n, ï + "a/2 .o/vn)

1.96x5.1
Now -Fa2 21.6- 20.6
10

and Tta/2- 22.6

Hence (20.6, 22.6) is the confidence interval for the population mean 4.
Example 16: The mean of random sample is an unbiased estimate of the mean of the
population 3, 6,9, 15, 27.
(i) List ofall possible samples of size 3 that can be taken without replacement from the
finite population.
(ii) Calculate the mean of each of the samples listed in (a) and assigning each sample a
probability of /10. Verify that the mean of these + is equal to 12. Which is equal
to the mean of the population 0 i.e E (%)
0 ie., prove that
=
is an unbiased
estimate of e. UNTU 2004s, 2008S, (K) Nov. 2009, (A) Nov. 2010 (Set No. 2)
Solution: () The possible samples of size 3 taken from 3, 6, 9, 15, 27 without
replacement, are °C=10 samples ie,(3,6,9), (3,6,15), (3,6,27), (6,9,15), (6, 9, 27), (3,9.
15), 3, 9, 27), (9, 15, 27), (6, 15, 27), (3, 15, 27).
3+6+9+15+27
(i) Mean of the population e =

5
= 12

Means of the samples are 6, 8, 12, 10, 14, 9, 13, 17, 16, 15.

Probability assigned to each one is each.

10

6 812 1014913 17 16 15
P() /10 1/10
y10 y0
1/10 /10 1/10 /10 1/10 10 1/10 1/10
1/10
 319
Estimation

1
AT)08 12 1 0130
+1710+16.015101 120 12=0
ET)=
i s an unbiased estimate of 0.
ie. the mean ofa random sample is an unbiased estimator of the mean ofthe populatio
Example 17: Suppose that we observe a random variable having the binomial distributio
successes in n trials.
and get r
(a) Show thatis an unbiased estimate ofthe binomial parameter p.
n

(b) Show that

X is not an unbiased estimate of the binomial parameter p.
nt2
Solution: 0 is an unbiased estimator of e if E (0)= 0

(a)
=
E(x) = =p [: Elx) = np]

i s an unbiased estimator of p.

El)+ : E(ar + b) =
a Elx) + b]

n+2 n+2 E) =p]

i s not an unbiased estimator of p.

Example 18: Find 95% confidence limits for the mean of a normality distributed
population from which the following sample was taken
15, 17, 10, 18, 16, 9, 7, 11, 13, 14.
JNTU (H) Nov. 2009 (Set No. 4), (A) Nov. 2010 (Set No. 4), (K) May 2010S, May 2013 (Set No. 4)]
15+17+10+18+16+9+7+11+13+14 = 13
Solution: We have =

10

-
n-1
I(15 13) + (17-13) + (10 - 13P + (18 13 + (16 13)

+(9-13) + (7- 13 + (11 - 13+ (13 - 13) + (14 - 13 ] = 40

3
Since a/21.96,
2al2 we have
 S
= 1.96 V40 = 2.26
Zal/2n
Confidence limits are i t 13 +2.26
za/2 = =
(10.74, 15.26)
 dard
deviation 16.
100 and standard
22A population random variable has
mean
Example
What are the mean and standard deviation of the sample
JNTUApril
mean for
2004,(4)
the random
Apr.2012
samples
(Set of
No.21
s
s

4 drawn with
replacement?
Solution: Given = 100, a = 16, n =4
Since the sampling is done with replacements, the population may be considered
as
infinite. We have to find u and og.
16
100 and 8
H 4
o
=

Example 23:A random sample of 400 items is found to have mean 82 and S.D. of 18
Find the maximum eror of estimation at 95% confidence interval. Find the confidence limits
for the mean if f =82 UNTU Nov. 2008, (A) Nov. 2011, (H) May 2011 (Set No.2
Solution: Given standard deviation=o =18
Sample size = n = 400

Z2for 95% confidence = 1.96 (from tables)

Sample mean = ï = 82

1.96x 188
Maximum error, E= ap = 1.764
20

The limits for the confidence are

-Zn <u<+Zn
Confidence limits are 80.236 and 83.764
Estimaton 323

ample 24 Measurements of the weights of a

random sample of 200 ball be
a certain
« machine during one weck showed a mean of 0.824 and a standard deviation
made by
of0.042.
Findmaximum error at 95% confidence interval ? Find the confidence limits for the
32.
if x
Nov. 2011 (Set No.4)]
=

mean JNTU Nov. 2008,(A)

Solution: We are given
Mean of the sample = x = 0.824

Z2 Zvalue for 95% level =

1.96 (From Tables)
g Standard deviation = 0.042

n size of the sample = 200

Maximum error, E= Z -1.96x0.042

- -
0.0059
n 200
Now -Zan n 0.824 0.0059 0.8181
and
+ZanJ
+Zan 0.824 +0.0059

Hence the limits for the confidence are

=
0.8299

F-Zal2 <p<+*Za/2
Confidence limits are 0.8181 and 0.8299
Example 25: A sample of size 300 was taken whose variance is 225 and mean 54.
Construct 95% confidence interval for the mean. JNTU (H) Nov. 2010 (Set No. 1)]

Solution: Since the sample size 300 is large ( 30), normal distribution is used as the
sampling distrubution.
Here n = 300, 7 = sample mean = 54, o = v225 =15

S.E. of = 150.866
300
95% confidence limits for the population mean are

t 1 . 9 6 (S.E. o f x ) = 5 4 +1.96 (0.866) = 54t1.697 = 55.697 and 52.3

The required confidence interval is (52.3, 55.7).

Example 26 In a random sample of 100 packages shipped by air freight 13 had

some damage.
Construct 95% confidence interval for the true proportion of damage package.

Solution: Here p =
Sample proportion of damage packages- 100 130.13
=1-p =1-0.13 =0.87
 PQ takep for P)
S.E. of p- n
( P is not known,
n
we

0.13x0.87 = 0.034
100
95% confidence limits for the population proportion of P of damage packa
nage packages ate
Ptl96 (S.E. of p) = 0.13#1.96 (0.034) = 0.13+0.067

ie., the confidence limits are 0.063 and 0.197

Hence the 95% confidence interval for the true proportion of damage packaoa
ages is
(0.063,0.197).
Example 27: It is desired to estimate the mean number of hours of continuOu
use
until a centain computer will first require repairs. If it can be assumed that d= 48 hom
ours,
howlarge the sample will be needed so that one will be able to assert with 90% confiden
ce
that the sample mean is off by at most 10 hours. JNTU(H)Nov. 2010 (Set No.3
Solution: Sampling error, E =10 hours, a = 48 hours, a/2 = 1.645

ZO
We have E =- Vn E

(1.645x 48
'n= ==(7.896) =62.35
10

Hence, sample size, n = 62

Example 28: Among 100 fish caught in a large lake, 18 were inedible due to the
pollution of the environment. With what confidence can we assert that the error of this
estimate is at most.065? JNTU(H) Dec. 2011 (Set No. 3)
Solution: We are given
n= Sample size = 100

and Max. error of estimate, E = 0.065

Here p= sample proportion of

inedible fish= =0.18

q=l-p =1-0.18=0.82

Maximum error of estimate for true proportion E= za/2 -

=

:P is not known, we take p for P)

0.065- Za/2 0.18x0.82=Zn0.038)

100
 0.065
a2 0.03871

when=1.71, the probability =0.4564

PE (from normal tables)
=0.4564=>2xP(z>
Hence with 91% contidence z,)=2(0.4564) =
0.065. we can 0.9128-91.28% =91%
assert that the
error of this
ostimate is at
mO
Fample 29:A random sample of 500
nerature of73.54 degrees Fahrenheit points ona heated plate resulted in an
with a standard average
tind 99% confidence interval for the deviation of 2.79 degrece Fahrenheit.
average temperature of the plate.
Solution:We are given JNTU(H) Dec. 2011 (Set No. 4)

n-500, =73.54 and o=2.79

We have a/2 =2.58 (for 99%)
We know that 99% confidence interval is
-Za/2 +
~u/2
Now al2 2.58x 2.79 7.1982
=0.32
500 V500500
99% confidence interval =(73.54-0.32,73.54
+0.32)
= (73.22,73.86)

Example 30 In a study of an automobile insurance a random sample of 80 body

repair costs had a mean of Rs.472.36 and a standard deviation of Rs.62.35. If x is used as

pount estimate to the true average repair costs, with what confidence we can assert that the
maximum error does not exeed Rs. 10? JNTUGH) Apr. 2012 (Set No. 4)

Solution: We are given

n= sample size = 80, o= S. D. = 62.35

Maximum error, E = 10
62.35
We have 10=
E=Fa/2 Za/20

. Za/2 10/80143
62.35
to za/2 1.43 is 0.4236.
From normal table, the area corresponding
=

2 0.8472
between z_a/2 to za/2 is (0.4236)
=

Then the area

hus we can ascertain with

84.7% confidence.
CHAPTER-8

TESTS OF HYPOTHESIS
8.1 INTRODUCTION
(FORLARGE SAMPLES
In the previous chapler we huve scen how u
parameter can be estimated from sampe
dall. We can find either a single number for the parameter (a point estimate) or an interval or
values (an interval estimate). However there are many problems, in which, rather than estimating
the value of a parameter we need to decide whether to accept or reject a statement about the
naramcter. This statement is called a hypothesis and the decision-making procedure about the
bypothesis is called hypothesvis testing. This is one of the most useful aspects of statistical
inference, since many types of decision-making problems, tests or experiments in the engineering
world can be formulated as hypothesis - testing problems.

8.2 TESTOF HYPOTHESIS

The main object of the Sampling Theory is the study of the Tests of Hypothesis or Tests
of Significance.
In many cireumstances, we are to make decisions about population on the basis of only
sample information. For example, on the basis of sample data,
(i) a drug chemist is to decide whether a new drug is really effective in curing a disease,
(i) a quality control manager is to determine whether a process is working properly,
(ii) a statistician has to decide whether a given coin is biased, etc.
Such decisions are called statistical decisions (or simply decisions)
Definition : Statistical Hypothesis |JNTU (K) Dec. 2013 (Set No. 3)|
In many circumstances, to arrive at decisions about the population on the basis
of sample information, we make assumptions (or guesses) about the population
parameters involved. Such an assumption (or statement)
is called a Statistical
which not be true. The procedure which enables us to decide
Hypothesis may or may
on the basis of sample results whether a hypothesis is true or not, is called Test of
Hypothesis or Test of Significance.
Since we use probability distributions to represent population, a statistical hypothesis
distribution of a random variable.
thought of as a statement about the probability
may also be
he hypothesis will usually involve one or more parameters of this distribution.

the city are smokers.

Example: 1. The majority of men in
both the schools effective.
2. The teaching methods in
are

HYPOTHESIS |JNTU Nov. 2008 (Set No. 2)

8.3 PROCEDURE FOR TESTING A
the following steps:
Test of Hypothesis involves
Step 1: Statement (or assumption) of Hypothesis
There are two types of hypothesis
Null Hypothesis (ii) Alternative Hypothesis.
(i) 335
336 Computer Orlented Statistical Method
) Null Hypothesis : For applying the tests of signilicance, we first set up a hypothesis
a definite statement about the population paramcter. Such a hypothesis is usually a hypothesis
of no-difference, is called Null Hypothesis.
It is in the form Ho:H =Ho
where Ho is the value wihich is assumed or claimed for the population characteristic. It is the
reference point against which the Alternative Hypothesis is set up, as explained in the next step.
Definition: A null hypothesis is the hypothesis which asserts that there is no significant
difference between the statistic and the population parameter and whatever observed difference
is there, is merely due to fluctuations in sampling from the same population. It is always
denoted by H,. To test whether one procedure is better than another, we assume that there is
no difference between the procedures. Similarly to test whether there is a relationship between
two variates, we take H, that there is no relationship.
For example, in case of a single statistic, H, will be that the sample statistic does not
differ significantly from the hypothetical parameter value and in the casc of two statistics (H)
will be that the sample statistics do not differ significantly.
(i) Alternative Hypothesis: Any hypothesis which contradicts the Null Hypothesis is
called an Alternative Hypothesis, usually denoted by H, It is set in such a way that the
rejection ofnull hypothesis implies the acceptance of alternative hypothesis. The two hypothesis
For example, if we
H and H are such that if one is true, the other is false and vice versa.

otest the null hypothesis thatthe population has a specified mean Ho (Say) i.e., H, :H=

Ho, then the Altemative Hypothesis could be

() H H (i.e., either >Po or P)
or (ii) H,H>Ho
or (ii) H,:Ho the
The Alternative Hypothesis (i) is known as a two-tailed alternative and
is known as
Alternative Hypothesis in (ii) is known as right-tailed and in (iii)
left-tailed.
hypothesis is very important to decide whether we
The setting of alternative
have to use a single-tailed (right or left) or two-tailed test.

Alternate Hypothesis is in one of the following forms:

H #Ho
or Hj:p> Ho

or Hj:< Ho
One has to choose from the above three forms depending on the situation posed, as
explained below.
In the example relating to the heights of students, the situation involved only testing
the statement made about the average height of a class of students. Therefore, the Alternative

Hypothesis was of the form u #Ho .

 ts of Hypothesis (For Large
Samples) 337

In the example about the life of electric bulbs, the company manager want
whether the average litfe was more than 560 hrs. Therefore the Alternative Hypotnesis wa

the form >Ho.

Consider another
example, the manufacturer ofcigarettes ordered that the nicotine content
in cigarette should not exceed the stipulated level of 25 mgs. In order to ccheck tmis, ne
selected a sample of 100 cigarettes from the lot to be packed, and found that the average
nicotine content was 23.5 mgs. Could he be reasonably sure that the stipulations laid down Dy
him were being met? In this case, the manufacturer wanted the nicotine content to be less than

a particular valuc, and therefore, the Alternative Hypothesis is of the form <
Ho
Siep 2: Specification of the Level of Significance
The level of significance denoted by a is the confidence with which we rejects or
accepts the Null hypothesis H i.e., it is the maximum possible probability with which we

are willing to risk an error in rejecting H, when it is true. The level of significance is
generally specified before a test procedure so that the results obtained may not influence
our decision. In practice, we take either 5% (i.e., 0.05) or 1% (i.e., 0.01) level of
significance, although other levels such as 2%, 1/2% etc. may also be used. 5% Level of
significance in a test procedure indicates thatthere are about 5 cases in 100 that we would
reject the null hypothesis Ho when it is true ie., we are about 95% confident that we
have made the right decision. Similarly, in 1% Level of significance, there is only 1 case
in 100 that the null hypothesis H,is rejected when it is true .e., we are about 99% confidnt
that we have made the right decision. Level of significance is also known as the size of
the test.

Step 3: Identification of the Test Statistic

There are several tests of significance, viz., z,t,F etc. First we have to select the

right test depending on the nature of the information given in the problem. Then
construct the test criterion and select the appropriate probability distribution.

Step 4: Critical Region: The critical region is formed based on following factors.
(a) Distribution of the Statistic i.e., whether the statistic follows the normal, '",

r or "F* distribution.
(6) Form of Alternative Hypothesis:
If the form has sign, the critical region
is devided equally in the left and right tailes, sides of the distribution.

