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## Chapter 13: inference about two populations

*s p*2 (1 /

*n *1 + 1 /

*n*2 )

*s p *(1 /

*n*1 + 1 /

*n *2
2. In problems where the population variances are unknown and unequal, the test statistic is
1 −

*x *2 ) − ( µ1 − µ 2 ) d.f. =
(

*x *1 −

*x *2 ) ±

*tα*/ 2

*s*1 /

*n *1 +

*s*2 /

*n*2
A study of the scholastic aptitude test (SAT) revealed that in a random sample of 100 males the mean
SAT score was 431.5 with a standard deviation of 93.7. A random sample of 100 female SAT scores
produced a mean of 423.9 with a standard deviation of 88.6. Can we conclude at the 1% significance level
The problem objective is to compare male and female SAT scores, a quantitative variable.
Therefore, the parameter of interest is (µ1 − µ2 ). The sample standard deviations are similar, making it reasonable to assume that σ 2
1 = σ 2 . Thus, we employ the equal-variances

*t*-test. The complete test
Rejection region:

*t* >

*t α */ 2 ,

*n *1+

*n *2– 2 =

*t*.005 ,198 ≈ 2 . 601 or

*t* < -2.601
(

*x *−

*x *) − ( µ − µ )
There is not enough evidence to conclude that male and female SAT scores differ.
Estimate the difference between male and female mean SAT scores (in Example 13.1) with 90%
A statistician wants to compare the relative success of two large department store chains. She
decides to measure the sales per square foot. She takes a random sample of five stores from chain 1 and
five stores from chain 2. The gross sales per square foot are shown below. Do these data provide
suffompent evidence to indicate that the two chains differ? (Test with
The problem objective is to compare the population of sales per square foot of one chain with the
population of sales per square foot of another chain. As a result, the parameter of interest is µ1 − µ2 . The sample variances are quite different; we can assume that σ 2
1 ≠ σ 2 . The appropriate technique is the
unequal variances

*t*-test of µ1 − µ2 .
(

*x *−

*x *) − (

*µ *−

*µ *)
(

*s *2 /

*n *+

*s *2 /

*n *)2
Rejection region:

*t* >

*tα/2,d. f . *≈

*t.05, *=
(

*x *−

*x *) − (

*µ *−

*µ *)

*p *(1 /

*n*1 + 1 /

*n*2 )
There is not enough evidence to indicate that the mean sales per square foot differ between the two
Test to determine if there is enough evidence to indicate that µ1 exceeds µ 2 for the following situation. (Use α = .05)
Estimate µ1 − µ2 with 95% confidence using the results of Exercise 13.1.
A statistician found that in a random sample of 43 cans of paint produced by one manufacture
the mean drying time was

*x *1 = 185 minutes, with a standard deviation of 20 minutes. In a random sample of 40 cans of paint produced by another manufacturer, the mean drying time was

*x *2 = 201 minutes, with a standard deviation of 57 minutes. Do these data allow us to conclude
at the 10% significance level that the mean drying times of the two kinds of paints differ?
Estimate the difference between mean drying times with 99% confidence for the data in Exercise
Find the 95% confidence interval estimate of µ1 − µ2 for Example 13.3.
An automobile parts manufacturer has been experimenting with a new type of spark plug
designed to improve gas mileage. In order to determine if the new spark plug is effective, eight
cars are randomly selected. The conventional spark plug is installed in four of the cars, and the
new experimental spark plug is installed in the other four. The gas mileage is measured and
reported below. Can we conclude at the 5% significance level that the new experimental spark
plug is effective in increasing gas mileage?
Find the 99% confidence interval estimate of ( µ1 − µ 2 ) for Exercise 13.6.

**13.4 Inference About the Difference Between Two Means: Matched Pairs **

Experiment
The greatest difficulty encountered by students in this section is recognizing when an experiment
has been conducted using matched pairs. If you are such a student, bear in mind that a matched pairs
experiment has been conducted when, because of the way in which the samples were selected, there is a
direct connection between an observation in one sample and an observation in the second sample. That
direct connection may be very strong, such as when one group of people is used to measure the
effectiveness of a sleeping pill or when one set of cars is used to measure the gas mileage of each of two
different brands of gasoline. The connection can be fairly weak, such as comparing the amount of traffic
at two intersections when the times of the day or the days of the week are matched. In any situation
where a direct connection exists between each pair of observations (one from the first sample and one
from the second sample), the experiment is matched pairs.

