Engineering Statistics

Chapter 7
Introduction to
Hypothesis Testing

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What is a Hypothesis?
A hypothesis is a claim
(assumption) about a
population parameter:
population mean
Example: The mean monthly cell phone bill
of this city is = $42

population proportion
Example: The proportion of adults in this
city with cell phones is p = .68
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The Null Hypothesis, H0

States the assumption (numerical) to be
tested
Example: The average number of TV sets in U.S.
Homes is at least three ( H0 : 3 )

Is always about a population parameter,

not about a sample statistic
H0 : 3

H0 : x 3
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The Null Hypothesis, H0

(continued)

Begin with the assumption that the null

hypothesis is true
Similar to the notion of innocent until
proven guilty
Refers to the status quo
Always contains = , or sign
May or may not be rejected.
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The Alternative Hypothesis, HA

Is the opposite of the null hypothesis
e.g.: The average number of TV sets in
U.S. homes is less than 3 ( HA: < 3 )

Challenges the status quo

Never contains the = , or sign
May or may not be accepted
Is generally the hypothesis that is
believed (or needs to be supported)
by the researcher.
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Hypothesis Testing Process

Claim: the
population
mean age is 50.
(Null Hypothesis:

Population

H0: = 50 )

Is x 20 likely if = 50?
If not likely,
REJECT
Null Hypothesis

Now select a
random sample

Suppose
the sample
mean age
is 20: x = 20

Sample
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Reason for Rejecting H0

Sampling Distribution of x

20
If it is unlikely that we
would get a sample
mean of this value ...

= 50
If H0 is true
... if in fact this were
the population mean

x
... then we reject the
null hypothesis that
= 50.

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Level of Significance,
Defines unlikely values of sample
statistic if null hypothesis is true
Defines rejection region of the sampling
distribution

Is designated by , (level of significance)

Typical values are .01, .05, or .10

Is selected by the researcher at the

beginning
Provides the critical value(s) of the test.
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Level of Significance
and the Rejection Region
Level of significance =

H0: 3
HA: < 3
H0: 3
HA: > 3
H0: = 3
HA: 3

Represents
critical value

Rejection
region is
shaded

Lower tail test

Upper tail test

/2
Two tailed test

/2
0

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Errors in Making Decisions

Type I Error
Reject a true null hypothesis
Considered as a serious type of error.
The probability of Type I Error is
Called level of significance of the test
Set by researcher in advance.
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Errors in Making Decisions(continued)

Type II Error
Fail to reject a false null hypothesis
The probability of Type II Error is

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Outcomes and Probabilities

Possible Hypothesis Test Outcomes
State of Nature

Key:
Outcome
(Probability)

Decision

H0 True

H0 False

Do Not
Reject
H0

No error
(1-)

Type II Error
()

Reject
H0

Type I Error
()

No Error
(1-)

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Type I & II Error Relationship

Type

I and Type II errors can not happen at

the same time

Type I error can only occur if H0 is true

Type II error can only occur if H0 is false

If Type I error probability ( )

, then

Type II error probability ( )

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Factors Affecting Type II Error

All else equal,

when the difference between
hypothesized parameter and its true value

when

when

when

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Critical Value Approach to Testing

Convert sample statistic (e.g.: x ) to test
statistic ( Z or t statistic )
Determine the critical value(s) for a specified
level of significance from a table or
computer
If the test statistic falls in the rejection region,
reject H0; otherwise do not reject H0.
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Lower Tail Tests

H0: 3

The cutoff value,

-z or x , is called a
critical value

HA: < 3

Reject H0

-z
x

x z

Do not reject H0

n
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Upper Tail Tests

The cutoff value,

z or x , is called a
critical value

H0: 3
HA: > 3

Do not reject H0

x z

Reject H0

n
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Two Tailed Tests

H0: = 3
HA:
3

There are two cutoff

values (critical values):

z/2
or

x/2
x/2

Lower
Upper

/2
Reject H0

/2
Do not reject H0

-z/2
x/2

0
Lower

x /2 z /2

Reject H0

z/2
x/2

Upper

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Critical Value Approach to Testing

Convert sample statistic ( x ) to a test statistic
( Z or t statistic )
Hypothesis
Tests for
Known

Unknown

Large
Samples

Small
Samples
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Calculating the Test Statistic

Hypothesis
Tests for
Known

Unknown

The test statistic is:

x
z

Large
Samples

Small
Samples

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Calculating the Test Statistic

(continued)
Hypothesis
Tests for
Known
The test statistic is:

t n1

s
n

But is sometimes
approximated
using a z:

x
z

Unknown

Large
Samples

Small
Samples

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Calculating the Test Statistic