Left sided Right sided

Two sided

If the form of alternative hypothesis

has <
sign, the entire critical region is taken in
the left tail of the distribution.
 Statistical Methods
338 Computer Oriented

the entire critical region is taken on

ofalternative hypothesis has sign,
>
form
the
the right side of the distribution.
Step 5: Making Decision
ascertain whether the computed
We compute the value of the appropriate statistic and
on the specified Level ofsignificance.
value falls in acceptance or rejection region depending critical values given in Statistical
the acceptance or rejection region we have to use
n nding value and the critical value decision
is taken for
Tables. By comparing the computed
critical value, we accept H, otherwise
accepting or rejecting H. Ifthe computed value <

we reject H
8.4 ERRORS OF SAMPLING JNTU (H) Dec. 2009 (Set No. 2)1
The main objective in sampling theory is to draw valid inferences about the
we decide to
population parameters on the basis of the sample results. In practice
sample from it. As such we have two
accept or to reject the lot after examining a

types of errors.
() Type I error: Reject H, when it is true.
JNTU (A) Dec. 2009, (H) Dec. 2014, May20171
It is the error ofrejecting Null hypothesis Ho, when it is true. When a null
hypothesis is true, but the difference (of means) is significant and the hypothesis is
rejected, then a Type I Error is made. The probability of making a Type I error is
denoted by a , the level of significance. The probability of making a correct decision
is then (1-a).
(ii) Type Il error : Accept H, when it is wrong i.e., accept H, when H, is true.
It is the error of accepting the null hypothesis Ho when it is false.
In other words, if the Null Hypothesis is false but it is accepted by test, then
error committed is called Type 11 error or B error.
If we write
P(Reject H, when it is true) = P(Type I error) = a
and P(Accept H when it is wrong) = P(Type Il error) = B
then a and ß are called sizes of Type I and Type II errors respectively
i.e., a = P (Rejecting a good lot)

B -P(Accepting a bad lot)

The sizes of Type I and Type ll errors are also known as producer 's risk and
consumer 's risk respectively.
The statistical testing of hypothesis aims at limiting the Type I error to a
preassigned value (say: 1% or 5%) and to minimize the type Il error. The only way
to reduce both types of errors is to increase the
sample size, if possible.
1. Critical Region JNTU (K) March 2014 (Set No.1)].
Under a given hypothesis let the sampling distribution of a
statistic r is approximately
a normal distribution with mean E) and S. D. = o, = S. E. of .

-E)_Observed value -Expected value is called the standardized

S.E. ofr S.E.of
Tests of Hypothesis (For Large Samples)
339
normal variate ortest statistic or
z
aietribution with mean O and S.D. score and its
I when the distribution is the standard normal
sample is large.

Critical
region Critical
025 or region
2.5% 95 o 95%
025 or
2.5%
Acceptancelregion
Z =1.96
Z0 Z=1.96

From above figure, we see that if lies between -1.96 and

z
confident that the hypothesis is true, since the area under the
1.96 then we are 95%
between z=-1.96 and standard normal curve
=1.96 is 0.95 i.e., 95%
z
of the total
P(-1.96S zs1.96) =0.95. But if for a simple random area i.e.

the range -1.96 to 1.96 i.e., if

sample we find that z lies outside
|z|>1.96, then we say that such an event occurs with
probability of only 0.05 when the given hypothesis is true.
score differ
significantly from the value expected under theIn this case, we say that z -

hypothesis H,is to be rejected at 5% level of hypothesis and hence the

significance. Thus if |z|>1.96, the the
hypothesis Ho is rejected at 5% level of significance. The set of
range -1.96 to 1.96 i.e., |z|>1.96 constitutes
z
scores outside the -

the critical region or

region of significance. On the otherhand, the set of z of rejection
the hypothesis or the region of
the range -1.96 to 1.96 is scores inside -

called the region of

-1.96 and 1.96 are acceptance of the hypothesis. The values
called critical values at 5% level of
Similarly, we can define critical
significance.
region any other level of
at
significance.
Definition. A region
corresponding to a statistic P, in the sample space S which leads
to the rejection of
the acceptance of
H, is called Critical Region or
Rejection Region. Those region which lead to
H give us a region called Acceptance Region.
In other words, the
critical
region in which a sample value falling rejected is known as the
is
region. In general we take two critical regions which cover 5% and 1 % areas of the
normal curve.
2. Critical Values or
The value of the test
Significant Values : JNTU 2004S|
statistic, which
acceptance region is called the critical separates
the the critical region (or rejection
value region) and
on or significant value. This value is dependent
(i) the level of significance used, and
(ii) the alternative hypothesis, whether it is one-tailed or two-tailed.
For
larger samples, corresponding to the statistic t, the variable z = -E (t) IS
S.E of t
normally distributed with mean 0 and variance 1.
 340 Computer Oriented Statistical Metho
hods
n e value of z given above under the null hypothesis is known as test statistic
The critical value z, of the test statistic at level of significance a for a two-tailed test
is given by
P(1z> z)= a .(1)
That is, z, is the value of z so that the total area of the critical region on both
a s 1s a. Since the normal curve is a symmetrical one, equation () implies,

P )+ P(z<-,)a
i.e. 2P(z>z,) = a

or P(z>z)=a/2
That is, the area of each tail is a/2.
The critical value z, is that value such that the area to the right ofz, is a/2 and

the area to the left of -z, is a/2. (Refer figure).

3. Two-Tailed test at 1level of significance 'a':

Acceptance
Region

Lower
Upper Critical
Rejection Critical
Region (a/2) Value Value Rejection
Region (a/2)

Z =-2 Z 0

In the case of one-tailed alternative,

P(Z> Z,)=a ifit is one-tailed (right)
P(Z < - Z) =
a
if it is one-tailed (left)
For Level of Significance *a
Right Tailed Test Left Tailed Test

Acceptance
Acceptance Region
Region

Rejection Rejection
Region (a) Region (a)

Z = 0
z-a
Tests of Hypothesis (For Large Samples) 341

From the above figures, it is clear, that the critical value of Z for ed
test (right or lett) at level of significance 'a' is same as the critical
a
singetaZ for
value
value oof 2Z for
i s same as the critical
wO-tailed test
two-tailed of sionif
at level of
test at significance 2a. ance a
The critical values of Z at different level of significance (a) for both singe
ailed and two-tailed tests are calculated from equations
P(Z> Z) -a
P(Z Z) =a
P(Z-2.) = a

using the normal tables. They are listed below.

Critical values (Z,) of Z:

Level of Significance

1% (.01) 5% (.05) 10%(.1)

Two-tailed test 2 2.58 Ho 1Z= 1.96 1Z 1.64s
Right-tailed test 2, 2.33 2. 1.645 Z= 1.28
Left-tailed test Z-2.33 2-1.645 2 . - 1.28
8.5 ONE-TAILED AND TWO-TAILED TESTS
JNTU (A) Dec. 2009, (K) Nov. 2011, H) May 2012 (Set No. 1), May 2017, Sept. 20171
If we have to test whether the population mean has a specified value Ho, then the
Null Hypothesis is H :=ko and the Altermative Hypothesis may be () H, :u * Ho (Le.,
HH or <%) or (i) H:u>kor (ii) H, :p<Ho. The Alternative Hypothesis in () is
known as a two-tailed (i.e., both right and left tail) alternatives and the alternative
in (i) and (iii) are known as right tailed and left tailed alternatives
hypothesis
respectively.
-
-

Critical region Critical region

at a level at a level

Z =0 2 a Z= 0

One-Tailed Test
Ifthe Alternative Hypothesis H, in a test of a statistical hypothesis be one tailed
i.e., either right tailed or left tailed but not both), then the test is called a one-tailed
- -

test. For
example, to test whether the population mean u=Ho, we have H:= Hoagainst
the alternative hypothesis H, givén by
) Hu> Ha (right tailed) or (i) H,:p<,(left tailed) and the corresponding
-

test is a single- tailed or one-tailed or one sided. In the right tail test
H:4>ko» the
-
 Computer Oriented Statistical Methods
342
critical region (or rejection region) z > Z, lies entirely in the right tail of the sanpling
distribution of sample mean with arca equal to the level of significance u (see figure)

z < -z, lies entirely in the

Similarly, in the left- tailedtest (H, :u<po), thecritical region
tail of the sampling distribution of the sample mean
? with area cqual to the level of
ert
significance a (see figure).
Two-Tailed Test:
against the Alermative Hypothesis
Suppose we want to test the Null Hypothesis H, :p Ho
=

H:HH

Po Z = 2g
z = -Za

the critical region under the curve is

Since H is two -
tailed alternative hypothesis,
equally distributed on both sides of the mean
tail The critical area under the left -

tail
Thus, the critical area under the right =

= Half of the total area

1
probability of rejection

with critical statistic Za/2, where a is the level of significance.

is then, s or Za/2 Sz
The critical region z -2a/2 .

one-tailed tests at 1%, 5% and 10% level

Critical values ofx for both two-tailed and
are given in the following
table:
ofsignificance
Table: Critical values of z

1% 5% 10%
Level ofsiginificancea
Critical values for a=2.58 =1.96 z1=1.645
two-tailed test

Zu 2.33 2 1645 1.28

Critical values for
Right-tailed test

Zu-1.645 Z 1.28
Critical values for -2.33

Left-tailed test
 343
Tests of Hypothesis (For Large Samples)
e
Applying one-tailed or two-tailed test for a particular oroblem depends
two-tailed test
st
nature ofthe Altermative Hypothesis. Ifthe alternative test is two-tailed we appiy two-a
and if Altemative Hypothesis is one-tailed, we apply one-tailed test.
and

For example, to test whether a coin is biased or not, two - tailed test should
Dlaseu com gives either more number of heads than tails (which corresponds to n g

tail) or more number of tails than heads (which corresponds to left tail)
EXAmpIe Consider two population brands of bulbs one manufactured by
routine process (mean p,) and the other manufactured by new technique (mean H)"
ant to
test if the bulbs differ significantly then the hypothesis is H,: H
and the alternative hypothesis will be H, H# H. This gives us a two-tailed test.
Suppose if we want to test if the bulbs produced by new process (,) have higher
average life than those produced by standard process (,), then we have

H HH, and H,: P A

In this case we have to adopt a left-tailed test.
If we want to test whether the product of new process (¥,) is inferior to that
of standard process (u,), then we have
H: , P ,and H, : , H
which gives a right-tailed test.
Hence the decision about applying a two-tail test or a single-tail (left or right)
test will mainly depend on the problem under study.
PROCEDURE FORTESTINGOF HYPOTHESIS:
UNTU (H) Nov. 2009 (Set No.2,4), (K) Nov. 2011, Dec. 2013 (Set No. 3)]
Various steps involved in testing of Hypothesis are given below: Infact, the
same steps are followed for conducting all tests of significance.
Step 1:Null Hypothesis : Define or set up a Null Hypothesis H, taking into
consideration the nature of the problem and data involved.
Step 2: Alternative Hypothesis: Set up the Alternative Hypothesis H, so that we
could decide whether we should use one-tailed or two-tailed test.
Step 3: Level of Significance : Select the appropriate level of significance(a)
depending on the reliability of the estimates and permissible risk. That is, a
suitable a is selected in advance if it is not given in the problem.
(Usually we choose 5% level of significance)

Step4: Test Statistic: Compute the test statistic Z= under the null
.E of
hypothesis.
Here t is a sample statistic and S.E. is the standard error of t.
Step 5: Conclusion : We compare the computed value of the test statistic Z with
the critical value Z, at given level of significance (a).
If |Z< , (that is, if the absolute value of the calculated value of Z is less
than the critical value 2,) we conclude that it is not
the null hypothesis.
significant. We accept
If Z> Z, then the difference is significant and hence the null
rejected at the level of significance a.
hypothesis is
 Statistical Methods
344 Computer Oriented

Clearly
For two-tailed test
Ifjz< 1.96 accept H, at 5% level of significance.
If|Z>1.96 reject H, at 5% level of significance.
If |Z< 2.58, accept H, at 1% level of significance.
If |Z> 2.58 reject H, at 1% level of significance.
left testi
Forsingle-tailed (right or5% level of significance.
If|Z< 1.645, accept H, at
If |Z> 1.645, reject H, at 5% level of significance.
If |Z <
H, at 1% level of significance.
2.33 accept
level of significance.
If|Z>2.33 reject H, at 1%
SAMPLES
8.6 TESTOF SIGNIFICANCE FOR LARGE
as large samples. The tests
size, > 30, then we consider such samples
Ifthe sample n
different from those used in small samples
because
of significance used in large samples are is done. Ifn is
under which large sample analysis
small samples fail to satisfy the assumptions
Poisson, Chi-square etc. are closely approximated by
large, the distributions, such as Binomial, the sampling distribution of a statistíc is
normal distributions. Therefore, for large samples,
approximately a normal distribution.
in such trial isp.
wish to test the hypothesis that the probability of success
Suppose we
deviation o of the sampling distribution of
Assuming it to be true, the mean H and the standard

number of successes are np and npg respectively.

standard normal
Ifx be the observed number of successes
in the sample and Z is the

variate then Z = .

Thus we have the following test of significance:

i) If|Zj<1.96, the difference between observed and

the expected number of successes
is not significant.
at 5% level ofsigniticance.
(i) If |Z|>1.96, the difference is significant
at 1% level of significance.
(ii) If |Z|> 2.58, the difference is significant
Assumptions for Large Samples:
under which significance tests are applied
The following arethe assumptions
the properties of the normal curve.
1. The random sampling distribution ofstatistic has
This may not hold good in case of small samples.
close to the population
2. Values (ie., statistic) given by the samples are sufliciently
for calculating the standard
values (i.e., parameters) and can be used in its place
error (S. E.) of the estimate.
universe is not known, it can be
For example, for computing the S. E., ifthe S. D. of the
this is not possible.
replaced by the S. D. of the sample. In case of small samples,
Tests of Hypothesis (For Large Samples) 345

SOLVED EXAMPI.ES
Example l:A coin was tossed 960 times and returned heads 183 times. Iest e
hvpothesis that the coin is unbiased.
Use 0.05 level of
a
significance.
Solution: Here n =
960,p- Probability of getting head 1/2 =

91-p;: u=np =960480

G npgp)9 =480x =v240=15.49
x= number of successes =
183
1. Null Hypothesis Ho: The coin is unbiased
2. Alternative Hypothesis H: The coin is biased
3. Level of significance: a =0.05

4. The test statistic is Z H-183-480-297 -19.17

=-19.17
15.49 15.49

1Z=19,.17
As 1Z1>1.96, the null hypothesis H, has to be rejected at 5% level of significance and
we conclude that the coin is biased.
Example 2: A coin was tossed 400 times and returned heads 216 times. Test the
hypothesis that the coin is unbiased: Use a 0.05 Level of significance.

JNTU (K) Nov. 2009 (Set No. 3)

Solution:Here n = 400

p-Probability ofgettinghead

9=1-p=l-;-2. A=np=400-200

-npq nplg) =200x= v100 =10

number of successes 216 =

.Null Hypothesis H: The coin is unbiased

2. Alternative Hypothesis H: The coin is biased
Level of significance: a =0.05

4. The test statistic is 7

Z 216-200=1.6
10 10
 346 Computer Oriented Statistical Methodas

AS |z<1.96, the null hypothesis H has to be accepted and we conclude that the coin

is unbiased.
Example 3: A die is tossed 960 times and it falls with 5 upwards 184 times. Is the die

unbiased at a level of significance of 0.01 ?

