TESTS OF HYPOTHESIS
Z-test
TESTS OF HYPOTHESIS
Hypothesis testing helps to decide on the basis of a sample data, whether a hypothesis
about the population is likely to be true or false.
Statisticians have developed several tests of hypotheses (also known as the tests of
significance) for the purpose of testing of hypotheses which can be classified as:
(a) Parametric tests or standard tests of hypotheses; and
(b) Non-parametric tests or distribution-free test of hypotheses.
Parametric tests and Non-parametric tests
Parametric tests usually assume certain properties of the parent population from which we draw
samples. Assumptions like observations come from a normal population, sample size is large,
assumptions about the population parameters like mean, variance, etc.,
The important parametric tests are: (1) z-test; (2) t-test; (3) χ2 -test, and (4) F-test.
Non-parametric tests do not depend on any assumption about the parameters of the
population.
Remark : A hypothesis which does not specify completely ‘r’ parameters of a population is termed
as composite hypothesis with r degrees of freedom.
Z-Test
Population normal, population infinite, sample size may be large or small but
variance of the population is known, Ha may be one-sided or two-sided: In such a
situation z-test is used for testing hypothesis of mean and the test statistic z is
worked as under:
Test concerning a single mean
Suppose the mean of a population is not known. We want to test if the mean
value is a given value . The null hypothesis is : = and the alternative
hypothesis is :
Z=
If , for the samples, calculated > K , is rejected and if, calculated k, is
accepted.
What is a Hypothesis?
A hypothesis is an assumption (claim or a statement) about the
population parameter (say population mean (variance (, p )
which is to be tested.
Example: I assume the average weight of this class is 58 kg!
Example: Average age of all college students in a city is 23.
Procedure of Hypothesis Testing
Step 1:
Null hypothesis (H0) :
A population parameter (such as the mean, the standard deviation, and so on) is equal to a
hypothesized value.
The assumption we wish/ want to test is called the null hypothesis. The symbol for null hypothesis is
H0
According to null hypothesis, no difference exists between the population mean (µ) and sample
mean .( or hypothesised mean (µ0) )
H0 : µ =µ0 = 0
Example: The average weight of this class is 58 kg!
H0 : µ = 58
The Alternative Hypothesis (Ha ):
A population parameter is smaller, greater, or different than the hypothesized value in the null
hypothesis
eg. The average weight of the students is not equal to 58kgs. (H1 : ≠ 58)
Step 2: Set up a suitable significance level.
The confidence with which an experimenter rejects or accepts Null
Hypothesis depends on the significance level adopted. Level of
significance is the rejection region ( which is outside the confidence or
acceptance region).The level of significance, usually denoted by the α.
Though any level of significance can be adopted, in practice we either
take 5% or 1% level of significance . When we take 5% level of
significance(α= .05), In case we take the significance level at 5 per cent,
means that researcher is willing to take 5 per cent risk of rejecting the
null hypothesis when it (H0) happens to be true.
i.e. we are about 95% confident that we have made the right decision.
If our sample statistic(calculated
value) fall in the nonshaded
region( acceptance region), then it
simply means that there is no
evidence to reject the null
hypothesis.
It proves that null hypothesis (H0 ) is
true. Otherwise, it will be rejected.
A critical value is a point on the
distribution of the test statistic under the
null hypothesis that call for rejecting the
null hypothesis
Rejection Level of Significance, percent
Region
= 10 % =5% =1%
One- tailed 1.28 1.645 2.33
region
Two-tailed 1.645 1.96 2.58
region
Rejection Level of Significance, percent
Region
= 10 % =5% =1%
One- 1.28 1.645 2.33
tailed
region
Two-tailed 1.645 1.96 2.58
region
Step 3 Determination of suitable test statistic: For example Z, t Chi-Square
or F-statistic.
Step 4 Determine the critical value from the table.
Step 5 After doing computation, check the sample result. Compare the
calculated value( sample result) with the value obtained from the table.