(continued)
Hypothesis
Tests for
Known

Unknown

The test statistic is:

t n1

s
n

Large
Samples

Small
Samples

(The population must be

approximately normal)
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Review: Steps in Hypothesis Testing

1. Specify the population value of interest
2. Formulate the appropriate null and
alternative hypotheses
3. Specify the desired level of significance
4. Determine the rejection region
5. Obtain sample evidence and compute
the test statistic
6. Reach a decision and interpret the result.
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Hypothesis Testing Example

Test the claim that the true mean # of TV
sets in US homes is at least 3.
(Assume = 0.8)

1. Specify the population value of interest

The mean number of TVs in US homes
2. Formulate the appropriate null and alternative
hypotheses
H : 3
HA: < 3 (This is a lower tail test)
0
3. Specify the desired level of significance
Suppose that = .05 is chosen for this test
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Hypothesis Testing Example

(continued)

4. Determine the rejection region

= .05

Reject H0

-z= -1.645

Do not reject H0

This is a one-tailed test with = .05.

Since is known, the cutoff value is a z value:
Reject H0 if z < z = -1.645 ; otherwise do not reject H0
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Hypothesis Testing Example

(continued)

5. Obtain sample evidence and compute the

statistic

test

Suppose a sample is taken with the following results: n =

100, x = 2.84 ( = 0.8 is assumed known)
Then the test statistic is:

x
2.84 3 .16
z

2.0

0.8
.08
n
100
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Hypothesis Testing Example

(continued)

6. Reach a decision and interpret the result

= .05
z
Reject H0

-1.645
-2.0

Do not reject H0

Since z = -2.0 < -1.645, we reject the null

hypothesis that the mean number of TVs in US
homes is at least 3.
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Hypothesis Testing Example

(continued)

An alternate way of constructing rejection region:

Now
expressed
in x, not z
units

= .05
x
Reject H0

2.8684
2.84

Do not reject H0

0.8
x z
3 1.645
2.8684
n
100
Since x = 2.84 < 2.8684,

we reject the null

hypothesis.
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p-Value Approach to Testing

Convert Sample Statistic (e.g. x ) to Test
Statistic ( Z or t statistic )
Obtain the p-value from a table or computer
Compare the p-value with
If p-value < , reject H0
If p-value , do not reject H0

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p-Value Approach to Testing

(continued)

p-value: Probability of obtaining a test

statistic more extreme ( or ) than the
observed sample value given H0 is true
Also called observed level of significance
Smallest value of for which H0 can be
rejected.

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p-value example
Example: How likely is it to see a sample mean
of 2.84 (or something further below the mean) if
the true mean is = 3.0?
P( x 2.84 | 3.0)

2.84 3.0
P z

0.8

100

P(z 2.0) .0228

= .05
p-value =.0228
x
2.8684
2.84

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p-value example

(continued)

Compare the p-value with

If p-value < , reject H0
If p-value , do not reject H0.
= .05

Here: p-value = .0228

= .05
Since .0228 < .05, we reject
the null hypothesis

p-value =.0228

2.8684
2.84

3
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Example: Upper Tail z Test

for Mean ( Known)
A phone industry manager thinks that
customer monthly cell phone bill have
increased, and now average over $52 per
month. The company wishes to test this
claim. (Assume = 10 is known)
Form hypothesis test:
H0: 52

the average is not over $52 per month

HA: > 52

the average is greater than $52 per month

(i.e., sufficient evidence exists to support the
managers claim)

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Example: Find Rejection

Region

(continued)

Suppose that = .10 is chosen for this test.

Find the rejection region:

Reject H0
= .10

Do not reject H0

z=1.28

Reject H0

Reject H0 if z > 1.28

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Review:
Finding Critical Value - One Tail
What is z given =
0.10?

.90

.10

= .10
.50 .40

1.28

Critical Value
= 1.28

Standard Normal
Distribution Table (Portion)

.07

.08

.09

1.1 .3790 .3810 .3830

1.2 .3980 .3997 .4015
1.3 .4147 .4162 .4177
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Example: Test Statistic

(continued)

Obtain sample evidence and compute the test

statistic
Suppose a sample is taken with the following
results: n = 64, x = 53.1 (=10 was assumed known)
Then the test statistic is:

x
53.1 52
z

0.88

10
n
64
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Example: Decision

(continued)

Reach a decision and interpret the result:

Reject H0
= .10

Do not reject H0

1.28
z = .88

Reject H0

Do not reject H0 since z = 0.88 1.28

i.e.: there is not sufficient evidence that the
mean bill is over $52

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p -Value Solution

(continued)

Calculate the p-value and compare to

p-value = .1894
Reject H0
= .10
0
Do not reject H0

1.28
z = .88

Reject H0

P( x 53.1 | 52.0)

53.1 52.0
P z

10

64

P(z 0.88) .5 .3106

.1894

Do not reject H0 since p-value = .1894 > = .10

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Example: Two-Tail Test

( Unknown)
The average cost of a
hotel room in New York
is said to be $168 per
night. A random sample
of 25 hotels resulted in
x = $172.50 and
s = $15.40. Test at the
= 0.05 level.