1/6
=The probability of throwing 5 with one die
=

Solution: Here n =960,p

g=1 ; =np-960J=160
5
mpg=160x=11.55
x= number of successes 184
1. Null Hypothesis H: The die is unbiased

2. Alternative Hypothesis H:The die is biased

3. Level of significance: a = 0.01

4. The test statistic is Z=E_184-160 24

11.55
242.08
11.55

As |ZI<2.58, the null hypothesis Ho has to be accepted at 1 % level ofsignificance and

we conclude that the die is unbiased.
Note: As |Z]> 196, the null hypothesis H, has to be rejectd at 5% level ofsignificance and
we conclude that the die is biased.
Example 4: A die is tossed 256 times and it turns up with an even digit 150 times. Is
the die biased?

Solution : Here n = 256, P = The probability ofgetting an even digit (2 or4 or 6) = =

g=1-p=5 =np=256J=128

x = number of successes 150

1. Null Hypothesis H%: The die is unbiased

2. Alternative Hypothesis H: The die is biased

3. Level of significance: a =0.05
4. The test statistie is Z = =
ie, Z
8
=4=2.75
As |ZI>1.96, the null hypothesis H has to be rejected at 5% level of significance and
we conclude that the die is biased.
 347
sls of Hypothesis (For Large Samples)
Tests

lard
Evample S: Mean of population =0.700., mean of the sample=0.744: *
for population
e v i a t

ion
i o n

of the sample =
0.040, sample size = 10. Test the null hypothesis
Nov. 20151
nean = 0.700 JNTU (H) Ilyr.

Solution: Given = Mcan of population = 0.7

T= Mean ofsample =0.742
G Standard deviation of the sample = 0.040
and n= Sample size = 10

The test statistic is z =

Wn

ie., 0.04/10 0.742-0.-3.32

Since Iz|>1.96, the sample is not from the population whose mean is 0.1.

Under large sample tests, we will see four important tests to test the significance.
1. Testing of significance for single proportion.
2. Testing of significance for difference of proportions.
3. Testing of significance for single mean.
4. Testing of significance for difference of means.

8.7 TESTOF SIGNIFICANCE OF A SINGLE MEAN-LARGE SAMPLES

Let a random sample of size n (n>30) has the sample mean ï, and population
mean . Also the population mean has a specified value Ho

Working Rule:
1. The Null Hypothesis is H: i.e., "there is no significance difference
=u
between the sample mean and population mean" or the sample has been drawn

from the parent population.

2. The Alternative Hypothesis is (i) H :ï #p (u*Ho )or
(in) H:>u (4>fo ) or (ii) H:<u (4<<Ho)
Since n is large, the sampling distribution of f is approximately normal.
3. Level of Significance. Set the level of significance a.

4. The Test Statistic:

We have the following two cases.
Case I: When the standard deviation a of population is known.

In this case, standard Error of Mean, S. E. (F)=, where n= sample size,

G=s.d. of population.
 348 Computer Oriented Stafistical Metho
The test statistic is

= = *H. where is the sample mean.

S.E(F) a/n
Case : When the standard deviation a of population is not known.
n this case, we take s, the s.d. of sample to compute the S. E. of mean

S.E. (T)=
Hence the test statistic is

S.E.(T) s/Vn
5. Find the critical value z of z at the level of significance a from the normal
table.
6. Decision3

(a) If z]<zg , we accept the Null Hypothesis Ho

(b) If |z|>za, we reject the Null Hypothesis Ho.
The rejection rule for H,:ï =H (or u = #o) is given below:

Table: Critical values of z

Level of siginificance a 1% 5% 10%
Critical region for H* Ho z> 2.58
z>1.965 z >1.645
Critical region for H>Ho »2.33 z>1.645 z>1.28
Critical region for
<Hoz-2.33 z<-1.645 z-1.28
Note: 1. We reject Null Hypothesis Ho when |z |> 3 without
mentioning any level of
significance.

The test statistie is, z

= where a is the S.D of the population.
If the population S.D is not known, then use the statistic
Z
s/mwhere S is the sample S.D.

2. The values t 1.96 are called 95% fiducial limits

or confidence limits
for the mean of the population corresponding to the given sample.
Similarly, Ft2.58 (or) [ -2.58 (S.E.of F),I +2.58
n (S.E. of F)] are called
99% confidence limits and
f+2.33 are called 98% confidence limits.
vn
 349
Tests of Hypothesis (For Large Samples)

SOLVED EXAMPLES
lectrical)
Eampie d ACcording to established for a mechanical (or an
the norms
standard
est, persons who are 18 years old have an average height of 73.2 with a
aptitude test
of 8.6. TIf4 randomly selected persons of that age averaged 76.7, test the he sis
deviation
u =73.2 against the alternative hypothesis u> 73.2 at the 0.01 level of significance.

JNTU 2004, 2005S (Set No. 1), (H) Nov. 2012, Sept. 2017]
Solution: Given n = 4, u = 73.2, F = mean of the sample = 76.7

and o = S.D of population = 8.6

1. Null Hypothesis H,:4 = 73.2

2. Alternative Hypothesis H,: u> 73.2 (Right-tailed test)
3. Level of significance: a = 99% (or probability is 0.01)

4. The test statistic is z = =0.73.2

8.6
3.=
3.5 0.814
4.3
4
Tabulated value of z at 99% level of significance is 2.33
Hence calculated z < tabulated z

The null hypothesis H, is accepted. That is, I and u do not differ significantly.

Example 2: A sample of 64 students have a meanweight of 70 kgs. Can this be

regarded as a sample from a population with mean weight 56 kgs and standard
JNTU 2006, (A) Nov. 2010 (Set No. 2)]
deviation 25 kgs.
Solution
Given = mean of the sample = 70 kgs

m e a n of the population = 56 kgs

o S.D of population = 25 kgs

and n sample size

=
64

mean weight of 70 kgs can

Hypothesis H: A sample of students with
Null
64
1.
with mean weight 56 kgs and standard
be regarded as a sample from a population
deviation 25 kgs.

2. Alternative Hypothesis H,: Sample cannot be regarded as one coming from

the population.

3. Level of significance a =0.05 (assumption)

70-56
T h e test statistie is z
=
4.48
(25
V64
 350 Computer Oriented Statis Methods
5. The null hypothesis H, is rejected, since |Z .0
Note: The null 1% level
hypothesis can be rejected even at of significance
Evample 3:An oceanographer wants to check whether the depth of the occan in a
ertain region is 57.4 fathoms, as had previously been recorded. What can he conclude at the
0.03 levelofsignificance, if readings taken at 40 random locations in the given region yielded
a mean of 59.1 fathoms with a standard deviation of 5.2 fathoms.
JNTU 2003, 2003 S, 2004, (H) Sept. 2017
Solution: Given n = 40, T = 59.1 and o = 5.2

1. Null Hypothesis H,: =

57.4
2. Alternative hypothesis H,: 4 * 57.4
3. Level of
significance : a=0.05

7 - X-
9.1-57.4
44. The test statistic
statistie is
is Z-
= 5.2/V40 2.067

Tabulated value of Z at 5% level of significance is 1.96

Hence calculated Z> tabulated Z.
The null hypothesis H, is rejected.
Example 4: In
a random
sample of 60 workers, the average time taken by them to get
to work is 33.8 minutes with a standard deviation of 6.1
minutes. Can we reject the null
hypothesis 32.6 minutes in favour of
=

alternative null hypothesis H >32.6 at a 0.025 =

level of significance.
JNTU 2005 (Set No. 1
Solution: Given n =
60, 7 =
33.8, =32.6 and o =6.1
1. Null Hypothesis H: =
32.6
2. Alternative Hypothesis H, : > 32.6
3. Level of significance : a 0.025 =

4. The test statistic is Z = * 33.8-32.6

6.1/V60 2
0.7875 1.5238
n
Tabulated value of Z at 0.025 level of
significance is 2.58.
Hence calculated Z < tabulated z
The null hypothesis H, is accepted.
Example 5: A sample of 900 members has a mean of 3.4
this sample has been taken from a large cms and S.D 2.61 cms. Is
population
population is normal and its mean is unknown find of mean 3.25 cm and S.D 2.61 cms. If the
the 95% fiducial limits
of true mean.
JNTU (H) May 2011 (Set No.
1), (K) Nov. 2011 (Set No. 1
 pothesis (For Large Samples)
Tests of Hypo 351

Solution:
Given n=
900 , = 3.25,
3.4 cm, o =
2.61,
and S 2.61

1, Null Hypothesis H,: Assume that the sample has been drawn from tnc
population with mean u = 3.25.

2. Alternative Hypothesis H,: 3.25.

3. The test statistic is, Z = 3.4-3.25

2.61/9001.724
n
i.e., Z= 1.724< 1.96

We accept the null hypothesis Ho

i.e., The sample has been drawn from the population with mean ju= 3.25.

95% confidence limits are given by

.61
t1.96 J 3.4 + 1.96 x 3.4 0.1705
i.e., 3.57 and 3.2295
Example 6: Asample of 400 items is taken from a population whose standard deviation
is 10. The mean of the sample is 40. Test whether the sample has come from a population with
mean 38. Also calculate 95% confidence interval for the population.
JNTU 2005, (H) Dec. 2011 (Set No. 1)]

Solution: Given n =
400, F =
40, u 38 and o =
10
1. Null Hypothesis H,: 38
2. Alternative Hypothesis H,: # 38
3. Level of significance : a = 0.05

40-38
4.
The test statistic is, Z = 10/V400
4

i.e. Z= 4> 1.96

We reject the null hypothesis H

is not from the population whose mean is 38.
ie., The sample

95%confidence interval is -1.96., + 1.96.

e40 1.96(10) a0
40++1.96(10)
400 V400
 Computer Oriented Statistical
thods
or
40 , 40+ or (40 - 0.98, 40 + 0.98)
20
i.e., (39.02, 40.98)

Eample 7: An ambulancc service claims that it takes on the

minutes toreach its destination in average less than 10
minutes and the variance of 16 emergency calls. A sample of 36 calls has a mean of 1
minutes. Test the claim at 0.05 level significance.
IJNTU 2005, (H) May 2012 (Set No.
Solution : Given n = 36, F = 11, u = 10 and o = V16 = 4 4)1
1. Null
Hypothesis H,: u =10
2. Alternative Hypothesis
H,: 4 <10
3. Level of
significance : a =
0.0s

4. The 11-10
test statistic is, Z = .6 1.5
Tabulated value of Z at 5% level of significance is 1.645.
Hence calculated Z< tabulated Z
We accept the null hypothesis H
Example 8: It is claimed that a random sample of 49 tyres has a mean life of 15200
km. This sample was drawn from a
population whose mean is 15150 kms and a standard
deviation of 1200 km. Test the significance at 0.05 level.

JNTU 2005, 2006S, (K) May 2013 (Set No. 2)]

Solution : Given n = 49, F = 15200, u = 15150 and o = 1200
1. Null Hypothesis H,: u 15150 =

2. Alternative Hypothesis H, : 4 15150

3. Level of significance: a = 0.05
4. Critical region : Accept the null hypothesis if -1.96 < Z< 1.96

5200-15150
5. The test statistic is, Z = 1200/49
= 0.2917

Vn
Since Z| < 1.96 therefore, we accept the null hypothesis.

Example 9: An insurance agent has claimed that the average age of policy holders
who issue through him is less than the average for all agents which is 30.5 years. A random
sample of 100 policy holders who had issued through him gave the following age distribution

Age 16-20 21-25 26-30 31-35 36-40

No. of persons 12 22 20 30 16
 363
Tasts of Hypothesis (For Large Samples)
and
Calculate the Arithmetic and Standard deviation of this
mean dis
use these values to test his claim at 5% evel of
significanco.
Solution: Take 4 28, d - x,-4

A.M. = F =A + " 2 95x16 28.8

N 100

sD:S- 10
16416
0 100 h 51

=6.35

1. Null Hypothesis H,: The sample is drawn from a population with mean 1.e
F and do not differ significantly where u 30.5 years. =

2. Alternative Hypothesis H, H < 30.5 years (left tail test)

Now, F = 28.8, S = 6.35, = 30.5 years and n = 100

3. The test statistic is, Z= - 28.8-30.5 - 2.677

S/Vn 6.35//100
1Z1- 2.68.
Tabulated value of Z at 5% level of significance is 1.645 (left tail test)
Here calculated Z> tabulated Z.
The Null hypothesis H, is rejected.
i.e., and u differ significantly.
i.e., The sample is not drawn from a population with mean u = 30.5 years

Example 10:Themean life time of a sample of 100 light tubes produced by

a company
is found to be 1560 hrs with a population S.D of 90 hrs. Test the hypothesis for a =0.05 that
e mean life time of the tubes produced by the company is 1580 hrs.
UNTU(A) Dec. 2009 (Set No.1)
Mean ofthe sample =
1560 hrs
Solution: Given
=

Mean ofthe population =1580 hrs

n = Sample size = 100

Standard deviation =90 hrs

1. Null Hypothesis Ho:H = 1580

2. Alternative Hypothesis H : * 1580

3. Level of significance : a = 0.05

The test statistic is z= = 1D00-1580

o/Vn 90/V100 9
 354 Computer Orlented Statistioal Meh
hods

l 222
a/2 1or a = 0.05 is 1.96. Since lel> 1.96, the null Hypothesis Ho is rejected.

u * 1580,
Note: a/2 for a = 0.01 is 2.58

Since zl < 2.58, the null Hypothesis H can be accepted at 1% level ofsignificance

Example I1: The length of life X of certain computers is approximately normallv

distributed with mean 800 hours and standard deviation 40 hours. If a random sample of30
computers has an average life of 788 hours, test the null hypothesis that =800 hours against
the alternative that +800 hours at () 0.59% (i)1% (ii) 4% (iv) 5% (v)10%% (vi) 15%

level.

Solution: The NullHypothesis is H, 800 hours.

=

The Alternative Hypothesis is H:H * 800 hours.

Since H is two-tailed (i.e. two sided), we are to use two -tailed test.