*x D *±

*t*α / 2

*s D */

*n D *
A plant that operates two shifts would like to know if the two shifts differ in productivity. The
number of units that each shift produces in each of the 5 days is counted and recorded below. Can we
conclude that the two shifts differ in the mean number of units per shift? Test with α =.10.
The problem objective is to compare two populations where the data are quantitative. The experiment
is matched pairs, because each observation for shift 1 is matched by the day of the week with an
observation for shift 2. That is, sample 1 and sample 2 are connected by the days of the week. Hence, the
appropriate statistical technique is the

*t*-test of µ

*D *.
Rejection region:

*t* >

*t*α / 2 ,

*n D *−1 =

*t*.05, 4 = 2.132 or

*t *< -2.132

*D* =

*x*1 −

*x*2 = –2, 10, 13, 7, 11
There is enough evidence to indicate that the two shifts differ in the mean number of units per shift.
An accountant is in the process of investigating the consequences of switching to another
method of depreciating assets. In particular, she would like to know if the switch will result in a
decrease in after-tax profit. To help decide, she randomly selects six firms and calculates the
after-tax profits using both depreciation methods. The results (rounded to the nearest million) are
shown below. Do these data provide sufficient evidence to indicate that the adoption of method
2 results in a lower after-tax profit? Test with α = .05.
Find the 99% confidence interval estimate of the mean difference in Exercise 13.8.
13.10 Find the 99% confidence interval estimate of the mean difference in Example 13.4.
13.11 The automotive parts manufacturer referred to in Exercise 13.6 concluded that the reason his
product was not shown to be superior was because of too much variation between cars. To
eliminate that variation, he decided to redo the experiment by selecting four cars. Each is
operated for a set number of miles with the experimental spark plug and again with the
conventional spark plug. (The order is randomly determined.) The results are shown below. Do
these data provide sufficient evidence at the 5% significance level to conclude that the
experimental spark plug is effective in increasing gas mileage?
13.13 Find the 95% confidence interval estimate of the mean difference in Exercise 13.11.

**13.5 Inference About the Ratio of Two Variances **
1 / σ 2 in cases where we want to compare two populations of
quantitative data and the descriptive measurement is variability. The test statistic use to test σ 2
which is

*F*-distributed with

*v *1 and

*v *2 degrees of freedom where

*v *1 =

*n*1 − 1 and

*v *2 =

*n*2 − 1 . The interval estimator is
It is assumed that t he populations from which we’ve sampled are normally distributed.
A statistician wanted to perform a

*t*-test of µ1 − µ2 with

*n *1 = 16 and

*n *2 = 25 . One of the
requirements of this test is that the two unknown population variances are equal. She found that
18 . Can she conclude with α = .10 that the assumption of equal variances is not
Rejection region:

*F* >

*F*α
or

*F* <

*F*1− α / 2 ,

*v*
Value of the test statistic:

*F* = 56/18 = 3.11
There is sufficient evidence to indicate that the assumption of equal variances is invalid.
.025 , 24 ,15 = (3.11 )( 2. 70 ) = 8 . 40
13.14 In random samples of 8 and 10 from two normal populations, it was found that

*s *2 =
45 , respectively. Do these statistics allow us to conclude that the variance of population 1
exceeds the variance of population 2? Use α = .05.
13.16 An educational statistician wanted to determine whether the scholastic aptitude test (SAT)
scores achieved by children living in cities was less variable than scores achieved by children
living in the suburbs. A random sample of 25 SAT scores of city children yielded a variance of
7, 814 . A random sample of 25 SAT scores of suburban children produced a variance of
9, 258 . At the 5% significance level, can we conclude that the city SAT score variance is
less than the suburban SAT score variance?

**13.6 Inference About the Difference Between Two Proportions **
When the problem objective is to compare two populations and the data are qualitative, the
parameter we make inferences about is the difference between two population proportions. There are two
test statistics. The choice of which to use is made on the basis of the null hypothesis. If the null
hypothesis states that the two proportions are identical (

*H*0:

*p*1 −

*p*2 = 0 ), the test statistic is
If the null hypothesis states that the difference between the two proportions is a value other than zero
(

*H*0:

*p*1 −

*p*2 = .10 ), the test statistic is

*ˆ (*1 −

*pˆ ) / n *+

*pˆ (*1 −

*pˆ ) / n*
*ˆ *2

*) z / *2

*pˆ*1

*(*1

*pˆ*1

*) / *1

*ˆ *2

*( *1

*p *2

*) / n *2
How do I determine whether a particular problem is Case 1 or Case 2?
If the question asks if there is enough evidence to conclude that

*p*1 is not equal to

*p*2, greater than

*p*2, or less than

*p*2, the problem is Case 1. If we’re asked to show that the difference between

*p*1 and

*p*2 is not equal to, greater than, or less than a specific value, the problem is Case 2.
A police detective is examining the crime rates in two different areas of the city. One area is middle
class, while the second area is quite affluent. The detective believes that the proportion of burglarized
homes in the middle-class area is greater than the proportion of burglarized homes in the affluent area.
From random samples of 1,000 homes in each area, he finds that last year there were 32 burglaries in the
middle-class area and 14 in the affluent area. At the 1% significance level, can we conclude that the
The problem objective is to compare two populations—the population of homes in the middle-class
area and the population of homes in the affluent area. The data are qualitative since the values of the
variable are “the home was burglarized” and “the home was not burglarized.” Finally, because we want to
determine if one proportion is greater than a second proportion, the null and alternative hypotheses are

*H*0: (

*p*1 −

*p *2 ) = 0
where

*p*1 = proportion of burglarized homes in the middle-class area, and

*p*2 = proportion of burglarized homes in the affluent area
This is a situation described by Case l, so that
Rejection region:

*z* >

*z*α =

*z*.01 = 2. 33

*p *= ( 32 +14 ) / (1, 000 +1, 000 ) =. 023
( . 023 )( . 977 )(1 / 1, 000 +1 / 1, 000 )
There is sufficient evidence to conclude that the detective’s belief is correct.
In Example 13.7, estimate

*p*1 −

*p*2 with 90% confidence.
There is only one interval estimator of

*p*1 −

*p*2 . It is

*ˆ *2

*) z / *2

*pˆ*1

*(*1

*pˆ*1

*) / *1

*ˆ *2

*( *1

*pˆ *2

*) / n *2
= (. 032 − .014 ) ± 1.645 (. 032 )(. 968 ) / 1,000 + (. 014 )(. 986 ) / 1,000
Suppose that in Example 12.7 the detective believes that the proportion of burglarized homes in the
middle-class area is more than 1% higher than the proportion of burglarized homes in the affluent area.
Once again, test the detective’s belief at the 1% significance level.

*H*0: (

*p*1 −

*p *2 ) = .01
Because the null hypothesis in dicates a value other than zero, this situation is described as Case 2.
( − ˆ

*p *) /

*n *+ ˆ

*p *1
Rejection region:

*z* >

*z*α =

*z*.01 = 2.33
(. 032 )(. 968 ) / 1,000 + (. 014 )(. 986 ) / 1, 000
There is not enough evidence to conclude that the proportion of burglarized middle -class homes exceeds
the proportion of burglarized affluent homes by more than 1%.
13.18 Test with α = .05 to determine if the following data allow us to conclude that

*p*1 is less than

*p*2.
13.19 A random sample of 250 units from one production line produced 25 defectives. In a random
sample of 400 units from a second production line, 80 were found to be defective. Can we
conclude at the 1% significance level that the defective rate from the second production line
exceeds the proportion defective from the first production line by more than 5%?
13.20 For Exercise 13.19, estimate

*p*1 −

*p*2 with 90% confidence.
13.21 Recently, the Canadian parliament debated the reinstatement of the death penalty. One of the
factors in this debate was the amount of public support for the death penalty. In 1989, a sample
of 1,500 Canadians revealed that 70% favored the death penalty. In 1999, 61% in a sample of
1,500 supported the death penalty. Do these data provide sufficient evidence at the 1%
significance level to indicate that support has fallen between 1989 and 1999?
13.22 Referring to Exercise 13.21, can we conclude with α = .01 that support for the death penalty has
fallen by more than 5 percentage points?
13.23 In Exercise 13.21, estimate the difference in public support for the reinstatement of the death
penalty between 1989 and 1999 with 95% confidence.
13.24 A company that produces an insect repellant is constantly looking for ways to improve the
product. Each time a new formula is developed, it is tested by taking a random sample of 1,000
people. Five hundred people have the existing product sprayed on their arms, while the
remaining 500 have the new formula sprayed on their arms. Each person then sits for 5 minutes
with only their arms exposed in a room full of mosquitoes. The number of people who have at
least one mosquito bite is recorded. Suppose that for one such experiment it was found that 45
people who used the existing product and 29 people who used the new formula suffered at least
one mosquito bite. Do these results indicate at the 5% significance level that the new formula is
13.25 In Exercise 13.24, estimate

*p*1 −

*p*2 with 90% confidence.

**13.6 Market Segmentation **
This section introduced a common application of statistical inference. Market segmentation allows
managers to discover which segments of the market are buying their products and which are not. There
are several exercises in this section and in other chapters that you can work on yourself.

Source: http://web.sakarya.edu.tr/~adurmus/statistik/Statistics_Study_%20Guides/ch13.PDF

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