(tabulated or critical value)
Step 6 Making Decisions Making decisions means either accepting or
rejecting the null hypothesis. If computed value(absolute value) is more
than the tabulated or critical value, then it falls in the critical region. In that
case, reject null hypothesis, otherwise accept.
Type I and Type II Errors
When a statistical hypothesis is tested, there are 4 possible
results:
(1)The hypothesis is true but our test accepts it.
(2)The hypothesis is false but our test rejects it.
(3)The hypothesis is true but our test rejects it.
(4)The hypothesis is false but our test accepts it.
Rejecting a null hypothesis when it is true is called a Type I
error.
Accepting a null hypothesis when it is false is called Type II
error.
A principal at a school claims that the students in his school are above average
intelligence. A random sample of thirty students IQ scores have a mean score of 112.5. Is
there sufficient evidence to support the principal’s claim? The mean population IQ is 100
with a standard deviation of 15.
Step 1: State the null hypothesis.
The accepted fact is that the population mean is 100, so
H0: µ =100.
Step 2: State the alternative hypothesis.
The claim is that the students have above average IQ scores, so:
Ha: µ >100
Step 3: State the level of significance . If you aren’t given an alpha
level, use 0.05 , an alpha level of 0.05 is equal to a z-score of 1.645.
Rejection Level of Significance, percent
Region
= 10 % =5% =1%
One- tailed 1.28 1.645 2.33
region
Two-tailed 1.645 1.96 2.58
region
Step 4: Find the Z using the formula
Z=
Z= = 4.56
Step 5: Critical value at =0.05 is k= 1.645
Since Z calculated > k, reject
Conclusion: Principal’s claim is right.
A manufacturer of sports equipment has developed a new synthetic fishing line that he claims
has a mean breaking strength of 9 kilograms with a standard deviation of 0.5 kilogram. Test
the hypothesis that =9 kgs against the alternative 9 kgs if a random sample of 50 lines is
tested and found to have a mean breaking strength of 8.8 kilograms. Use a 0.01 level of
significance.
Solution: Given : = 8.8, =0.5 , n=50
Rejection Level of Significance, percent
Region
: = 9 kgs = 10 % =5% =1%
: 9kgs One- tailed 1.28 1.645 2.33
region
Two-tailed 1.645 1.96 2.58
region
Z=
Z= = 2.83
Critical value at =0.01 is k= 2.58
Since Z calculated > k, reject
Conclusion: The average breaking strength is not 9kgs.
In recent years, the mean age of all college students in city X has been 23. A random sample of
42 students revealed a mean age of 23.8. Suppose their ages are normally distributed with a
population standard deviation of =2.4. Can we infer at α =0.05 that the population mean age has
changed?
H0: µ =23
Ha: µ≠ 23
Z=
Z= = 2.16
Critical value at =0.05 is k= 1.96
Since Z calculated > k, reject
Conclusion: The mean age of the college students is not 23.
A stenographer claims that she can type at the rate of 120 words per minute. Can we reject her
claim on the basis of 100 trials in which she demonstrates a mean of 116 words with a standard
deviation of 15 words. Use 5 % level of significance.
: = 120
: 120
Z=
Z= = 2.67
Critical value at =5 % is k= 1.96
Since Z calculated > k, reject
Conclusion: The stenographer claim is not true.
Sugar is packed in 5- pound bags. An inspector suspects the bags may not contain 5
pounds . A sample of 50 bags. Produces a mean of 4.6 pounds. And a standard deviation of
0.7 pounds. Is there enough evidence to conclude that the bags do not contain 5 pounds as
stated.
: =5
: 5
Z=
Z= = 4.04
Critical value at =5 % is k= 1.96
Since Z calculated > k, reject
Conclusion: Bags do not weigh 5 pounds
A sample of 100 students is taken from a large population. The Mean height of these students
is 64 inches and the standard deviation is 4 inches. Can it be reasonably regarded that in the
population mean height is 66 inches.
: = 66
Rejection Level of Significance, percent
: 66 Region
= 10 % =5% =1%
Z= One- 1.28 1.645 2.33
tailed
region
Two-tailed 1.645 1.96 2.58
Z= = 5 region
Critical value at =5 % is k= 1.96
Since Z calculated > k, reject
Conclusion: The population mean height is not equal to 66 inches.