Let us assume that H, is true.

he test statistic is z =
o/yn
Here = sample mean =788 hours
n = sample size = 30

O standard deviation =40

788-800=-1.643 and |z|=1.643

40/30
Since z<2.81, we accept the Null Hypothesis H at 0.5% level ofsignificance.
(i) Since |z|<2.58, we accept the Null Hypothesis H, at 1% level ofsignificance
(ii) Since, |z|< 2.06, we accept H, at 4% level of significance.
(iv) Since, |z|<1.96, we accept H, at 5% level of significance.
() Since, 1=l<1.645, we accept H, at 10% level ofsignificance.
(vi) Since, |z|> 144, we reject H, at 15% level ofsignificance

Example 12: In 64 randomly selected hours of production, the mean and the standard

deviation of the number of acceptance pieces produced by an automatic stamping machine

are x=1.038 and o=.146
1.000
At the.05 level ofsignificancedoes this enable us to reject the null hypothesis
=

H>1.000? JNTU (H) Nov. 2010 (Set No.4)

against the alternative hypothesis
 Hypothesi (Fer args Sanpee

salution1etthe Null 1lypothesin he 11,

Thenthe Alternative 11yprnesis is 1(K
1, 4
Here Mean of the ample 10
Mean of the population =1,000
-S.D. of the population -0,146
and n Sample size h4

Thetest statistic is

1.038-1000
0.146/ 64 0.146/%
2.082
Thus we see that z =

2.082>1.645
Hence, we reject the Null
Hypothesis H, at 5% level of significance and conclude that
the mean of the population u1.000
Evample 13A trucking rm suspects the claim that average life of certain tyres 1
atleast 28,000 miles. To check the claim the rm
puts 40 of these tyres on its trucks and gets
a mean life time of 27463 miles with a
standard deviation of 1348 miles. Can the claim be
truc?
NTU (H)Apr.2012(Set No.2
Solution: The Null
Hypothesis is Hg :u=28,000 miles
TheAlternative Hypothesis is H,:u 28,000
Since H is two tailed (le., two
sided), we are to use two - tail test.
Let us assume that Ho is true.

The test statistic is z= 4

S.E.of a/yn
Here ? = sample mean =27463 miles,
n =40 ando=s=1348 miles
I: population S. D. a is not known]

27463-28000-537(v40)-2.52
1348/V40 1348
and 2 =|-2.52|=2.52>1.96

Hence, the Null Hypothesis H, is rejected at 5% level of significance and we conclude

that the mean life of tyres, cannot be taken as 28000 miles.
 356 Oriented
Computer statistical
8.8 IPLES
TEST FOR EQUALITY OF TWO MEANS-LARGE SAMPLE
Methoth
(rest of significance for difference of means of two large samples)
Let dom samples
, be the sample mcans of two independent large random
and samples sizes
nand n drawn from two ulations having means and H, and standar
lard deviatio
and o To test whetherthe two population means are cqual.
Let the Null Hypothesis be H:=H2
Then the Alternative
Hypothesis is H,:4 *H2

S. E. of
CE-+)=, , where a, and a, the S. D. the two
are
populations
To test whether there is any significant difference between x and , we hav
ave
to use the statistic

-)-8 -F)-8
S.E. of - )
.
where 8 4j H2 (=given constant.)
= -

If 8=0, the two populations have the same means.

If 80 the two populations are different.

Under Ho :H1 =H2, the test statistic becomes z

is approximately normally distributed with mean 0 and S. D1.

Note If the samples have been drawn from the population with common
S. D. o, then o,2 =

a,? g
=

Hencez =Ä --

is normally distributed with mean zero and standard deviation one.

tnm
Ifo is not known we can use an estimate of o given by o? " =

Rejection Rule for H:4 =k2

(i) If |z|> 1.96, reject Ho at 5% level of significane.

(i) If lz> 2.58, reject H, at 1% level of significance.
(ii) If lz|>1.645, reject H, at 10% level of significance.
(iv) If z>3 then either the samples have not been drawn from the same
population or the sampling is not simple.
Otherwise accept Ho
 357
Tests
of pothesis (For Large Samples) deviations
unknown
Standard
Vote : I f t If the two samples are drawn from two populations with
and S; Pro
o and o then of and o can be replaced bysample variances S
both the samples n and n are large.

In this case, the test statistic is z = 2)-0

n2
Means".
Write the formula for testing the hypothesis concerning "Two
No.1)]
JNTU Nov. 2008 (Set
a populauon
Solution Let x be the sample of size n, drawn from
mean of a random
be the mean of an independent random sample ofsize
variance o. Let
n
with mean H and
another population with mean , and variance o,.
To test the hypothesis for
drawn from
statistic is given by
difference of means, the

n2
have come from two populations with
If |z]1.96, it is accepted that the samples
<
we
level of significance. Otherwise, at this level of significance,
the same mean, at 5%
means is significant.
claim that the difference in

SOLVED EXAMPLES
and 2000 members are

Example 1: large samples of sizes 1000

The means of two
drawn from the same
Can the samples be regarded as
67.5 inches and 68.0 inches respectively.
S.D 2.5 inches.
JNTU (A) Nov. 2010 (Set No. 1))
population of
Solution: Let and u, be the ofthe two populations.
means

=
2000 and
= 67.5
inches, =
68 inches
Given n, 1000, n, =

=2.5 inches
Population S.D, o population
have been drawn from the same
1. Null Hypothesis H: The samples
of S.D 2.5 inches
and o =
2.5 inches
i.e., 4, =
4
2. Alternative Hypothesis H, : H, * P2
67.5-6
3. The test statistic is, z
2.5 L+L)
N (1000 2000

0.5 = - 5.16
Z0.0968 1.96 i.e., the calculated value of z> the table value of z'9
lzl =
5.16 >

1s at 5% level of significance and

Hence the null hypothesis , rejected we
conclude that the samples are not drawn Irom the same population of S.D. 2.5 inches.
 358 Computer Oriented Statistical Methods

210 pounds with S.D. 10

A
ENample 2: The mean yield of wheat from a district
was
pounds per acre from a sample of 100 plots. In another district the mean yield was
200 pounds with S.D. 12 pounds from a sample of 150 plots. Assuming that the S.D of yield
in the entire state was 11 pounds, test whether there is any significant difference between the

mean yield of crops in the two districts.

Solution: Let , and u, be the means ofthe two populations.

Given 210, =
200, and n,= 100, n, = 150,
Population S.D., o = 11
Null Hypothesis H,: =42 i.e. there is no difference between , and ,
1.

2. Alternative Hypothesis H,: H *42

210-200
= 7.041778
3. The test statistic isz =

T
100 150
z = 7.041> 1.96 i.e. the calculated value of z> the table value of z.
and conclude
Hence we reject the Null Hypothesis H, at 5% level of significance
in the two
that there is a significant difference between the mean yield of crops
districts.
Example 3: In a survey of buying habits, 400 women shoppers are chosen at random
in super market 'A' located in a certain section ofthe city. Their average weekly food expenditure
random in super market 'B'
is 250 with a S.D. of7 40. For 400 women shoppers chosen at
with a S.D. of ? 5.
in another section the city, the average weekly food expenditure is 220
of
Test at 1% level of significance whether the average weekly food expenditure
of the two
populations of shoppers are equal.

Solution: Let 4j and u^ be the means of the two populations.

Given n, = 400, = 7 250, S, = 740

n,
=
400, =
7 220, S, =
T 55
of the
1. Null Hypothesis H, : Assume that the average weekly food expenditure
two populations of shoppers are equal i.e., , : H =42

2. Alternative Hypothesis H,: Hh *H2

3. Test statistic is, z

. SiS
=
250-220
(40)(55)*
30
3.4
= 8.82

400 400

i.e. z = 8.82>2.58

Hence we reject the Null Hypothesis H at 1% level of significance and conclude

that the average weekly food expenditure of the two populations of shoppers are not
equal
 359
Tests of Hypothe (For Large Samples)
and
Example Samples of students were dran from two universities Make a
below.
weights in kilograms, mean and standard deviations are caleulated and shown
large sample test to test the significance ofthe difference between the means.

Mean S.D Size of the sample

UniversityA 55 10 400

UniversityB 57 15 100
No. 3))
|JNTU 2005 (Set

Solution 55, = 57, n, = 400, n, = 100,S, 10 and S, = 15

Given
1. Null Hypothesis H: = T, i.e., there is no difference

2. Alternative Hypothesis H, : #
3. Level of significance : a = 0.05 (assumed)

4. Critical region : Accept H, if -1.96 < Z <1.96

5. The test statistie is Z = y 55-57 -1.26

100
400 100

1Z= 1.26< 1.96

Hence, we accept the Null Hypothesis H, at 5% level of significance and conclude

that there is no significant difference between the means.

interested in studying whether there is

Example 5: The research investigator is a

significant difference in the salaries of MBA grades in two metropolitan cities. A random sample
ofsize 100 from Mumbai yields on average income of Rs. 20,150. Another random sample of
60 from Chennai results in an average income of Rs. 20,250. If the variances of both the
populations are given as o," = Rs. 40,000 and o,' = Rs. 32,400 respectively.

JNTU Nov. 2008 (Set No. 1)1

Solution: Given n, = 100, = 20,150 and n, = 60, 2 = 20250

and 40,000, a, = 32400

Letthe Null Hypothesis be H H That is, the difference of means is significant.

Then the Alternative Hypothesis is H, H# 4
20150-20250
The teststatistic is z -
40000 32400
n 100 60
100 100
V400+540 30.66 3.26

Since Z 3.26> 1.96=Z005 we reject the Null Hypothesis at 5% level ofsignificance

and conclude that there is a significant difference in the salaries of MBA grades in two cities.
 360 Computer Oriented Statistical Methods

ENample 6 : A researcher wants to know the intelligence of students in a school. He

selected two groups of students. In the first group there 150 students having mean 1Q of 75
with a S.D. of 15 in the second group there are 250 students having mean 1Q of 70 with S.D.
of 20. Is there a significant difference between the means of two groups ?
JNTU No. 2008 (Set No. 4), () Nov. 2015

Solution: Given n, = 150, =75, o, 15

and n, 250, 70, o, - 20

1. Null Hypothesis H The groups have been came from the same population

2. Alternative Hypothes is H, H, *
H

3. The test statistic is, z=

n n2
75 70
2.7116
225 400
V150 250
Tabulated value ofz at 1% level of significance is 2.33
Calculated z > tabulated z.
and conclude that the
reject the Null Hypothesis H, at 1% level of significance
Hence we

groups have not been taken from the same population.

in U.S.A. are tested for petrol mileage,
Example 7: Two types of new cars produced
one sample is consisting of 42 cars averaged
15 kmpl while the other sample consisting of 80
cars averaged 11.5 kmpl with population
variances as o,= 2.0 and o,'= 1.5 respectively. Test
whether there is any significance difference in the petrol consumption
of these two types of
cars. (use a = 0.01).
JNTU Apr. 2009 (Set No. 1)

named as A and B.
Solution: Let the types ofthe cars be
Number of cars of type A = 42

Average mileage for A

= =
15, Variance =
o,= 2.0
Number of cars of type B = 80

11.5
Average mileage for B 2
= =

Varianceo, = 1.5

1. NullHypothesis H,: HH
Alternative Hypothes is H, : # H2
2
|15-11.5
The test statistic is
2.1.5
=
3. z
 361
Tests of Hypothesis (For Large Samples)

3.5 3.5
13.587
v0.0476+0.01875 0.06635
Tabulated value of z at 1% significance level is 2.58 (Two-tailed)
Since 2alculiated2.58 (ztable), we reject Null Hypothesis H at 1% levelofsignificance
and conclude that there is a significant difference in petrol consumption.
Example 8A simple sample of the height of 6400 Englishmen has a mean of 61.85
inches and a S. D of 2.56 inches while a simple sample of heights of 1600 Australians has a
mean of 68.55 inches and S. D. of 2.52 inches. Do the data indicate the Australians are on the
verage taller than the Englishmen? (Use a as 0.01). JNTU (A) 2009 (Set No.2)
Solution: We are given
nSize of the first sample = 6400

n = Size of the second sample = 1600

= Mean ofthe first sample = 67.85

= Mean ofthe second sample = 68.55

o= Standard deviation of the first sample = 2.56

G = Standard deviation of the second sample =

2.52

1. Null Hypothesis Ho :4 =H2

2. Alterative Hypothesis H:4 <42

3. Level of significance: a=0.05
4. The test statistic is

67.85-68.55 -0.7
-
(2.56) (2.52) 6.5536 6.35

n 6400 1600 6400 1600

-0.7
-0.7 -9.9
V0.001+0.004 0.0707

1z=9.9> 1.96
Hence, we reject the Null Hypothesis Ho at 59% level of significance and conclude that

the Australians are taller than Englishmen.

Example 9 The mean life of a sample of 10 electric bulbs (or motors) was found to
be 1456 hours with S.D. of 423 hours. A second sample of 17 bulbs (motors) chosen from a
different batch showed a mean life of 1280 hours with S.D. of 398 hours. Is there a significant
between the means of two batches? JNTU (K) 2009, Nov. 2012 (Set No. 1)]
difference
Solution: It is given that
batch 10
Sample size of first
=

nSample size of second batch = 17

 Computer Oriented Statistical Methods
362

Mean life of first batch =

1456

1280
= Mean life of second batch =

Standard deviation of first batch = 423

of second batch
Standard deviation
=
398
G2

1. Null Hypothesis Ho H =42

22. Alternative Hypothesis H:4 *2
Level ofsignificance: a = 0.05
3.
1456-1280
3 -_14
The test statistic is
z=-
4.
, (423) (398
10 17

176 176
1.067
V17892.9+9317.88 164.96

hypothesis Ho ie., there is no difference between

Since z <z, =1.96, we accept thenull
the mean life of electric bulbs of two batches.
with a S.D. of 8. While
The marks scored by 32 boys is 72
Example 10: average
indicate that the boys perform better than
that for 36 girls is 70 with a S.D. of 6. Does this
0.05?
girls at level of significance
be the means ofthe two populations
Solution: Let 4 and 4,
Let the Null Hypothesis be H, :H =H2
is
Then the Alternative Hypothesis H, :4>H2
is difference between H and u
H, is true, i.e., there
no
Let us assume that
the test statistic
sample sizes are large
we use
Since the

Z =-
F-

72,5=70,o, =8,o, 6,n 32, n, =36

= =

Here =

72-70 2 =1.1547<1.96
.z=-

64 36
64 36 2+1
V32 36
value of z is less than the table value, we cannot reject the Null
Since the computed
and girls is the
at 5% level and conclude that the performance of boys
Hypothesis
same.
 363
T e s t so fHypothesis (For Large Samples)
Hvp

students
male
aple 11:At
Example 1 a certain
large university a sociologist speculates that hypothesis,
her
ansiderably more money on junk food than
do female students. To
test
do
students.
Of
randomly selects from the registrar's records the of 200
ologist names
on
amountsp
2 5 are
men and 75 are women. The
these,1125 sample mean of the average
women the
per week by the men is Rs. 400 and standard deviation is 100. For the
unkf o o dp e
difference
Test the
ean is Rs. 450 and he sample standard deviation is Rs.150.
means at .05 level. UNTU (H) Nov.
2010 (Set No. 2)
hetween
the

Solution: Let and k2 be the means of the two populations.

Let the Null Hypothesis be Ho 1 2

Then the
Alternative Hypothesis is H:
and H2
Let us assume
that Ho is true i.e., there is no difference between i

We are given
=125
m-Number of men
75
n-Number of women
=

400
Mean of men
=

450
Mean of women
=

100
S.D. of men
=

a
S.D. of women
=
150
o2
0.05
Level of significance, a =

400-450

The test statistic is

z = 2
(100)(100) -*(150)-
V 125 75
n2
-50 -50
50-2.5654
19.49
80+300 /380

2.5654>1.96 i.e., the

difference is highly significant.conclude that the
z=-2.56541 significance and
=

5% level of
the Null Hypothesis at
Hence, we reject are not equal. those of its main
two population
means
superior to bulbs are
claims that its time 647
A company have a mean life of
Example 12: of 40 of its bulbs
showed that a sample 40 bulbs made by
its main
Competitor. Ifa study of
While a sample
with a S.D of 27 hrs. S.D of 31 hrs. Test
the
of continuous use use with a
nrs 638 hrs of continuous substantiate
COmpetitor had a mean life time of 5% level. (OR Does this
at
two m e a n s
difference of No. 4), Ill yr.
Nov. 2015]
gnificance between
the Nov. 2010 (Set
JNTU (H)
significance)
e claim at 0.05 level of two populations.
and H be
the means
of the
olution: Let ,
 384 Computer Oriented 8tatistical Methods

Let the Null Hypothesis be H

Then the Altermative Hypothesis is 14, : , >
Since the sample sizes are large, the test statistic

F-5

is approximately nomally distributed with mean 0 and S.D. 1.

40
Here = 647,5 638,o, 27,o2 =31,n
= = =
n^ =

647-638 9 =1.38
27)31) 729+961

40 40 40

=1.38 <1.645
cannot reject the Null
Since the computed value of z is less than the table value, we
between the two sample means is
Hypothesis at 5% level and conclude that the difference
not significant.
intersections between 4 p.m and
Evample 13: Studying the flow oftraffic at two busy
6 p.m to determine the possible need for turn signals. It was found that on 40 week days
intersection from the south
there were on the average 247.3 cars approaching the first
which made left turn, while on 30 week days there were on the average 254.1 cars approaching
the first intersection from the south made left turns. The corresponding sample standard
deviations are 15.2 and 12. Test the significance between the difference of two means at

JNTU (H) Dec. 2011 (Set No,1)]

5% level.

Solution: Let the average cars in two places be Hj and p2 respectively.

Let the Null Hypothesis be Ho H =H2
Then the AJternative Hypothesis is Hj :41*#2
Let us assume that Ho is true i.e.,there is no significant difference between Hj and u2

Since the sample sizes are large, the test statistic is

Here m=40,n =30, =247.3,T =254.1, oj #s =15.2 and o2 =

s2 =12.

-6.8
247.3-254.1

15.2) (12) S.776+4.8

4040 30
 Oothesis (For Large Samples) 306

68 6.8
v10.576 3.252 09
2.091>1,6

the computed value of : is greater than the table value, we rejøot the Null
Mhesisat S 9
at level and conchude that the two avernge cars are significantly diflerent
are not the same m the two
busy intersections.
ample 14
Example In a certain factory there are two independent proce»ses for
afa turing the same tem. The average weight in a sample of 700 items produccu
s is found to be 250 gms with a
standard deviation of 30 gms while the onding
ursin sample ot 300 items from the other process are 300 and 40. Is there signicant
krnee between the mean at 1% level.
JNTU (H)Apr. 2012 (Set No.1
olution: Let the average weight in the two independent processes be uj and 2
Rspectively.

Letthe Null Hypothesis be Hg :H H2 =

Then the Alternative Hypothesis is H:H*42

Let us assume that Ho is true i.e., there is no significant difference between A
and
The test statistic is =

Here 2 5 0 , T = 300,o =30,32 = 40 and n = 700, n> =300

250-300 -50
Now 2 = -19.43
900 1600 9,16
700 300 V73
Thus |z=19.43> 2.58
Since the computed value of z is greater than the table value, we reject the Null
Hypothesis at 1% level and conclude that there is a significant difference between the
means.
Example 15: The mean height of 50 male students who participated in sports is 68.2
nches with a S.D of 2.5. The mean height of S0 male students who have not participated in

Sport is 67.2 inches with a S.D of 2.8. Test the hypothesis that the height of students who
participated in sports is more than the students who have not participated in sports.
JNTU (H)Apr. 2012 (Set No. 2)]
Solution: Letthe mean height in the two cases be Hi and H2 respectively.
 Computer Orlented Statietical Method

Let the Null Hypothen be 11

Then the Alternatlve Hypothesls is 14 2
e us assume that H, is true, ie., there is no significant difference between

y and A

The test statistic is z

Here (68.2, = 67.2,o 2.5,a2 2.8, m = 50,n2 = 50

68.2-67.2
Now2 1.88
6.25+7.84
(2.5) (2.8)
50 50 V 50
E=1.88 <1.96
Since the computed value ofz is less than the tabulated value ofz, we accept the Null
Hypothesis at 5% level ofsignificance and conclude that there is no significant difference
in the heights.

Example 16: The nicotent in milligrams of two samples of tobacco were found to be
as follows. Find the standard error and confidential limits for the difference between the

means at 0.05 level.

SampleA 24 27 26 23 25
JNTU (H) Apr. 2012 (Set No. 4)
SampleB 29 30 30 31 24 36
Solution:
Calculations for means and s,s
Sample A Sample B

-( - - (2 -%
=4-25 2-30
24 -1 1 29 -1
27 2 4 30 0 0

26 1 30 0

23 2 4 31 1

25 0 24 -6 36
36 6 36
125 10 180 =2*2|| 74
 Hypothesis (For Large Samples)
Tests

3 T 9.8
21-=25,
5
F, =22 180 = 30
6

Hence variance, s 2-7)

2.5
=

and variance, s2 =14.8

S. E.of G-7)- n 2- ,14.8 5 6

1.72

Hence the 95% confidence limits are

(-E)£1.96 (S. E. of 7 -

2)
ie. (25-30) £1.96 (1.72) i.e.,
-5t3.37 or (-8.37,-1.63)
 CHAPTER-98
TEST OF SIGNIFICANCE
ISMALL SAMPLE TESTS
9.1 INTROOUCTION
of significance based on the theory of
considered certian tests
In the earlier chapter. we
will be valid only for
made in deriving those tests
the normal distribution. The assumptions
normal distribution to test for a
small (n< 30 ), we can use
large samples. When the sample is tests only
of two population means as in large sample
specified population mean or difference whose S.D., o is known. If o is not
when the sample is drawn from a normal population
is normally distributed, the sampling
known, we cannot proceed as above. If a population whether o is
size is also normally distributed
distribution of the sample mean for any sample
known or not

9.2 TEST OF SIGNIFICANCE FOR SMALL SAMPLES

is the study of tests of
A very important aspect of the sampling theory
decide on the basis of the sample results, if
significance, which enable us to

statistic and the hypothetical

i) The deviation between the observed sample
parameter value is significant.
(ii) The deviation between two sample statistics is significant.
The following are some important tests for small samples

(i) Student's t e s t (ii) F-test (ii) 2-test

9.3 STUDENTS ' TEST
Let f= Mean of a sample
n Size ofthe sample
G Standard devíation ofthe sample
Mean ofthe population supposed to be normal.

Then the student's r is defined by the statistic =

s/n-1
E-F*
Ifsbe the sample variance, s =-
n

%-7* n
If s = i
* _ -, then
n-1 n-1

S/yn s/Vn-1
Sis called the unbiased estimate of population variance o*.

402
 significance (Small Sample Tests) 403

If andard deviation
the stane of a Sample then the
statistic i
Note.
sample is given
given directly,
given by t=
S.DVn-1
2.Ifthe calculated value of r exceeds the table value of t at 5% evc of
significance, the null hypothesis H,is rejected. If the calculated valueo
1 is less than the tabulated value of t at 5% level the null hypothesis
accepted.

mptions for t-test:

Student's

ollowing are made in student's t test

The foll
)Sample size, n<30
(2)The parent population from which sample is drawn is normal.

(3)The population standard deviation is unknown.

is random.
4)The sample observations are independent, i.e., sample
Usesoft-test:
This test is used

(1) to test for a specified mean

drawn from
equality
to test for of two means of two independent samples
2) and
two normal populations,
S.D. of the populations being unknown,
data.
of difference between the means of paired
(3) to test the significance
FOR
94 CONFIDENCE OR FIDUCIAL LIMITS
which the
find from the sample data the limits within
we want to
Suppose called the 95%
lie with a probability of 0.95. The limits are
population mean will mean for the given sample.
confidence limits of the population freedom at 5% level of
is the table value of t
for (n-1) degrees of
If aos
for are
given by F thas
95% confidence limits
Significance, then
For P(1|> 1,)=0.05
i.e., P(1t|s%0s) =0.95
95% confidence limits
for are given by

| 1 o o s Le.. s / n os or

S/n SpST +o0s S/Vn

-1o05 is thee
limits for areFt 0 S/Vn where o
Similarly 99% confidence
freedom at 1% level of significance.
for (n-1) degrees of
Dulated value of t
 404 Compiter Oriernted
Statistical Methoxth
.8 STUDENTS *'TEST FOR SINGILE MEAN
Suppose we want to test
( a ) i a random sample x,of size n haAN been drnWn from normal populasr
ion
with a specified mean .
(b) i fthe sample mean differs significantly from the hypothetical value.

the population mean.

In this case the statistic is given by (»-) Where F, H, S, n hav

e

usual meanings.
Let a random sample ofsize n (n < 30) has a sample nmean x. To test the hypothesi

that the population mean has a specified value H, when population S. D. a is not knou
OWn.
Let the Null Hypothesis be Ho:=Ho
Then the Alternative Hypothesis is H :* Po
-
that is true, the test statistic given by l=. =, where s is the
Assuming H, s/n-1

sample S. D. follows distribution with v=(n-1) d.f.

-

We calculate the value of |1| and compare this value with the table value of t at
a level of significance. If the calculated value of1> the table value of t, we reject H,at

a level. Otherwise we accept Ho.

S
In this case, 95% confidence limits for the population mean are t l g
n-
where a =0.025 for two-tailed test and s= sample S. D. and 99% confidence limits for

are xlan-1 where a =0.05.

For a two-tailed test at a level of significance, value of a/2 is taken for a

SOLVED EXAMPLES
Example 1: A mechanist is making engine parts with axle diameters of 0.700 inch.
A random sample of 10 parts shows a mean diameter of 0.742 inch with a S.D. of 0.040 inch.
Compute the statistic you would use to test whether the work is meeting the specificationa
0.05 level ofsignificance JNT (A) Apr. 2012 (Set No. 2)
Solution: Here the sample size n = 10 < 30

Hence the sample is small sample.

Also sample mean f = 0.742 inches, the population mean u = 0.700 inches and

S,D.= 0.040 inches are given.

 405
niRicance (Small Sample Tests
We use student's t.Test

wll Hypothesis ,
Null Hypothe The product is
confirming to specification.
ternative Hypothesis H,: u 0.700
Level of sigaificance, a = 0.0S
Lev

The test statistic is, t =

Swn-1
Here 0.742 inches, u = 0.700 inches, S.D. = 0.040 inches and n 10
negrees of freedom (d.) =n - 1 = 10 - 1 =9

0.742-0.700
= 3.15
. 0.040
V10-1
calculated value of r 3.15 =

Thus the
value oft at 5% level with 9 degrees of freedom is as22
The tabulated
value of t > tabulated value of t, therefore, H, is rejected.
Since calculated
product is not meetingthe specification.
The
hours with a SD. of 20
sample of 26 bulbs gives a mean life of 990
Eumple 2:
Example2:A is 1000 hours Is the sample
not
clams that the mean life of bulbs
hus
The manufacturer
t o the starndard.
NTU(A) Nov. 2010(Set Na. 2,4)]
size, n= 26 < 30
Solution: Here sample
The sample is small sample.
+ =
990
Also given, sample mean,
1000 and S.D..s =
20
Population mean, u=
=n-1 = 26 1 =
25
Degrees of freedom
and n.
Here we know ï, 4, S.D.
students T test.
So we use standard.
the
The sample is upto
1. Null Hypothesis H,: 1000
<

2. Alternative Hypothesis H: H
standard) (left-tail test)
(The sample
is below
: a =0.05
3. Level of significance 990-1000
- - 2.5
T- 20 25
is t s/Nn-1
4. The test statistic
| t | =2.5 = 2.5 left-tailed test
Calculated
value ofi of freedom for
i.e., level with 25 degrees
of r ' at 5%
Tabulated value conclude

is1.708. the null

hypothesis H, and
r, we reject
>
tabulated
calculated r
Since the
standard.

upto
that sample is
not
 406 Computer Oriented Statistical Methods

Evample 3: A machine is designed to produce insulating washers for electrical devices

0 average thickness of 0.025 em. A random sample of 10 washers was found to have a
thickness of 0.024 cm with a S.D of 0.002 cm. Test the significance of the deviation. Value of
for 9 degrees of freedom at 5% level is 2.262.
Solution: Here the sample size is 10< 30
The sample is small
Also given Sample mean, 7 = 0.024 cm
Population mean ,u =
0.025 cm

S.D. = 0.002 cm

Degrees of freedom (d.f) =

n -1 =10 -

1 =
9
1. Null Hypothesis H,: The difference between and u is not significant.
2. Alternative Hypothesis H,: H* 0.025
3. Level of significance: a =0.05
-H 0.024-0.025 - 1.5
4. The test statistic is f =

s/n-1 0.002
10-1
| | = 1.5
Calculated value oft = 1.5 for two tailed test.
Tabulated value of r for 9 degrees of freedom at 5% level = 2.262

Since calculated < tabulated t, we accept the null hypothesis and conclude
that the difference between F and u is not significant.
The mean life time of a sample of 25 fluorescent light bulbs produced by
Example 4:
a company is computed to be 157 hours with a S.D of 120 hours. The company claims that the
average life of the bulbs produced by the compa y is 1600 hours using the level of significance
of 0.05. 1s the claim acceptable?
Solution: Given sample size, n =
25

Sample mean, + 1570=

Population mean, u 1600 =

S.D.(s) 120

Degrees of freedom = n
-

1 =
24
1. Null Hypothesis H : The claim is acceptable. H = 1600 hrs

2. Alternative Hypothesis H, : H# 1600 hrs

3. Level of significance : a =0.05

- 1570-1600-30
4. Thetest statistie I s/n-i120//2424.49*44
is

= 1.22
i.e., Calculated1 = 1.22
 407
Test of

af Significance
Significane
(Small Sample Tests)
The tabulated value of t at 5% level with 24 degrees of freedom for two tale
Iest 1s 2.06

Since the calculated value of i < tabulated value of t, we accept

hypothesis.
nV
sis H i.e., the claim that the average life of the bulbs produced by the company
hrs 1s acceptable.
is1600
Example 5 : The average brcaking strength of the steel rods is specifieca to u 18.5
thousand
pounds. To test this sample of 14 rods were tested. The mean and standard deviations
ned were 17.85 and 1.955 respectively. Is the result of experiment significant?
obtained

UNTU (H) May 2011 (Set No. 3), (K) Nov. 2011 (Set No. 1))
Solution: We are given
Sample size, n = 14

Sample mean, f = 17.85

S.D. (s)= 1.955

Population mean, u = 18.5

Degrees of freedom =n-1 = 13

1. Null Hypothesis H,: The result of the experiment is not significant.

. Alternative Hypothesis H,: 18.5
3 Level of significance: a = 0.05

-H 17.85-18.5 0.65
- 1.199
4. The test statistic is, s
t =

/ n - i . 1.955/13 0.542
| t =1.199
i.e., Calculated t = 1.199

Tabulated t at 5% level of significance for 13 d.f. for two tailed test = 2.16

level
Since calculated t < tabulated t, we accept the Null Hypothesis H, at 5%
and conclude that the result of the experiment is not significant.
of size 16 values from a normal population showed a
A random sample
Example 6:
mean of 53 and a sum of squares of deviations from the mean equals to 150. Can this sample
Obtain 95% confidence limits
be regarded as taken from the population having 56 as mean?.
JNTU (K) May 2012 (Set No.2)]
of the mean of the population.

Given sample size, n =

16
Solution:
Sample mean, f= 53

Now 2(x, -F)* = 150

= -
1 =
10 » S =V10
n-1

=
n
-

1 = 16-l= 15
Degrees of freedom, v
 408 Computer Oriented Statistical Methods

() 1. Null Hypothesis H,: The sample is taken from the population having 56

as mean i.e.. t = 56

2. Alternative hypothesis H, : |l 56

3. Level of significance: a =0.05

4. The test statistic, r = -s/ INote that S.D. is not given dircctly]

53-56
V10/I6-3.79

1t|= 3.79
d.f for two tailed test
The tabulated value oft at 5% level of significance for 15
is 2.13.

Since calculated t > tabulated 1, the null hypothesis H, is rejected i.e., the
sample cannot be regarded.as taken from the population.

(b) The 95% confidence limits of the mean of the population are given by

1.6827.
Tt o.05=53t2.13x0.79' = 53 +

54.68 and 51.31

Hence 95% confidence limits are [51.31, 54.68]
compressive strength of
Example 7: A random sample of six steel beams has a
mean

this information
58,392 p.s.i (pounds per square inch) with a standard deviationof 648 p.s.i. Use
of
and the level of significance a =0.05 to test whether the true average compressive strength
the steel from which this sample came is 58,000 p.s.i. Assume normality.
2013 (Set No. 1)1
JNTU 2005S, 2006S, (H) Nov. 2010, (H) May & Dec. 2011 (Set No. 2), (K) May
Solution: We have
n Sample size (number ofsteel beams) = 6< 30. .. The sample is small.
= Sample mean (average compressive strength) = 58392 psi.

S = Standard deviation of six beams = 648 psi

1 6- 1 55
Degrees of freedom (d.f.) = n - =

In this problem o is known and n < 30. Hence we use t - distribution.

1. Nul Hypothesis i s H : = 58000

2. Alternative Hypothesis H,: 4 # 58000

Level of Significance : a = 0.05

3.
', the test is
4. Critical Region: Since Alternative Hypothesis is of the type #
two-tailed and the critical region is -3.365 < t< 3.365
st C signiflcance (Small Sample Tests) 409

The Test Statistic is fm 58392-58000

s/n-1 1.353
648//5
ince .353 <
3.365 =

a/2 we accept the null hypothesis H

Hence the average compressive strength of the 1 to
steel beam is not c
58000 psi.
Example 8:
8:AA sample of 100 iron bars is said to be drawn from a large number of
AuSwhose lengths are normally distributed with mean 4 feet and S.D. 6 f. If the sampie
bars

san
m e a n is 4.2 feet, Can the sample be regarded as a truly random sample?
UNTU (H) Nov. 2010 (Set No. 2)]
Solution: Let the Null Hypothesis be H,: The sample can be regarded as a truly

random sample.

Then the Altermative Hypothesis is H: The sample cannot be regarded as a random

sample.
Here n= Sample size =100
= Sample mean = 4.2

Population mean = 4
G= S.D. of population =0.6
Let us assume that H, is true. The test statistic is

f- 42-42_10
o/Vn 0.6/10 0.6 3

3.33 >3
cannot be
conclude that the given sample
Hence we reject the Null Hypothesis and
random sample from the given
population.
regarded as a truly
is not given, we may reject H at 5%
or

Note: Since |z|>3 and level ofsignificance

is rejected without
In the above problem, the Null Hypothesis
1% level of significance.
significance.
mentioning any level of mean 67 and
S.D.5.2. Is this sample
of 155 members has a
Example 9:A sample JNTU (H) Nov. 2010 (Set
No. 3)1
of mean 70?
nas been taken from a large population
takenfrom population
be The sample has been
Hypothesis H:
Solution: Let the Null been taken from population.
is The sample has not
Hypothesis H:
Thenthe Alternative

Here n Sample size =155

 410 Computer Oriented 8tatisticel Methods

= Mean ofthe sample = 67

a= S.D. of sample = 5.2

70
Mcan of the population =

67-70
is
Thetest statistic S.D/Vn-I 5.2/155-1

-3 -7.16

5.2/154
1 =7.16>3
Hypothesis at 5% level ofsignificance and conclude that the samole
We reject the Null
the given population.
has not been taken from

machine has produced washers

having a mean thickness
Example 10:In the past a
the machine is in proper working order a sample of 10
of 0.050 inch. To determine whether
washers is chosen for which the mean thickness is 0.053 inch and the standard deviation is

0.003 inch. Test the hypothesis that the machine is in proper working order using a level of

significance of 0.05?
Solution: We have = 0.050,n = 10,7 = 0.053,s = 0.003, where & is the sample S.D.

Since the sample size is small and population S.D a is not known, we uset- test.
Let the Null Hypothesis be Ho:u = 0.050

Then the Alternative Hypothesis is H,:p:0.050

0.053-0.050 0.003
The test statistic is = X-H =3
s/Nn-1 0.003/V10-1 0.001

Degrees of freedom =n-1=10-1=9.

The tabulated value oft for 9 d.f. at 59% level = 2.262
Since the calculated value of t> the tabulated value, we reject the Null Hypothesis at
5% level and conclude that the machine is not in proper working orde.

Example 11: The following are the times between six calls for an ambulance in a
city and the patient's arrival at the hospital: 27,15,20,32,18 and 26 minutes. Use these
on the
figures to judge the reasonableness of the ambulance services claim that it takes
ambulance and patient's arrival at the hospital ?
average 20 minutes between the call for an
JNTU (A) Nov. 2011 (Set No. 1)

Sample size 6
Solution We have n
=
=

27+15+20+32+18 +26 138

T mean of the sample =- 23
6 6
 of ificance (Small Sample Tests) 411

u=meanofthe population - 20
and

s samplevariance 24-*
n-1

-(27-23) +(15-23+ (20-23) +(32-23 +(18-23 +(26-25)

(16+64+9+81+25+9)==40.8
5
SSample S. D. = V40.8 =6.4

Since the sample size is small and population S. D. o is not known, we use f- test.

Letthe Null Hypothesis be Ho:u=20

Then the Alternative Hypothesis is H :H# 20
The test statistic is

T23-20 36 7.348 1,148

S/n 6.4/V6 6.4 6.4
Degrees of freedom =n--1=6-1=5
The tabulated value of t for 5 d.f. at 5% level = 2.571

for two -tail test, a/2 is taken for a]

|t|=1.148<2.571

Since the calculated value of |t|< the tabulated value, we accept the Null Hypothesis
at 5% level and conclude that the average time between the call for an ambulance and
patient's arrival at the hospital is signifiant.
Problems related to Student's t Test (When S.D of the sample is not given directly)

SOLVED EXAMPLES
Example 1Arandom sample of 10 boys had the following LQ's : 70, 120, 110, 101,
8,83,95, 98, 107 and 100.
(a) Do these data support the assumption ofa population mean I.Q of 100?
6) Find a reasonable range in which most ofthe mean 1.Q. values of samples of 10 boys
lie. UNTU2008,(A)Apr.2012,(H) May 2011 (Set Na.4)(K) May2013(Set No.2)
Solution: (a) Here S.D. and mean of the sample is not given directly.
We have to determine these S.D. and mean as follows.

972
Mean, F L-4=
10
97.2
 Statistical Methods
Computer Orlented
412

739.84
70 -27.2
519.84
120 22.8
163.84
110 12.8
14.44
3.8
101
-9.2 84.64
88 201.64
-14.2
83 4.84
-2.2
95 0.64
0.8
98 96.04
9.8
107 7.84
2.8
100
1833.60
972

2%-F} - 1833.60
We know that S? -F }
n-l
9

14.27
Standard deviation, S= v203.73 mean
assumption of a population
The data support the
1. Null Hypothesis H:
the population.
L.Q of 100 in
100
2. Alternative Hypothesis H, : 4

a = 0.05
3. Level of significance,
97.2-100
- 0.62
-0.62

4. The test statistie

is
s/n 14.27//10
Calculated value of =0.62
t = 0.62 i.e., significance is
9 d.f. at 5% level of
(10 1) d.f. i.e.,
Tabulated value of t for

2.26 (two-tailed list).

the null hypothesis
value ofi <
tabulated value of1, we accept
Since calculated of 100 in the population.
the assumption of mean 1.Q
H. ie., the data support
confidence limits are given by thos S/vn
(b) The 95%
10.198 = 107.4 and 87
4.512 = 97.2
97.2 2.26
x
=
of
the mean I.Q values ofsamples
confidence limits within which
T h e 95%

10 boys will lie is (87, 107.40). be 70, 67, 62, 68,

are found to
males ofa given locality
The heights of 10
the average height greaterthan 64
2: is
Fample reasonable to believe that at
61,68,70,64,64,66 inches.
Is it freedom (= 1.833
for 9 degrees of
level assuming that
inches? Test at 5% significance 2011 (Set No. 2)]
UNTU (H) May
a 0.05).
Test Significance (Small Sample Tests)

Solution: Mean, += 2 660

66
10
Calculation for sample mean and S.D.
(r-7)

70 4 16
67 1
62 16
68 2 4
61 5 25
68 2
70 16
64
64 -2
66 0
660 90

We know that S= E-R}-90=10

Sample S.D.(S) = V10 =3.16

1. Null Hypothesis H, : The average height is not greater than 64 inches
ie, 4 = 64 inches.

2. Alternative Hypothesis H,: > 64 inches

3. Level of significance: a =0.05
4. Teststatistie is t = - 66-64 =1.9
S/n-1 3.16/9 L.051.3
ie., Calculated value = 1.9
Tabulated for 10- 1 = 9 d.f at 5% level of significance for single tail test is
1.833 (given).

Since calculated value of t > tabulated value of i, we reject the null

hypothesis H i.e., The average height is greater than 64 inches.
Example 3 A random sample from a company's very extensive files shows that the
rders for a certain kind of
machinery were filled, respectively in 10, 12, 19, 14, 15, 18, 1l and
15
days. Use the level of significance a =
0.01 to test the claim that on the average such
Oorders filled in 10.5 days. Choose the Altemative Hypothesis
are
so that rejection of Null
Hypothesis =10.5 days implies that it takes longer than
indicated.
JNTU 2004S, (K) May 2010S (Set No. 3)]
 Statistical Methods
414 Computer Oriented

Solution: We have
112
14
15 + 18 + 11 +13)
8, F (10 + 12+ 19 4 14

and sample S.D, S is given by

S2-7*
n-1
14)'+. + (13 14)1
14) +(19
-

(10 14) + (12 -

+4+ 25 +0 + 1 + 16
+9 + 1)
(16
(72) 10.286

S V10.286 3.207

1. Null Hypothesis H, : H
=
10.5 days
10.5 days
Alternative Hypothesis H, >
:
2.
: a =
0.01
3.
3. Level of significance
4. Critical region :1> t%o
2.998
Reject H, ift> 1,o1
5. The Test Statistic is t sJn-i

14-10.5
3.207/7 3.087

at 1% level of significance is 2.998 (two

Tabulated value of t for (8
-

1) d.f.
tailed test). the null hypothesis H
tabulated value of 1, we reject
Since calculated value of t>
are filled in more
than 10.5 days.
i.e., The orders on average from a large
for a random sample of 10
The life time of electric bulbs
Example 4:
consignment gave the following data. 9 10
Item 2 34 5 6 7 8
52 3.8 3.9 43 44 5.6
Life in1000 hrs: 12 46 39 4.1 of bulbs is 4000 hrs. Use a 0.05
the that average life time
Can we accept the hypothesis JNTU Nov. 2008 (Set
No. 2)1
level of significance.
10.
Solution: Here n
=

4000
1. Null Hypothesis H:=
4000
2. Alterntive Hypothesis H, :H
: a
=
0.05
3. Level of significance

4. The test statistic is I S/yn

 Gignificance (Small Sample Tests) 415
TeSta

From given data, x =

average life time of bulbs

.2+4.6+ 3.9+4.1+5.2+3.8+ 3.9 +4.3+4.4+0
10

10(41)-4.1 4100
10

Now S =

(-}
10.2-4.1) +(4.6-4.1)? +(3.9-4.1)' +(4.1 -4.1) +(5.2-4.1)
+(3.8-4.1) +(3.9-4.1) +(4.3-4.1? (4.4-4.1) (5.6-4,1)"J
8.41 +0.25 +0.04+ 1.21 +0.09 +0.04+0.03+0.09+ 2.25)

12.42 1.38 1380 1380

=
=
S =

4100-4000
Hence S/yn V1380/1o8.512.
t=8.512
No. of degrees of freedom = 9

a22.262 (Fromtables)
Thus we see that |t |> la/2

Hence, we reject the Null Hypothesis Ho at 5% level and conclude that the average life
time bulbs is not equal to 4000 hrs.

Example 5: The manufacturer ofa certain make of electric bulbs claims that his bulbs
have a mean life of 25 months with a S. D. of 5 months. A random sample of 6 such bulbs gave
the following values. Life of months: 24, 26, 30, 20, 20, 18. Can you regard the producers
claims to be valid at 1% level JNTU (A) 2009 (Set No. 2)
ofsignificance?
Solution: We have n =6,
24+26+30+20+20+183823
Samplemean, F=4
6

and sample S.D, S is given by S2=24-X)

n-1

ie., S =[(24-23) +(26-23) +(30-23) +(20-23) +(20-23)' +(18-23)

=[1+9+49+9+9+25] ==20.4

S=V20.4=4.52
416 Computer Oriented Statistical Methods

1. Null Hypothesis Hg u= 25

2. Alternative Hypothesis H:25

3. Level of significance: a - 0.01

4. The test statistic isI=

S/n

ie,, ft 2 - 2 5 -24.52 -1.08

452-108
i.e.,4.52/6
t=1.08

ie., calculated value of =1.08

level ofsignificance is 3.365.
Tabulated value of for (6-1) degrees offreedom at 1%
tabulated value of r1, we acceptthe
null hypothesis H, and
Since calculated value oft <
valid.
conclude that the claim is in his 'gutkha' on
that the nicotine content
Producer of 'gutkha', claims
Example 6: of 8 'gutkha' of this type
Can this claim accepted if a random sample level of
the average is 1.83mg. 1.6mg? Use a 0.05
2.1, 1.9, 2.2, 2.1, 2.0,
contents of 2.0, 1.7,
have the nicotine 2009, Apr. 2012 (Set No. 1)
UNTU (K) (A)
significance.
8 and u
= 1.83 mg
Solution: Given
=
n

1. Null Hypothesis Ho: u=1.83

H:4*1.83
2.Alternative Hypothesis
a = 0.05
3. Level of significance:
-
4. The test
statistic is 1= n-1)
S/Vn
the sample values gutkha.
to be computed from
where and S are

x- (r-
0.05 0.0025
2.0
-0.25
0.0625
1.7
0.0225
2.1 O.15
-0.05
0.0025
1.9
0.0625
2.2 0.25
0.0225
2.1 0.15
0.0025
2.0 0.05
-0.35
0.1225
1.6
15.6 0.3
Total
 Signlficance (Small Sample Tests) 417
Test

15.6
Now F--1.95 and s.2-_0.3 or S-0.21
S 0.21
7

1.95-1.83 (0.12V8).
0.21/8 0.21
1.62
Tabulated o0s for (8-1) i.e., 7d.f is 1.895.
Since calculatedi< tabulated fos. the mull hypothesis H, may be accepted at 5% level
ofsignificance
significance and we may conclude that the data are consistent with the assumption of gutkha
is 1.83me
on the average
Example 7: Eight students were given a test in STATISTICS and after one month

aching they were given another test of the similar nature. The following table gives the
ercase I their marks in the second test over the first.
incr

Student No. 23456 7 8

Increase of Marks 4-2 6 -8 12 5 -7|2
Do the marks indicate that the students have gained from the coaching.
JNTU 2008,(A) Nov. 2011 (Set No. 2)]
follows.
Solution:We shall compute the mean and S.D. of the increase of marks as
Zx 4-2+6-8+12+5-7+2 12-15

We have S E-F}
[4-1.5? +(-2- 1.5? +(6-1.5 +(-8- 1.5)
+(12 1.5) +(5 1.5) +(-7- 1.5) (2-1.5)
+ -

110.25+ 12.25+ 72.25 +0.25]

=
[6.25+ 12.25+ 20.25 +90.25+

324 46.2857
S.D.S =/46.2857= 6.8
by coaching, it implies that the
mean
benefitted
have not been
Assuming that the students
O1the difference of the two tests is zero i.e., =0.
1.5-0 0.3830

hen S/n-1(6.8)//7
8 1=7
of degrees of freedom
= -

No.
Tabulated os 2.36
at 5% level significance
the value oft is not significant
As the calculated value oft <tos benefítted by coaching,
students have been
test evidence tht the
h e provides no
 418 Comnterrietad Stafietica Mehod

A random sample of 10 hags of pesticide are taken whese weights are 5

49. 52. 44, 45, 48, 46. 45, 40, 45 (in kgs) Test whether the average packing cam he teken to he

50 kes. UNTU H)Apr. 2012 (SerNe.

Solution: We have u- 50,n- 10

Let the Null Hypothesis he Bo -so

50
Thenthe Aternative Hypothesis is H,
is I=
Thetest statistic s/Vn-1

47.2
Now (50+ 49+ 51+ 44+45+48+46+45+49+45)= 10

Variance, -
n-1

= i(50-472+(49-47.2 +(51-47.2) +(44-47.2)

(45-47.2) +(48-47.2) +(46-47.2)

(45-47.2) ]
+(45-47.2) +(49-47.2) +

+4.84
=17.84 +3.24+14.44+10.24
=6.1777
4.84+3.24+ 4.84]=
+0.64 + 1.44 + 9

= V6.18 = 2.48
Standard deviation, s

47.2-50 -8.43.39
Hence 248/3 2.48
=n-1 = 10-1 =9.
Degrees of freedom at 5% level
=2.262
ofrfor 9 d.f.
Thetabulaed value test, a/2
is taken for a ]
echanges in
two-tailed
[: for
t =3.39>2.262 a d m i n i s t e r e d to
12 patientsresulted
in the following
stimulus will in
stimulus that the
Example 9:A
concluded
Can it be (Set No.1)
No.)}
UNTUÇK) May.2012
5 , 2 , 8 , - 1 , 3 , 0 , - 2 , 1 , 5 , 0 , 4 , 6 .

JNTUK)
the blood pressure: increase in blood pressure? the m e a n
accompanied by an
We shall compute
general be increment in blood
pressure.
the
Letr denote
Solution: as
follows:

standard
deviation of blood pressure
and 2.58
F- (5+2+8-1+3+0-2+1+5+0+4+6) 2.58
12 12
 Test of significance (Small Sample Tests) 419

Also S 2 2%-7*

144 +(2-2.58)? +(8-2.58)2+ (-1-2.58)2 +(3-2.58)" +

(0-2.58) +(-2-2.58)2 +(1-2.58? +(5 -2.58)2 +(0-2.58)

(4-2.58) +(6-2.58)1
9.53
S 3.08.
1. Null Hypothesis: Ho:4 =H2.
2. Alternative Hypothesis: H # H2.
3. Level of significance: a = 0.05

2.58-0 2.58
4. The test statistic is t= 2.89
S/n 3.08/V12 0.89

Tabulated lo.05 for (12-1) degrees of freedom =

2.2
level of
Since calculated tabulated t, the Null hypothesis is rejected at 5%
1 >
be accompanied by an
significance and we may conclude that the stimulus will in general
increase in blood pressure.
diamonds can be operated at a
Example 10: A new process of producing synthetic is To test
level only if the average weight of the
diamonds greater than 0.5 carat.
profitable 0.46,0.60,0.52,0.49,0.58
with
diamonds are produced weights
the profitability ofthe process, 6 evidence to indicate
0.54 carat respectively. Do the 6 measurements present sufficient
and in excess of 0.5 carat ?
of the diamonds produced by the process is
that the average weight JNTU(A) Dec. 2014]

3.18
*=0.53
+0.60+0.52 +0.49 +0.58+0.54) 6
=

Solution: Here ï =(0.45

6
Mean and S. D.
Calculation for Sample

(r-F
-0.08 0.0064
0.45
0.07 0.0049
0.60
- 0.01 0.0001
0.52
-0.04 0.0016
0.49
0.05 0.0025
0.58
0.01 0.0001
0.54
0.0156
3.18
 420 Computer Oriented Statistical Methods

:.S -F}=x0.0156 =0.00312

n-l

S=v0.00312 0.056
1. Null Hypothesis H : The average weight of
the diamond is not greater than 0.5
carat ie.u=0.5
2. Alternative Hypothesis H:4>0.5
3. Level of significance: a =0.05
4. Test statistic is t=
S/Vn

ie. t=- 0.53-0.5(0.0363122

0.056/6 0.056
5. Critical Region :
Tabulated value of t for (6-1)d.f. at 5% level of significance for one tail test is
2.015
Decision:Since calculated value oft < tabulated value of t, we accept the null
hypothesis Ho. ie., the average weight ofthe diamond is not greater than 0.5
carat.
The calculated does not fall within the rejection region.Therefore, we cannot reject
Ho. Hence the data do not give sufficient evidence to indicate that the mean diamond
weight exceeds 0.5 carat.
Evample 11:A random sample of 8 envelopes is taken from the letter box of a post
ofice and their weights in grams are found to be: 12.1, 11.9, 12.4, 12.3, 11.5, 11.6, 12.1 and
12.4.
() Does this sample indicate at 1% level that the average weight of envelopes received at
their post ofice is 12.35 grms.
()Find 95% confidence limits for the mean weight of the envelopes received at that
post office.
Solution:
( ) 1 . Nall Hypothesis is Ho:u =12.35 grms i.e., the samples have been drawn
from the population of mean 12.35 grms.
same

2. Alternative Hypothesis is H:u*12.35 grms

3. Level of significance: a =0.01

4. The test statistic is

t= S/Vn
Here n= 8.
 significance (Small Sample Tests) 421
TaST

Now, sample mean,

F-(12.1+l1.9+12.4+12.3+11.5+11.6+12.1+12.4) =96.3 12.03

Sample variance, S =

i n-1

a2.1-120375) +(11.9-12.0375) +(124-12.0375+(12.3-12.0375)*

11.5-12.0375) +(01.6-12.0375)2 +(12.1-12.0375)2 +(12.4-12.0375)]
[0.0039+0.0189+0.1314+0.0689 +0.2889+0.1914 +0.0039+ 0.1314]

0.83870.1198
7

S=0.1198 =0.3461
Thus 112.0375-12.35 2.8284(-0.3125)-0.883875 = -2.5538
0.3461/8 0.3461 0.3461

2.5538 ie., calculatedvalue of t=2.5538

Tabulated value of f for (8-1) degrees of freedom at % level of significance is

3499 ie., a/n =o.005 =3.499

Since calculated value of t < tabulated value oft, we accept the Null Hypothesis Ho
and conclude that the sample istaken from the population whose mean is 12.35.
S
(i)The 95% confidence limits are given by Xtla/2

Ow a/2o.025=2.365 [From Tables]

a/2 Z.305X0.5461)0.818502R04
0.2894
8 2.8284

ience confidence limits (12.0375-0.2894,12.0375 +0.2894)

= (11.7481, 12.3269).
 Statistical Methods
Computer Orientecd
422

OF MEANS
9.6 STUDENT'S TEST FOR DIFFERENCE
and n,
samples of sizes n
Let a n d be the means of two independent
To test
two normal populations having
means
, and 4
from
(30,n, <30) drawn whether the difference -4, is
mcans are equal (i.e., to test
whether the two population
significant).

the Null Hypothesis be Ho:4

42
Let

Hypothesis is H, 4 2
Then the Altermative
common
variance g is given
then an unbiased
estimate S* of the
lf = o,
o, =
o,
variances.
the two sample
by S?=j FU;d2 where si,s3 are

t",S2
Also S. E. of
(T-5) =S+ where where S=
+n2
given by
1=. ,follows
test statistic
that Ho is true, the
Assuming

1-distribution with (n +n,

-

2) d.f.

and
Here f= i=l

s E-+20-J
S +n-2

level
value at a
with the table
tn-2
this value
and compare at a level.
calculate the value of | | we reject Ho
We the table value,
of |t|>
calculated
value
If the
ofsignificance. means are
accept H. two population
Otherwise we difference of
limits for the
confidence
95%

where a =0.025.
F-)tla difference of two
population
means
are

limits for the

confidence
and 99%

F - F ) 4 S + w h e r e a = 0 . 0 0 0 5
 Significance (Small 423
Test o f
Sample Tests)

SOLVED EXAMPLES
Exampic amples of two types of electric light bulbs were tested for leng
A following data were obtained

Type
Type I1
Sample number, n, = 8
n,-7
Sample mean, f = 1234 hours
= 1036 hrs

Sample S.D., 3, =36 hrs s 4 0 hrs

Is the difference in the means sufficient to warrant that type I is erior

ype
II regarding length of life. UNTU (A)June 2016]
Solution: Sncethe sample sizes are small and g,,O, are not known, weusef-test. Let

and u be the two population means.

Nall Hypothesis H, : The two types I and II of electric bulbs are identical

Alternative Hypothesis H, : H, *
Since two sample means 7 and 5 are given and also sample standard deviations

S. and S, are given, we use the statistic

t
-

where
+n-2
1
+ 7(40)] =
1659.08
8+7-28(36)*

1234 1036
= 9.39

659.08
8 + 7-2 13
n, + n, -2
=

of freedom (d.f)
=

Degrees
Tabulated value of t for 13 d.f at 5% level is 2.160 (tw0-tail test)

Since calculated > tabulated , we reject the null hypothesis H, and conclude
Unat the two types I and II of electric bulbs are not identical.

random samples of sizes 9 and 7 are 196.42 and 198.82

Example 2:The means of two
deviations from the mean are 26.94 and 18.73
Espectively. The sum of the squares of the been
pectively. Can the sample be considered to
have drawn from the same normal population.
7, F= 196.42, j
= 198.82 =

Solution: Given n, 9, n,
=

and -F?= 26.94, - 5 = 18.73

 Computer Oriented Statistical Methos
424

+n2 -2

26.94+18.75 - 3.26
9+7-2
= S 1.81 population
drawn from the
same
are
The two samples
1. Null Hypothesis H,:

2. Alternative Hypothesis H, 4, H
: a = 0.05
3. Level of significance
I - = 196.42-198.82

is i==
4. The test statistic
(1.81)
n2

-2.4 = - 2.63
0.912
2.663
of | t|
=

calculated value 2.15.

ie,
5% level
significance is
of
of t for 9
+7-2 14 d.f
=
at the null hypothesis H,
and
Tabulated value w e reject
tabulated value of 1,
value oft > population.
calculated from the same
Since are not drawn
conclude that the two samples of pigs fed
o n two

in weights (in 1bs)

are given the gain
Example 3:Below
B.
diets A and
30 31| 35 25 14 32 24
34 24
25 32 30 35 18 21 35 29 22
21 35|
Diet A 30
31 40 30 32
22 10 47 in increase

44 34
on
Diet B 44| regards
their effect

significantly
as

diets differ
Test, if the t w o

weight. mean
between the
difference
Solution: is no
significant
There

Null
Hypothesis H and B i.e., F,
=

H2
1. diets A
due to
increase in weight
H,: , 4
Hypothesis
Alternative
2. = 0.05
a
significance,
Level of
3. means
and S,D's.
for sample = 450
4.
C a l c u l a t i o n

336, Ey
15, Ex
=

=
12, n,
Here n,
450 30
336 28, F 15
. 12
 425
sianificance (Small Sample Tests)
o

(-T ( F y -F}
3 9 44 14 196
16 34 4 16
2 22 8 64
6 36 10 20 400
4 16 47 17 289
-14 196 31 1

32 4 16 40 10 100

24 16 30 0 0

30 4 32 2 4

9 35 25
31
49 18 -12 144
35
-3 9 21 -9 81
25
35 25
29 -1
22 -8 64

380 450 1410

336
15
--380, - il
=1410
i

S=
+n-2

1
12+15-7380
+ 1410] =
71.6

5. The test statistic is

30-28 2
10.74 = 0.609
-j 1

12 + 15 -2 25
n, tn, -2
=

Degrees of freedom
Tabulated t for 25 d.fat 5% level 2.06
< tabulated 1, therefore, accept the null hypothesis H
we
Since calculated their effect on increase in
differ significantly as regards
That is, the two diets do not
weight. 60 and
of 5 patients treated with medicine A weigh 42, 39, 48,
Example4: A group with medicine B weigh 38,
of 7 patients from the same hospital treated
4 kgs. Second group Do agree with the claim that medicine B
increases the
62 kgs. you
2, 56, 64, 68, 69 and No. 3)]
UNTU (K)Nov. 2011 (Set
weight significantly.
 426 Compter Oriented Statistieal Methonh

Solutlon: Calculation for sample means nnd 5.D's

42 16 38 19 361
39 49 42 15 225
48 4 S6
60 14 196 64 49
41 25 68 21
69 12 44
62 5 25

230 0 290 399 0 926

230 399
Now T 46, 57,
and --290, Z0-5} =926
S=
2 -}+ 2 ( -YP]= E_ 1290 +
926] 121.6

S= 11.03
1. Null Hypothesis H: There is no significant difference between the medicines
A and B as regards their effect on increase in weight i.e., H,: H= H
2. Alternative Hypothesis H,: HA2
3. Level of significance, a =0.05

46-57
4. The test statistic is 1 = - -11=-1.7
6.46
s a1.03),+
Calculated value of| = 1.7

Tabulated value of for 5 + 7 - 2 10 d.f at 5% level of significant is 1.81

(right-tail test).
Since calculated < tabulated 1, we accept Hi.c., The medicines A and B do
not differ significantly as regards their effect on increase in weight.

