MTH120_Chapter11

11- 1

Chapter Eleven
Two Sample Tests of Hypothesis
GOALS
When you have completed this chapter, you will be able to:

ONE
Understand the difference between dependent and independent samples.

TWO
Conduct a test of hypothesis about the difference between two independent
population means when both samples have 30 or more observations.

THREE
Conduct a test of hypothesis about the difference between two independent
population means when at least one sample has less than 30 observations.

McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.

11- 2

Chapter Eleven continued
Two Sample Tests of Hypothesis
GOALS
When you have completed this chapter, you will be able to:

FOUR
Conduct a test of hypothesis about the mean difference between paired or
dependent observations.

FIVE
Conduct a test of hypothesis regarding the difference in two population
proportions.


11- 3

Comparing two populations
We wish to know whether the distribution
of the differences in sample means has a
mean of 0.

If both samples contain at least 30
observations we use the z distribution as
the test statistic.


11- 4

Comparing two populations
No assumptions about the shape of the
populations are required.
The samples are from independent populations.

The formula for computing the value of z is:

X1 − X 2
z=
2 2
s1 s 2
+
n1 n2


11- 5

EXAMPLE 1

Two cities, Bradford and Kane are separated only by the
Conewango River. There is competition between the two
cities. The local paper recently reported that the mean
household income in Bradford is $38,000 with a standard
deviation of $6,000 for a sample of 40 households. The same
article reported the mean income in Kane is $35,000 with a
standard deviation of $7,000 for a sample of 35 households.
At the .01 significance level can we conclude the mean
income in Bradford is more?


11- 6

EXAMPLE 1 continued

 Step 1: State the null and alternate hypotheses.
H0: µB µK ; H1: µB > µK
Step 2: State the level of significance. The .01
significance level is stated in the problem.
Step 3: Find the appropriate test statistic. Because both

samples are more than 30, we can use z as the test statistic.


11- 7

Example 1 continued
 Step 4: State the decision rule.
The null hypothesis is rejected if z is greater than
2.33.

Step 5: Compute the value of z and make a decision.

$38,000 − $35,000
z= = 1.98
($6,000) 2 ($7,000) 2
+
40 35


11- 8

Example 1 continued
The decision is to not reject the null hypothesis.
We cannot conclude that the mean household
income in Bradford is larger.

The p-value is:

P(z > 1.98) = .5000 - .4761 = .0239


11- 9

Small Sample Tests of Means
 The t distribution is used as the test statistic if one
or more of the samples have less than 30
observations.
The required assumptions are:
1. Both populations must follow the
normal distribution.
2. The populations must have equal
standard deviations.
3. The samples are from independent
populations.


11- 10

Small sample test of means continued
 Finding the value of the test statistic requires two steps.
1. Pool the sample standard deviations.
2 2
(n1 − 1) s1 + (n2 − 1) s 2
s2 =
p
n1 + n2 − 2

Determine the value of t from the following formula.

X1 − X 2
t=
2 1 1 
sp
 +  
 n1 n2 

11- 11

EXAMPLE 2
A recent EPA study compared the highway fuel
economy of domestic and imported passenger
cars. A sample of 15 domestic cars revealed a
mean of 33.7 mpg with a standard deviation of
2.4 mpg. A sample of 12 imported cars
revealed a mean of 35.7 mpg with a standard
deviation of 3.9. At the .05 significance level
can the EPA conclude that the mpg is higher on
the imported cars?


11- 12

Example 2 continued

Step 1: State the null and alternate hypotheses.
H0: µD µI ; H1: µD < µI
Step 2: State the level of significance.The .05
significance level is stated in the problem.
Step 3: Find the appropriate test statistic. Both

samples are less than 30, so we use the t
distribution.


11- 13

EXAMPLE 2 continued

Step 4: The decision rule is to reject H0 if t<-1.708.
There are 25 degrees of freedom.

Step 5: We compute the pooled variance:
2 2
(n1 − 1)( s1 ) + (n 2 − 1)( s 2 )
s2 =
p
n1 + n 2 − 2
(15 − 1)(2.4) 2 + (12 − 1)(3.9) 2
= = 9.918
15 + 12 − 2


11- 14

Example 2 continued
We compute the value of t as follows.
X 1 −X 2
t =
1 1 
s2 
p + 
 n1 n2 

33.7 −35.7
= =− .640
1
1 1 
8.312 + 
15 12 

H0 is not rejected. There is insufficient sample evidence
to claim a higher mpg on the imported cars.


Hypothesis Testing Involving Paired
11- 15

Observations
 Independent samples are samples that are not
related in any way.
Dependent samples are samples that are paired or
related in some fashion. For example:
If you wished to buy a car you would look at the

same car at two (or more) different dealerships and
compare the prices.
If you wished to measure the effectiveness of a new

diet you would weigh the dieters at the start and at the
finish of the program.

Hypothesis Testing Involving Paired
11- 16

Observations
Use the following test when the samples are dependent:
d
t=
sd / n
 where d is the mean of the differences
 s is the standard deviation of the differences
d
 n is the number of pairs (differences)


11- 17

EXAMPLE 3
 An independent testing agency is comparing the daily
rental cost for renting a compact car from Hertz and
Avis. A random sample of eight cities revealed the
following information. At the .05 significance level
can the testing agency conclude that there is a
difference in the rental charged?


11- 18

EXAMPLE 3 continued

City Hertz ($) Avis ($)
Atlanta 42 40
Chicago 56 52
Cleveland 45 43
Denver 48 48
Honolulu 37 32
Kansas City 45 48
Miami 41 39
Seattle 46 50


11- 19

EXAMPLE 3 continued

Step 1: H0 : µd = 0 H1: µd ≠ 0

Step 2: H0 is rejected if t < -2.365 or t > 2.365.
We use the t distribution with 7 degrees of
freedom.


11- 20

Example 3 continued
City Hertz Avis d d2
Atlanta 42 40 2 4
Chicago 56 52 4 16
Cleveland 45 43 2 4
Denver 48 48 0 0
Honolulu 37 32 5 25
Kansas City 45 48 -3 9
Miami 41 39 2 4
Seattle 46 50 -4 16


11- 21

Example 3 continued
Σd 8.0
d= = = 1.00
n 8

Σd 2 −
( Σd ) 2 78 −
82
sd = n = 8 = 3.1623
n −1 8−1

d 1.00
t= = = 0.894
sd n 3.1623 8


11- 22

Example 3 continued
Step 3: Because 0.894 is less than the critical
value, do not reject the null hypothesis. There is
no difference in the mean amount charged by
Hertz and Avis.


11- 23

Two Sample Tests of Proportions
 We investigate whether two samples came from
populations with an equal proportion of successes.

 The two samples are pooled using the following formula.
X1 + X 2
pc =
n1 + n2

where X1 and X2 refer to the number of successes in the
respective samples of n1 and n2.


11- 24

Two Sample Tests of Proportions continued
The value of the test statistic is computed from
the following formula.

p1 − p 2
z=
pc (1 − pc ) pc (1 − pc )
+
n1 n2


11- 25

Example 4
Are unmarried workers more likely to be absent
from work than married workers? A sample of
250 married workers showed 22 missed more
than 5 days last year, while a sample of 300
unmarried workers showed 35 missed more than
five days. Use a .05 significance level.


11- 26

Example 4 continued

The null and the alternate hypothesis are:

H0: : U U M H1: : U > >M

The null hypothesis is rejected if the computed
value of z is greater than 1.65.


11- 27

Example 4 continued
The pooled proportion is
35 + 22
pc =
300 + 250

The value of the teat statistic is
35 22
−
z= 300 250 = 1.10
.1036(1 −.1036) .1036(1 −.1036)
+
300 250


11- 28

Example 4 continued
The null hypothesis is not rejected. We cannot
conclude that a higher proportion of unmarried
workers miss more days in a year than the
married workers.

The p-value is:

P(z > 1.10) = .5000 - .3643 = .1457


MTH120_Chapter11

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