A battery manufacturer claims that the mean reserve capacity of a certain battery is 1.5 hours.
To test his claim a random sample of 50 batteries were selected and find the mean reserve
capacity to be 1.55 hours. It had previously been determined that the population standard
deviation is 0.32 hours. If α =1%, is there enough evidence to support the claim?
: = 1.5 hours
: 1.5 hours
Rejection Level of Significance, percent
Z= Region
= 10 % =5% =1%
One- 1.28 1.645 2.33
Z= = 1.1049 tailed
region
Two-tailed 1.645 1.96 2.58
region
Critical value at =1 % is k= 2.58
Since Z calculated < k, a
Conclusion: The mean reserve capacity of a battery is 1.5 hours.
A random sample of 400 items gives the mean 4.45 and the variance as 4. Can the sample be
regarded as drawn from a normal population with mean 4 at 5% level of significance?
: =4
: 4
Z=
Z= = 4.5
Critical value at =5 % is k= 1.96
Since Z calculated > k,
Test on two means (Test for equality of means)
Suppose we intend testing whether the means of two populations are equal.
The null hypothesis is : = (the means of two populations are equal) the alternative
hypothesis is :
Let and be the standard deviations of the two populations.
Let a random sample of size , be drawn from the first population. Let the mean of this
sample b . Also, let a random sample of size , be drawn from the second population. Let
the mean of this sample be .
Then , the test statistic is Z=
If , for the samples, calculated > K , is rejected and if, calculated k, is accepted.
It is known that IQ of boys has standard deviation 10 and that IQ of girls has standard deviation 11.
Mean IQ of 100 randomly selected boys is 95 and mean IQ of 80 randomly selected girls is 97. Can it
be concluded that on an average, boys and girls have the same IQ? (Use 1% level of significance)
Solution: Here = 10, =10, =95
=11, =80, = 97
Let and represent the population means of the IQ’s of boys and girls, respectively.
: =
: Rejection Level of Significance, percent
Region
the test statistic is Z= = 10 % =5% =1%
and the critical value is k=2.58 at =0.01 One- 1.28 1.645 2.33
tailed
Z cal= = 1.26 region
Two-tailed 1.645 1.96 2.58
cal < k, is accepted. region
Conclusion: Boys and girls have the same IQ.
A random sample of 1200 men from one state gives the mean pay of Rs. 4000 per month with SD
of Rs. 60 and a random sample of 1000 men from another state gives the mean pay of Rs. 5000
per month, with S.D. of Rs. 80. Discuss whether the mean levels of pay of men from the two states
differ significantly at 1% level of significance.
Solution: Here = 60, = 1200 , =4000
=80, =1000 = 5000
Let and represent the population means of the pay of men from the two states respectively.
: =
:
the test statistic is Z=
and the critical value is k=2.58 at =0.01
Z cal = =326
> k, is rejected..
cal
Conclusion: payment of men differ in the states.
Test for proportion
Suppose the proportion of an attribute in a population is not known and we want to test whether the
proportion is given value .
Let a random sample of size n be drawn from the population and let x units possess the attribute.
Then the sample proportion is p=
the test statistic is Z=
If , for the samples, calculated > K , is rejected and if, calculated k, is accepted.
Note: 𝑠𝑎𝑚𝑝𝑙𝑒 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜n = p
𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 =
𝑛 = 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧e
= (1 − )
The manufacturers of a certain brand of pens believes that 35% of the pen users in
Bangalore used their brand of pens. To verify this claim, a survey of pen users was
conducted. Among 347 of them 107 people said they used the particular brand. Does this
figure support the manufacturers claim.
Solution: =0.35, n=347, x=107
p= = =0.308
: p = =0.35
:p
test statistic is Z=
Q0= 1-P0
Z cal = = 1.64
and the critical value is k=1.96 at =0.05
cal < k, is accepted. Thus, manufacturer claim is true.
