MATH 110 Exam 8 Statistics Questions and
Answers- Portage Learning
1. . Independent random samples were selected from population 1 and population 2. The
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following information was obtained from these samples:
,
.1 .1, .1, .1, .1, .1, .1,
a) Find the 95% confidence interval for estimating the difference in the population means(µ - µ ).
.1 , .1, .1, .1, .1, .1 , .1, .1, .1, .1, .1 , .1, .1, ,
.1 1 .1 , .1 , . 1 ,
.1,
2
Solution. When we look back at table 6.1, we see that 95% confidence corresponds to z=1.96. .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1,
a) Notice that the sample sizes are each greater than 30, so we may use eqn. 8.1: .1, .1, .1 , .1, .1, .1, .1 , .1 , .1, .1, .1 , .1, .1, .1, .1 ,
b) Notice that the 95% confidence interval covers both positive and negative values. Therefore,we
.1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, ,
.1
cannot be 95% confident that there is a difference in the two population means.
.1, .1, .1 , .1 , .1 , .1 , .1, .1 , .1 , .1, .1 , . 1, .1, .1 ,
2. 2.Acompany would like to determine if there is a difference in the number of days that
,
.1 ,
.1 .1 , .1, .1 , .1 , .1, .1 , .1 , .1 , .1 , .1 , .1 , .1 , .1 , .1, .1 ,
employees are absent from the East Side Plant compared to the West Side Plant. So, the
,
.1 .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1 , .1,
company takes a sample of 54 employees from the East Side Plant and finds that these
.1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1 ,
people missed an average of 5.3 days last year with a standard deviation of 1.3 days. A
.1, .1, .1, .1, .1, .1, .1, .1 , .1, .1, .1, .1, .1, .1, .1, .1, .1,
sample of 41 employees from the West Side plant revealed that these people were absentan
.1, .1, .1 , .1, .1, .1, .1, .1 , .1, .1, .1 , .1, .1, .1, .1 , ,
.1
average of 6.8 days last year with a standard deviation of 1.8 days.
.1, .1, .1, .1, .1, .1, .1, .1 , .1 , .1, .1, .1 , .1,
a) Find the 96% confidence interval for estimating the difference in the population means(µ - µ ).
.1 , .1, .1, .1, .1, .1 , .1, .1, .1, .1, .1 , .1, .1, ,
.1 1 .1 , .1 , . 1 ,
.1,
2
Solution. When we look back at table 6.1, we see that 96% confidence corresponds to z=2.05.If we
.1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, ,
.1 .1,
say that the East Side Plant corresponds to population 1 and the West Side Plant corresponds to
.1, .1 , . 1, .1, .1 , .1, .1 , .1 , .1, .1 , .1, .1, .1 , .1 , .1, .1 , .1 ,
population 2, then:
.1 , .1 , .1 ,
n =54, n =41, s =1.3, s =1.8, xぁ= 5.3, xあ= 6.8,
1 .1 ,
2 .1 ,
1 .1 ,
2 .1 , ,
.1 .1, .1 , .1 , ,
.1 .1, .1,
a) We will use eqn. 8.1: .1 , .1, .1 , .1 ,
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.1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1,
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, b) Notice that the entire 96% confidence interval is negative (it is never positive or zero). Therefore, .1, .1, .1 , .1, .1, .1, .1, .1, .1, .1 , .1, .1, .1, .1, .1 ,
we can say that we are 96% confident that there is a difference in the two populationmeans.
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.1
c) Since the entire confidence interval is negative, we can be 96% confident that (µ1 - µ2) is .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1,
negative. This means that on average, people from the West Side Plant will be absent moredays
,
.1 .1, .1, .1, .1, .1, .1, .1, .1, .1, .1 , .1, .1, .1, .1, .1, ,
.1
than people from the East Side Plant..
.1 , .1 , .1 , .1, .1 , .1 , .1 ,
3. The mayor of a city would like to know if there is a difference in the systolic blood pressure
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of those who live in her city compared to those who live in the rural area outside the city. So,
.1 , .1 , .1, .1 , .1 , .1 , .1 , .1, .1, .1 , .1, .1, .1, .1, .1, .1 , .1, .1, .1, .1 ,
77 city dwellers are selected and it is found that their mean systolic blood pressure is 142 with
.1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1 , .1 , .1 , .1,
a standard deviation of 10.7.Also, 65 people are selected fromthe surrounding rural area and it
.1, .1 , .1, .1 , .1, ,
.1 .1, .1, .1, .1, .1 , ,
.1 .1, .1, .1, .1, .1,
is found that their mean systolic blood pressure is 129 with a standard deviation of 8.6.
.1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1 ,
a) Find the 98% confidence interval for estimating the difference in the population means(µ - µ ).