H0: = 168
HA:
168

(Assume the population distribution is normal)

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Example Solution: Two-Tail Test

H0: = 168
HA:
168
= 0.05

/2=.025

Reject H0

-t/2
-1.96

n = 25
is unknown, so
use a t statistic
Critical Value:
t24 = 1.96

t n1

/2=.025

Do not reject H0

1.46

Reject H0

t/2
1.96

x
172.50 168

1.46
s
15.40
n
25

Do not reject H0: not sufficient evidence that

true mean cost is different than $168
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Hypothesis Tests for Proportions

Involves categorical values
Two possible outcomes
Success (possesses a certain characteristic)
Failure (does not possesses that
characteristic)

Fraction or proportion of population in the

success category is denoted by p.
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Proportions

(continued)

Sample proportion in the success category is

denoted by p
x
number of successes in sample
p n
sample size

When both np and n(1-p) are at least 5, p

can be approximated by a normal distribution
with mean and standard deviation
p(1 p)

P p
p
n
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Hypothesis Tests for Proportions

The sampling
distribution of p is
normal, so the test
statistic is a z
value:

pp
p(1 p)
n

Hypothesis
Tests for p
np 5
and
n(1-p) 5

np < 5
or
n(1-p) < 5
Not discussed
in this chapter
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Example: z Test for Proportion

A marketing company
claims that it receives
8% responses from its
mailing. To test this
claim, a random sample
of 500 were surveyed
with 25 responses. Test
at the = .05
significance level.

Check:
n p = (500)(.08) = 40
n(1-p) = (500)(.92) = 460

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Z Test for Proportion: Solution

Test Statistic:

H0: p = .08
HA: p .
08= .05

pp

p(1 p)
n

.05 .08
2.47
.08(1 .08)
500

n = 500, p = 25/500 = .05

Decision:

Critical Values: 1.96

Reject

Reject

.025

.025
-1.96

-2.47

1.96

Reject H0 at = .05

Conclusion:
There is sufficient
evidence to reject the
companys claim of 8%
response rate.

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p -Value Solution

(continued)

Calculate the p-value and compare to

(For a two sided test the p-value is always two sided)
Do not reject H0

Reject H0

Reject H0

/2 = .025

/2 = .025

.0068

.0068
-1.96

z = -2.47

p-value = .0136:

P(z 2.47) P(z 2.47)

2(.5 .4932)
2(.0068) 0.0136

1.96
z = 2.47

Reject H0 since p-value = .0136 < = .05

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Type II Error
Type II error is the probability of
failing to reject a false H0

Suppose we fail to reject H0: 52

when in fact the true mean is = 50

50

52

Reject
H0: 52

Do not reject
H0 : 52
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Type II Error

(continued)

Suppose we do not reject H0: 52 when in

fact the true mean is = 50
This is the range of x where
H0 is not rejected

This is the true

distribution of x if = 50

50
Reject
H0: 52

52
Do not reject
H0 : 52
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Type II Error

(continued)

Suppose we do not reject H0: 52

when in fact the true mean is = 50
Here, = P( x cutoff ) if = 50

50
Reject
H0: 52

52
Do not reject
H0 : 52
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Calculating
Suppose n = 64 , = 6 , and = .05

6
52 1.645
50.766
n
64

cutoff x z
(for H0 : 52)

So = P( x 50.766 ) if =
50

50

50.766

Reject
H0: 52

52
Do not reject
H0 : 52
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Calculating

(continued)

Suppose n = 64 , = 6 , and = .05

50.766 50
P( x 50.766 | 50) P z
P(z 1.02) .5 .3461 .1539

64

Probability of
type II error:

= .1539
50

Reject
H0: 52

52
Do not reject
H0 : 52
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Chapter Summary
Addressed hypothesis testing
methodology
Performed z Test for the mean ( known)
Discussed pvalue approach to
hypothesis testing
Performed one-tail and two-tail tests . . .
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Chapter Summary

(continued)

Performed t test for the mean (

unknown)
Performed z test for the proportion
Discussed type II error and computed its
probability.

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Thank You

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Homeworks:

Problems no. x-x, x-x, x-x, x-x.

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