Example 5 Two horses A and B were tested according to the time (in seconds) to run
a particular track with the following results.
HorseA 28 30 32 33 33 29 34
Horse B 29 30 30 24 27 29
Test whether the two horses have the same running capacity.
UNTU 2006S, (A) Nov. 2010 (Set No. 1), (H) Dec. 2019 (R18))
 427
nitioance (Small Sample Tests)
Solutton: Given n, 7, n,
T i n s

efirst compule the sample means and standard deviations.

We

F Mean offirst sample » (28+ 30 1233 33 29 + 4)

219) 31.286

F Mean of second sample-(29+ 30+ 30+ 24+ 27 + 29)

(169) 28.16

x-F(x-F) - -P
-3.286 10.8 29 0.84 0.7056
28
-1.286 1.6538 30 1.84 3.3856
30
0.51 30 1.84 3.3856
32 0.714
1.714 2.94 24 4.16 17.3056
33
-1.16 1.3456
33 1.714 2.94 27
29 0.84 0.7056
29 -2.286 5.226
4 2.714 7.366

31.4358 169 26.8336

219

Now S = m tn-224-7 +2o-F*]

5.23
131.4358 +26.8336]
=
(58.2694)

2.3
S =V5.23
Pa
1. Null Hypothesis H,: H
Alternative Hypothesis H, : H, # P2
2.
a
= 0.05
3. Level of significance:
- 31.286-28.102.443
4. The test statistic is
=
1
sL 2
of significance is
2.2.
-2=11 d.f. at 5% level
t for 7 +6
Tabulated value of conclude
we reject the null hypothesis H, and
>
Tabulated t,
Since Calculated do not have
the same running
capacity.
B
th both horses A and
 428
Computer Orlented Statietical Methods

Evample : To examine the hypothesis that the husbands are more intelligent than the
wives, an investigator took a
the 1.Q. The
sample of 10 couples and administered them a test which measures
results
are as follows
Husbands 17 105 97 105123 109 86 78 103 107
Wives 106 9887 104 116 9590 69 108 85
Test the hypothesis with
reasonable test at the level of significance of 0.05.
a

JNTU 2004, 2005, 2007s, (A) Nov. 2010, (K) May 2013 (Set No. 3)
Sohution: We have m,- 10, n, - 10 and

F 117 + 105 +97 + 105 + 123 109 + 86 + 78+ 103 + 107)

161030)= 103

1(106+98 + 87 + 104+ 116+95 +90+69 + 108 + 85)

T0 958) -
95.8

Now we compute the standard deviations of both samples

- Fr-F - -
117 14 196 106 10.2 . 104.04
105 2 4 98 2.2 4.84
97 6 36 87 -8.8 77.44
105 2 4 104 8.2 67.24
123 20 400 116 20.2 408.04
109 6 36 95 -0.8 0.64
86 -17 289 90 -5.8 33.64
78 -25 625 69 -26.8 718.24
103 0 108 12.2 148.84
107 4 16 85 -10.8 116.64
1030 1606 958 1679.6

Now S= t - 22G- +2,-D1

1606+1679.6]3285.6) 182.53

S= 13.51
1. Null Hypothesis H,: 4, = 4, (i.e., no difference in 1.Q.)
 429
significance mall Sample Tests)

ern ative Hypothesis H, P,> P

(i.e., husbands are more intelligent than wives) (one tailed test, ng ht)

Level of significance -0.05

: a

3.

is = F- 103-95.8-1.19168
The test statistic
(13.5110 There
1.19168 < 1.734, we accept the null hypothesis H, i.e.,
since
difference
in 's. 0.95
no with probability
Evample7 7: Find the maximum difference that we can expect if their standard
means
herween t h e m e of samples of sizes 10 and 12 from a normal population
are found to be 2 and 3 respectively. No. 2)}
Apr. 2012 Set
viations
JNTU 2004S, (A) Nov. 2010,

12, s, 2 and s, =
3
Solution: We
have n, =
10, n, = =

1 *12(3}] =
7.4

+-2
10+12-2

S=74=2.72

1. Nall Hypothesis H,:

H
. Aternative Hypothesis H, ,
: a
=
0.05
significance
3. Level of

statistie is t=
-
-
4. The test

=243

7-T|= l4.S
= (2.086) (2.72)
2.086
10+12-2-20 d.f. at 5% level of significance is
Tabulated value of1for
(two-tailed) difference
between the means is 2.43.
the following
items respectively had
maximum
Hence the and 7
samples of8
Two independent
Aample 8:
values.
11 13 1115 9 12 14
11 10-
Sample I 1 1 1 0 | 1 3 9
8
9 significant ?
SampleI1 the means
of samples
between
No. 1), (A) Dec. 20171
Is the difference
JNTU 2005S (Set
 430 Computer Oriented Statistical Methods

Solution: Given n, = 8, n, = 7 and

F ( 1 1 +11 + 13 + 11+ 15 +9 +12 + 4) = 12

70
9 +11 10+ 13 +9 +8+10) =
+ 10 =

(r-F) - F O - )
11 -1
11 11
13 10 0
11 -1 13 3 9
15 9 -1
9 -3 9 8
12 0 0 10 0
14 2 4

96 26 70 16

Now S =
+-2Et,-7)' E«,-5*]
8+7-2 26 +
16) =
3.23
13
S = 1.8
1. Null Hypothesis H,: H, 2
2. Alternative Hypothesis H,: 4, # H, (two tailed test)
3. Level of significance : a =
0.05

4. The test statistic is t = - 12-10

2.15
T 0.9316
(1.3)
7
Tabulated value of t for 8 +7 -2 13 d.f. at 5% level
of significance is 2.16.
Since calculated tabulatedt, we accept H, i.e., the difference between the
r <
means of samples is not
significant.
Example 9: Measuring specimens of nylon yam, taken from two machines, it was
found that 8 specimens from first machine had a mean denier of
9.67 with a standard deviation
of 1.81 while 10 specimens from second machine had
a mean denier of 7.43 with a standard
deviation of 1.48. Assuming that the
proportions are normal, test the hypothesis
H: -H, 1.5 against H: #,-
=
1.5 level P> at 0.05 of significance.
JNTU 2005S, 2006S (Set No. 1,4)]
Solution: We have n, =
8, n,= 10, 7 =
9.67, 5 =
7.43, =
1.81 and s,
s, =
1.48
 431
fiCance (Small Sample Tests)

N o wS

nsi+ns
=4 + n , s :

80.812+ 10(1.48)
+n-2 8+10-2

26.2088+21.901 48.123.o07
16 16

S- V3.007 =1.734

Hypothesis H, : H, - 4= 1.5 (= 8)
Null
.
Alteraative Hypothesis H, : H , 1.5
>

Level
of significance: a =0.05
s.

is = -)-6
test statistic
4.The
S.

(9.67-1.43)-1.5. 0.74 0.8997

60.8225
.134) 10
1.746
Tabulated value of t for 8 + 10-2= 16 d.f. at 5% level of significance is
(one-tail test).
Since calculated t < tabulated 1, we accept H
first week. After
Example10: Ten soldiers participated in a shooting competition in the
Their scores before
in the competition in the second week.
mtensive training they participated
as follows:
and aftertraining are given
43
Scores before 67 24 57 55 63 5456 6833
Scores after 70 38 58 58 56 67 68 75 42 38
Do the data indicate that the soldiers
have been benefited by the training.
[JNTU 2005s (Set No. 2)]

have =
10, n,= 10
Solution: We n,
Let F = Mean of the seores before training

+ 24+ 57 + 55 +63 + 54 + 56 + 68 + 33 +43)

67

520) 52
10
and = Mean of the scores after training

(70 + 38 + 58 + 58 + 56 + 67 + 68 + 75 +42 + 38)

10

1570)57
 432
Computer Oriented Statistical Methods
NOW we
compute the standard deviations of the two samples.

(r-F (y ) ( r - F
67 15 225 70 13 169
24 -28 784 38 -19 361
57 25 58
$5 9 58 1

63 11 121 56 -1
54 4 67 10 100

6 4 16 68 11 121
68 16 256 75 18 324
33 -19 361 42 -15 225
43 -9 81 38 -19 361

520 1882 570

1664
Nows =
n +n2
- +2o:-*|
10+10-2Il882
+ 1664) =(3546) =197
S =14.0357
1. Null Hypothesis H,: H =H2 i..e, there is no benefit of training
2. Alternative Hypothesis H, : H, 4, ie, there is a benefit oftraining
3. Level of significance : a = 0.05

4. The test statistic is r = ) 52-57

s. (14.0357).,
n2

-0.796
6.277
Tabulated value of t for 10 + 10- 2-= 18 d.f. at 59% level ofsignificance (lett
tailed test) is -1.734
Since | =0.796 < 1.734 = |tlWe accept the null Hypothesis H i.e., the
soldiers have been benefited by the training.

Example 11: To compare two kinds of bumper guards, 6 of each kind were mounted
on a car and then the car was run into a concrete wall. The following are the costs of repairs.

Bumper Guard 1107 148 123 165 102 119

Bumper Guard2 134 115 112 151 133 129
 ance (Small Sample Tests 433

.Use
the 0.0 level of
signific between
tne
whether lest the difference
hese two
m e a n s o ft h e
samples signit.O
Samples is
significant.
NU2005, (H) Nov. 2010, (A1 Nav, 2011 (Set No.
SolutionWehave 3), une
6,n 6, r (107 764 = 127.33
+ 148+ 123+ 165 + 102+ 119)= 6
and
F(134 + 115 + 112+ 151 133 129) = 774 =129
=

129
6
We compute the standard deviations of
two samples.
-7
107-20.33 413.31 134 25
148 20.67 427.25 115 -14 196

123 -4.33 18.75 112 -17 289

165 37.67 1419.03 151 22 484

102-25.33 641.61 133 4 16

119-8.33 69.39 129 0

764 2989.34 774 1010

S Z(4-7}
n +n-2
+Z(y-*1
1 399.934
16-6-72989.34
+
1010] =(3999.34)

S 19.998

1. Null Hypothesis H,: H,H2

2. Alternative
Hypothesis H,: 4, # 4, (twotailed test)
: a
=
0.01
3. Level of significance

4. The test statistic is

127.33-129 -1.67
- 0.1446
11.546
s (19.998)
"m
0.1446
Calculated value of || =

3.169
10 d.f. at 1% level of significance is
Tabulated value of t for 6 + 6 - 2 there is
Null Hypothesis H. i.e.,
tabulated 1, we accept the
Since calculated r< means.
between the sample
n0 Significant difference A and B are drawn
of coal from two mines
samples ofspecimens yielding the following
Example 12:Random were measured

d their heat-producing capacity (in

millions ofcalories/ton)
Tesults:
 434 Computer Oriented Statistical Methods

Mine A 8350 8070 8340 8130 8260

Mine B 7900 8140 7920 7840 7890 7950
Isthere significant difference between the means of these two samples at 0.01 level of
signiticance. JNTU 2005, (K) Nov. 2009 (Set No. 3)1

Solution: Here 5, n^ =6
=

Calculation for mean's and S.D's of the samples:

Mine A Mine B
r-F (-)
(-8230) (r-F}
(y-7940) (y-5
8350 120 14400 7900 -40 1600
8070 -160 25600 8140 200 40000
8340 110 12100 7920 -20 400
8130 -100 10000 7840 100 10000
8260 30 900 7890 -50 2500

7950 10 100
41150 63000 47640 54600

F 41150 8230, F = 2 - * 0 = 7940

5
47640
6
7940
Now $ [4-7*+Z-*]
+n-2

16300+54600a17600)= 13066.67
6-2

S=114.31.
1. Null Hypothesis Ho :H 2

2. Alternative Hypothesis H :*H2

3. Level of significance, a = 0.01
4. The test statistic is

- 8230-7940 290
= 4.19
L 14.3 (114.31)0.6055)

ie., calculated value of I=4.19

Tabulated value ofr for 5+6-2=9 d.o.fat a =0.01 is 2.821
Since calculated t>o,05, we reject the null hypothesis Ho
Thus we conclude that there is significant difference between the means of the two
samples.
 ignificance mall Sample Tests)
ofacity
ci
from one
area
Evaple 13f :The IQs (intelligence quotients) of 16 students students from
from
h o w e da mean107 with a standard deviation of 10, while the 1Qs of 14 a
deviation of 8.
Is there
area of the city showed a mean of 112 with a standard
0.05 level of significance?
enificant difference between the 1Qs of the two groups at a
significantdiffere

UNTU (K) May 2010

(Set No.2)

solution: Let and u2 be the means of the two populations.

Let the Null Hypothesis be H :H =42
Then the Altermative Hypothesis is HH #42
We have =16,ï 107,5 =10
=
and n =14,5=112,5, =8
we use f- test.
Since the sample sizes are small and o,,O, are not known,
An unbiased estimate S of the common variance is given by

Ms 16x100+14x64249689,14
+n-2 16+14-2 28

S=V89.14 =9.44
there is difference between and p2
Let us assume that H,is true, ie., no

The test statistic is

107-112 -5
-1.447
s 944,.94x0.36
Also d.f. =16+14-2 =28
calculated value of |||< the tabulated value
=1.447 <1.701 for 28 d.f. ie., the
at 5% level with 28 d.f.
between the
and conclude that the difference
Hence, we accept the Null Hypothesis
1Qs of the two groups is not significant.
from 6 cow's milk and 6
14: The table gives the biological values of protein
Example
di"erences are significant.
butfalo's milk. Examine whether the

Cow's milk 1.8 2.0 1.9 1.6 |1.8 1.5 JNTU (H) Dec. 2011 (Set No. 3)]
Buffalo's milk 2 1.8 1.8 2.0 2.1 |1.9|
values of cow's milk and buffalo's milk
Solution: Let r and y represent the protein

respectively.
Let theNull Hypothesis be Ho:H1 =H2
Then the Alternative Hypothesis is H :41 *H2
436 Computer Oriented Statistical Methods

Calculations for Sample Means and S. D.s

(x-F2 y
- = y-1.93 (-
=x-1.77
1.8 0.03 0.0009 2 0.07 0.0049
2.0 0.23 0.0529 1.8 -0.13 0.0169
1.9 0.13 0.0169 1.8 -0.13 0.0169
1.6 -0.17 0.0289 2.0 0.07 0.0049
1.8 0.03 0.0009 2.1 .17 0.0289
1.5 -0.27 0.0729 1.9 -0.03 0.0009
| 10.6 z; 0.1734 11.6=Ey;| 0.0734

F =1.77 and F - 2 l 1 6=1.93

6 n 6

Now S -+20-5]
t-2
6+6-20.1734+0.0734]=2468_0 0.2468
6+6-2 10
S=V0.0247 =0.157
The test statistic is

- 1.77-1.93
t-

(0.157)
--0.16 = 0.2771-1.7651
0.157 0.157
t=1.7651
df. =n +n -2 6+6-2 10
Tabulated value oft for 10 d.f. at 5% level of significance is 2.228 (two -tailed test).

Since the calculated value of |t|< the table value of t at 5% level with 10 d.f. for two
-

tailed test, we accept Null Hypothesis H and conclude that there is no significant differencein

their means.