.1 , .1, .1, .1, .1, .1 , .1, .1, .1, .1, .1 , .1, .1, ,
.1 1 .1 , .1 , . 1 ,
.1,
2
Solution. When we look back at table 6.1, we see that 98% confidence corresponds to z=2.33.If we .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, ,
.1 .1,
say that the city residents corresponds to population 1 and the rural corresponds to population 2,
.1, .1 , .1 , .1 , .1, .1 , .1, .1, .1 , .1, .1 , .1, .1, .1, .1 , .1 ,
then:
.1 ,
n =77, n =65, s =10.7, s =8.6, x̄ = 142, x̄ = 129
1 .1 ,
2 .1 ,
1 .1 ,
2 . 1,
1 .1, .1, . 1 ,
.1 , .1 ,
2 .1, .1, . 1 ,
.1 ,
a) We will use eqn. 8.1: .1, .1 , .1, .1,
b) Notice that the entire 98% confidence interval is positive (it is never negative or zero). Therefore, .1, .1, .1 , .1, .1, .1, .1, .1, .1, .1 , .1, .1, .1, .1, .1 ,
we can say that we are 98% confident that there is a difference in the two populationmeans.
.1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, ,
.1
c) Since the entire confidence interval is positive, we can be 98% confident that (µ1 - µ2) is .1, .1 , .1, .1, .1, .1, .1, .1, .1 , .1, .1 , . 1, .1, .1, .1, .1 ,
positive. This means that on average, people from the city have higher systolic blood pressurethan
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.1
those from the rural area.
.1 , .1 , .1 , .1 , .1 ,
ProblemSet 8.2 Solutions .1, .1, .1,
1. Suppose we have independent random samples of size n = 780 and n = 700. The
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1 .1, .1, . 1 ,
.1, .1, .1,
2 .1, .1, . 1 ,
.1, .1,
number of successes in the two samples were x = 538 and x = 434. Find the 95%
.1, .1, .1 , .1, .1, .1, .1, .1 , .1,
1 .1, .1, .1,
2 .1, .1, .1,
.1, .1 , .1, .1,
confidence interval for the difference in the two population proportions. Solution. From
.1,
.1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1,
table 6.1, we see that 95% confidence corresponds to z=1.96.
,
.1 . 1, .1 , .1 , .1 , .1 , .1, .1 , .1 , .1,
Recall p = x /n = 538/780= .6897 and p = x /n = 434/700= .62. .1 ,
1 .1, .1, . 1 ,
.1 ,
1 1 .1, .1, . 1 ,
.1 , .1 , .1 , .1 ,
2 .1, .1, . 1 ,
.1 ,
2 2 .1, .1 , . 1 ,
.1 , .1 ,
Notice that the sample sizes are each greater than 30, so we may use eqn. 8.2:
.1, .1 , .1, .1, .1 , .1, .1, .1, .1, .1 , .1, .1, .1, .1 , .1,
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.1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1,
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Answers- Portage Learning
1. . Independent random samples were selected from population 1 and population 2. The
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following information was obtained from these samples:
,
.1 .1, .1, .1, .1, .1, .1,
a) Find the 95% confidence interval for estimating the difference in the population means(µ - µ ).
.1 , .1, .1, .1, .1, .1 , .1, .1, .1, .1, .1 , .1, .1, ,
.1 1 .1 , .1 , . 1 ,
.1,
2
Solution. When we look back at table 6.1, we see that 95% confidence corresponds to z=1.96. .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1,
a) Notice that the sample sizes are each greater than 30, so we may use eqn. 8.1: .1, .1, .1 , .1, .1, .1, .1 , .1 , .1, .1, .1 , .1, .1, .1, .1 ,
b) Notice that the 95% confidence interval covers both positive and negative values. Therefore,we
.1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, ,
.1
cannot be 95% confident that there is a difference in the two population means.
.1, .1, .1 , .1 , .1 , .1 , .1, .1 , .1 , .1, .1 , . 1, .1, .1 ,
2. 2.Acompany would like to determine if there is a difference in the number of days that
,
.1 ,
.1 .1 , .1, .1 , .1 , .1, .1 , .1 , .1 , .1 , .1 , .1 , .1 , .1 , .1, .1 ,
employees are absent from the East Side Plant compared to the West Side Plant. So, the
,
.1 .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1 , .1,
company takes a sample of 54 employees from the East Side Plant and finds that these
.1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1 ,
people missed an average of 5.3 days last year with a standard deviation of 1.3 days. A
.1, .1, .1, .1, .1, .1, .1, .1 , .1, .1, .1, .1, .1, .1, .1, .1, .1,
sample of 41 employees from the West Side plant revealed that these people were absentan
.1, .1, .1 , .1, .1, .1, .1, .1 , .1, .1, .1 , .1, .1, .1, .1 , ,
.1
average of 6.8 days last year with a standard deviation of 1.8 days.
.1, .1, .1, .1, .1, .1, .1, .1 , .1 , .1, .1, .1 , .1,
a) Find the 96% confidence interval for estimating the difference in the population means(µ - µ ).