A tech company believes that the percentage of residents in town XYZ that own a cell phone is
70%. A marketing manager believes this value to be different. He conducts a survey of 200
individuals and found that 130 responded yes to owning a cell phone. (a) State the null
hypothesis and alternate hypothesis. (b) At a 95% confidence level, is there enough evidence to
reject the null hypothesis?
Solution: Solution: =0.70, n=200, x= 130
p= = =0.65
: p = 0.70
: p 0.70
test statistic is Z=
Z cal = = 1.54
and the critical value is k=1.96 at =0.05
cal < k, is accepted. Thus, company claim is true.
From a large consignment of mangoes, a random sample of 50 mangoes was examined and
among them 4 were found to be bad. Test at 5% level of significance whether it is reasonable to
assume that 10% of mangoes were bad.
Solution: =10 %, n=50, x=4
p= = = 0.08
:p=
:p
test statistic is Z=
Z cal = = 0.47
and the critical value is k=1.96 at =0.05
cal < k, is accepted.
Among 80 potatoes which are randomly pulled out from the field, 11 are infected. Test the
hypothesis that 25% potatoes are infected.
Solution: =25 %, n=80, x=11
p= = = 0.1375
:p=
:p
test statistic is Z=
Z cal = = 2.3486
and the critical value is k=1.96 at =0.05
cal > k, is rejected.
Zark’s Burger, a fast food restaurant claims that 85% of the burger fanatics prefer to eat in their
place. To test this claim, a random sample of 90 burger customers are selected and asked what
they prefer. If 76 of the 90 burger fanatics said they prefer to eat at Zark’s, what conclusion do
we draw? Use a 0.05 level of significance.
Solution: =85 %, n=90, x=76
p= = = 0.84
: p = 0.85
:p
test statistic is Z=
Z cal = = 0.29
and the critical value is k=1.96 at =0.05
cal < k, is accepted.
Conclusion: 85% of the burger fanatics prefer to eat in their place.
Test for equality of proportion
Suppose there are two populations with unknown proportions and we wish test whether the
proportions (of a certain attribute) in the two populations are equal.
The null hypothesis is H0 : P1 =P2 (the proportions are equal )
The alternative hypothesis is H1: P1 ≠ P2
Let P be the common proportion. Let a sample of size n1 be drawn from the first population.
Among the n1 units let x 1 units posses the attribute, so that the sample proportion is p1= . Let
a random sample of size n2 be drawn from the second population and let x2 among n2 units
posses the attribute, so the sample proportion is p2=
Test statistic is,
Z = where P= , Q= 1-P
For a sample, cal ≤ k, is accepted.
In a random sample consisting 326 teenagers, 143 claimed to watch National Geographic channel
regularly. Among a random sample of 213 adults 137 watch it regularly. Test whether the
proportion of teenage viewers differ from the adult viewers.( use 5% level of significance).
Solution: Let P1 be the population proportion of teenage viewers of NGC
Let P2 be the population proportion of adult viewers of NGC
n1= 326, x1= 143, p1= = = 0.4387
n2 = 213, x2= 137, p2= = = 0.6432
P= = = 0.5195 Q= 1-P = 1-0.5195 = 0.4805
H0 : P1 =P2
H1: P1 ≠ P2 and α =5%
Z = = Z = = 4.646
and the critical value is k=1.96 at =0.05
cal > k, is rejected.
Conclusion: The proportion of teenage viewers of NGC differs from the proportion of adult
viewers.
In a sample of 600 high school students from a state, 400 are found to use ball pens. In another
sample of 900 from a neighboring state 450 are found to use ball pens. Do the data indicate that
the states are significantly different with respect to the habit of using ball pens among students?
Solution: n1= 600 , x1= 400, p1= = = 0.66
n2 = , x2= , p2= = 0.5
P= = 0.56 Q= 1-P = 1-0.56 = 0.44
H0 : P1 =P2
H1: P1 ≠ P2 and α =5%
Z = = =6.4
and the critical value is k=1.96 at =0.05
cal > k, is rejected.
Conclusion:
