MATH 110 MODULE 8 STATISTICS EXAM QUESTIONS
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1. . Independent random samples were selected from population 1 and population 2. The
following information was obtained from these samples:
a) Find the 95% confidence interval for estimating the difference in the population means (µ1 -
µ2).
Solution. When we look back at table 6.1, we see that 95% confidence corresponds to z=1.96.
a) Notice that the sample sizes are each greater than 30, so we may use eqn. 8.1:
b) Notice that the 95% confidence interval covers both positive and negative values. Therefore,
we cannot be 95% confident that there is a difference in the two population means.
2. 2. A company would like to determine if there is a difference in the number of days that
employees are absent from the East Side Plant compared to the West Side Plant. So, the company
takes a sample of 54 employees from the East Side Plant and finds that these people missed an
average of 5.3 days last year with a standard deviation of 1.3 days. A sample of 41 employees
from the West Side plant revealed that these people were absentan average of 6.8 days last year
with a standard deviation of 1.8 days.
a) Find the 96% confidence interval for estimating the difference in the population means (µ1 -
µ2).
Solution. When we look back at table 6.1, we see that 96% confidence corresponds to z=2.05.If
we say that the East Side Plant corresponds to population 1 and the West Side Plant corresponds
to population 2, then:
n1=54, n2=41, s1=1.3, s2=1.8, x ぁ= 5.3, x あ= 6.8,
a) We will use eqn. 8.1:
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, b) Notice that the entire 96% confidence interval is negative (it is never positive or zero).
Therefore, we can say that we are 96% confident that there is a difference in the two population
means.
c) Since the entire confidence interval is negative, we can be 96% confident that (µ1 - µ2) is
negative. This means that on average, people from the West Side Plant will be absent more
days than people from the East Side Plant..
3. The mayor of a city would like to know if there is a difference in the systolic blood pressure of
those who live in her city compared to those who live in the rural area outside the city. So, 77 city
dwellers are selected and it is found that their mean systolic blood pressure is 142 with a standard
deviation of 10.7. Also, 65 people are selected fromthe surrounding rural area and it is found that
their mean systolic blood pressure is 129 with a standard deviation of 8.6.
a) Find the 98% confidence interval for estimating the difference in the population means (µ1 -
µ2).
Solution. When we look back at table 6.1, we see that 98% confidence corresponds to z=2.33.If
we say that the city residents corresponds to population 1 and the rural corresponds to
population 2, then:
n1=77, n2=65, s1=10.7, s2=8.6, x̄1 = 142, x̄2 = 129
a) We will use eqn. 8.1:
b) Notice that the entire 98% confidence interval is positive (it is never negative or zero).
Therefore, we can say that we are 98% confident that there is a difference in the two population
means.
c) Since the entire confidence interval is positive, we can be 98% confident that (µ1 - µ2) is
positive. This means that on average, people from the city have higher systolic blood pressure
than those from the rural area.
Problem Set 8.2 Solutions
1. Suppose we have independent random samples of size n1 = 780 and n2 = 700. The
number of successes in the two samples were x1= 538 and x2 = 434. Find the 95%
confidence interval for the difference in the two population proportions. Solution. From table
6.1, we see that 95% confidence corresponds to z=1.96.
Recall p1 = x1/n1 = 538/780= .6897 and p2 = x2/n2 = 434/700= .62.
Notice that the sample sizes are each greater than 30, so we may use eqn. 8.2:
This study source was downloaded by 100000848594376 from CourseHero.com on 01-16-2023 21:26:16 GMT -
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https://www.coursehero.com/file/81393958/MATH-110-Module-8-ANSWERSdocx/
AND CORRECT ANSWERS- PORTAGE LEARNING| LATEST
AND VERIFIED EDITION ALREADY
GRADED A+| BRAND NEW!!! ASSURED PASS
1. . Independent random samples were selected from population 1 and population 2. The
following information was obtained from these samples:
a) Find the 95% confidence interval for estimating the difference in the population means (µ1 -
µ2).
Solution. When we look back at table 6.1, we see that 95% confidence corresponds to z=1.96.
a) Notice that the sample sizes are each greater than 30, so we may use eqn. 8.1:
b) Notice that the 95% confidence interval covers both positive and negative values. Therefore,
we cannot be 95% confident that there is a difference in the two population means.
2. 2. A company would like to determine if there is a difference in the number of days that
employees are absent from the East Side Plant compared to the West Side Plant. So, the company
takes a sample of 54 employees from the East Side Plant and finds that these people missed an
average of 5.3 days last year with a standard deviation of 1.3 days. A sample of 41 employees
from the West Side plant revealed that these people were absentan average of 6.8 days last year
with a standard deviation of 1.8 days.
a) Find the 96% confidence interval for estimating the difference in the population means (µ1 -
µ2).
Solution. When we look back at table 6.1, we see that 96% confidence corresponds to z=2.05.If
we say that the East Side Plant corresponds to population 1 and the West Side Plant corresponds
to population 2, then:
n1=54, n2=41, s1=1.3, s2=1.8, x ぁ= 5.3, x あ= 6.8,
a) We will use eqn. 8.1:
This study source was downloaded by 100000848594376 from CourseHero.com on 01-16-2023 21:26:16 GMT -
06:00
https://www.coursehero.com/file/81393958/MATH-110-Module-8-ANSWERSdocx/
, b) Notice that the entire 96% confidence interval is negative (it is never positive or zero).
Therefore, we can say that we are 96% confident that there is a difference in the two population
means.
c) Since the entire confidence interval is negative, we can be 96% confident that (µ1 - µ2) is
negative. This means that on average, people from the West Side Plant will be absent more
days than people from the East Side Plant..
3. The mayor of a city would like to know if there is a difference in the systolic blood pressure of
those who live in her city compared to those who live in the rural area outside the city. So, 77 city
dwellers are selected and it is found that their mean systolic blood pressure is 142 with a standard
deviation of 10.7. Also, 65 people are selected fromthe surrounding rural area and it is found that
their mean systolic blood pressure is 129 with a standard deviation of 8.6.
a) Find the 98% confidence interval for estimating the difference in the population means (µ1 -
µ2).
Solution. When we look back at table 6.1, we see that 98% confidence corresponds to z=2.33.If
we say that the city residents corresponds to population 1 and the rural corresponds to
population 2, then:
n1=77, n2=65, s1=10.7, s2=8.6, x̄1 = 142, x̄2 = 129
a) We will use eqn. 8.1:
b) Notice that the entire 98% confidence interval is positive (it is never negative or zero).
Therefore, we can say that we are 98% confident that there is a difference in the two population
means.
c) Since the entire confidence interval is positive, we can be 98% confident that (µ1 - µ2) is
positive. This means that on average, people from the city have higher systolic blood pressure
than those from the rural area.
Problem Set 8.2 Solutions
1. Suppose we have independent random samples of size n1 = 780 and n2 = 700. The
number of successes in the two samples were x1= 538 and x2 = 434. Find the 95%
confidence interval for the difference in the two population proportions. Solution. From table
6.1, we see that 95% confidence corresponds to z=1.96.
Recall p1 = x1/n1 = 538/780= .6897 and p2 = x2/n2 = 434/700= .62.
Notice that the sample sizes are each greater than 30, so we may use eqn. 8.2:
This study source was downloaded by 100000848594376 from CourseHero.com on 01-16-2023 21:26:16 GMT -
06:00
https://www.coursehero.com/file/81393958/MATH-110-Module-8-ANSWERSdocx/