.1 , .1, .1, .1, .1, .1 , .1, .1, .1, .1, .1 , .1, .1, ,
.1 1 .1 , .1 , . 1 ,
.1,
2
Solution. When we look back at table 6.1, we see that 96% confidence corresponds to z=2.05.If we
.1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, ,
.1 .1,
say that the East Side Plant corresponds to population 1 and the West Side Plant corresponds to
.1, .1 , . 1, .1, .1 , .1, .1 , .1 , .1, .1 , .1, .1, .1 , .1 , .1, .1 , .1 ,
population 2, then:
.1 , .1 , .1 ,
n =54, n =41, s =1.3, s =1.8, xぁ= 5.3, xあ= 6.8,
1 .1 ,
2 .1 ,
1 .1 ,
2 .1 , ,
.1 .1, .1 , .1 , ,
.1 .1, .1,
a) We will use eqn. 8.1: .1 , .1, .1 , .1 ,
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.1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1,
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, .1, .1, .1 , .1, .1, .1, .1, .1, .1, .1 , .1, .1, .1, .1, .1 ,
we can say that we are 96% confident that there is a difference in the two populationmeans.
.1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, ,
.1
c) Since the entire confidence interval is negative, we can be 96% confident that (µ1 - µ2) is .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1,
negative. This means that on average, people from the West Side Plant will be absent moredays
,
.1 .1, .1, .1, .1, .1, .1, .1, .1, .1, .1 , .1, .1, .1, .1, .1, ,
.1
than people from the East Side Plant..
.1 , .1 , .1 , .1, .1 , .1 , .1 ,
3. The mayor of a city would like to know if there is a difference in the systolic blood pressure
.1, .1, .1 , .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1 , .1, .1 ,
of those who live in her city compared to those who live in the rural area outside the city. So,
.1 , .1 , .1, .1 , .1 , .1 , .1 , .1, .1, .1 , .1, .1, .1, .1, .1, .1 , .1, .1, .1, .1 ,
77 city dwellers are selected and it is found that their mean systolic blood pressure is 142 with
.1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1 , .1 , .1 , .1,
a standard deviation of 10.7.Also, 65 people are selected fromthe surrounding rural area and it
.1, .1 , .1, .1 , .1, ,
.1 .1, .1, .1, .1, .1 , ,
.1 .1, .1, .1, .1, .1,
is found that their mean systolic blood pressure is 129 with a standard deviation of 8.6.
.1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1 ,
a) Find the 98% confidence interval for estimating the difference in the population means(µ - µ ).
.1 , .1, .1, .1, .1, .1 , .1, .1, .1, .1, .1 , .1, .1, ,
.1 1 .1 , .1 , . 1 ,
.1,
2
Solution. When we look back at table 6.1, we see that 98% confidence corresponds to z=2.33.If we .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, ,
.1 .1,
say that the city residents corresponds to population 1 and the rural corresponds to population 2,
.1, .1 , .1 , .1 , .1, .1 , .1, .1, .1 , .1, .1 , .1, .1, .1, .1 , .1 ,
then:
.1 ,
n =77, n =65, s =10.7, s =8.6, x̄ = 142, x̄ = 129
1 .1 ,
2 .1 ,
1 .1 ,
2 . 1,
1 .1, .1, . 1 ,
.1 , .1 ,
2 .1, .1, . 1 ,
.1 ,
a) We will use eqn. 8.1: .1, .1 , .1, .1,
b) Notice that the entire 98% confidence interval is positive (it is never negative or zero). Therefore, .1, .1, .1 , .1, .1, .1, .1, .1, .1, .1 , .1, .1, .1, .1, .1 ,
we can say that we are 98% confident that there is a difference in the two populationmeans.
.1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, ,
.1
c) Since the entire confidence interval is positive, we can be 98% confident that (µ1 - µ2) is .1, .1 , .1, .1, .1, .1, .1, .1, .1 , .1, .1 , . 1, .1, .1, .1, .1 ,
positive. This means that on average, people from the city have higher systolic blood pressurethan
.1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, ,
.1
those from the rural area.
.1 , .1 , .1 , .1 , .1 ,
ProblemSet 8.2 Solutions .1, .1, .1,
1. Suppose we have independent random samples of size n = 780 and n = 700. The
.1 , .1 , .1, .1, .1, .1 , .1, .1 , .1,
1 .1, .1, . 1 ,
.1, .1, .1,
2 .1, .1, . 1 ,
.1, .1,
number of successes in the two samples were x = 538 and x = 434. Find the 95%
.1, .1, .1 , .1, .1, .1, .1, .1 , .1,
1 .1, .1, .1,
2 .1, .1, .1,
.1, .1 , .1, .1,
confidence interval for the difference in the two population proportions. Solution. From
.1,
.1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1,
table 6.1, we see that 95% confidence corresponds to z=1.96.
,
.1 . 1, .1 , .1 , .1 , .1 , .1, .1 , .1 , .1,
Recall p = x /n = 538/780= .6897 and p = x /n = 434/700= .62. .1 ,
1 .1, .1, . 1 ,
.1 ,
1 1 .1, .1, . 1 ,
.1 , .1 , .1 , .1 ,
2 .1, .1, . 1 ,
.1 ,
2 2 .1, .1 , . 1 ,
.1 , .1 ,
Notice that the sample sizes are each greater than 30, so we may use eqn. 8.2:
.1, .1 , .1, .1, .1 , .1, .1, .1, .1, .1 , .1, .1, .1, .1 , .1,
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.1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1, .1,
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