MATH 110 Exam 8 Statistics Questions and Answers- Portage Learning
1. . Independent random samples were selected from population 1 and population 2.
ws ws ws ws ws ws ws ws ws ws ws
Thefollowing information was obtained from these samples:
ws w
s ws ws ws ws ws ws
a) Find the 95% confidence interval for estimating the difference in the population me
ws ws ws ws ws ws ws ws ws ws ws ws ws
ans(µ - µ ).w
s 1wswsws ws 2
Solution. When we look back at table 6.1, we see that 95% confidence corresponds to z=1
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
.96.
a) Notice that the sample sizes are each greater than 30, so we may use eqn. 8.1:
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
b) Notice that the 95% confidence interval covers both positive and negative values. Ther
ws ws ws ws ws ws ws ws ws ws ws ws
efore,we cannot be 95% confident that there is a difference in the two population means.
sw ws ws ws ws ws ws ws ws ws ws ws ws ws ws
2. 2. A company would like to determine if there is a difference in the number of da
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
ys thatemployees are absent from the East Side Plant compared to the West Side
ws w
s ws ws ws ws ws ws ws ws ws ws ws ws ws
Plant. So, the company takes a sample of 54 employees from the East Side Plant a
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
nd finds that these people missed an average of 5.3 days last year with a standard
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
deviation of 1.3 days. A sample of 41 employees from the West Side plant revealed
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
that these people were absentan average of 6.8 days last year with a standard devi
ws ws ws ws ws w
s ws ws ws ws ws ws ws ws ws ws
ation of 1.8 days. ws ws ws
a) Find the 96% confidence interval for estimating the difference in the population me
ws ws ws ws ws ws ws ws ws ws ws ws ws
ans(µ - µ ).w
s 1wswsws ws 2
Solution. When we look back at table 6.1, we see that 96% confidence corresponds to z=
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
2.05.If we say that the East Side Plant corresponds to population 1 and the West Side Pl
sw ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
ant corresponds to population 2, then:
ws ws ws ws ws
n =54, n =41, s =1.3, s =1.8, x ぁ= 5.3, x あ= 6.8,
1 ws 2 ws 1 ws 2 ws ws ws ws ws ws s
w ws
a) We will use eqn. 8.1: ws ws ws ws
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, b) Notice that the entire 96% confidence interval is negative (it is never positive or zero).
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
Therefore, we can say that we are 96% confident that there is a difference in the two pop
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
ulationmeans. w
s
c) Since the entire confidence interval is negative, we can be 96% confident that (µ1 -
ws ws ws ws ws ws ws ws ws ws ws ws ws ws
µ2) isnegative. This means that on average, people from the West Side Plant will be
ws ws sw ws ws ws ws ws ws ws ws ws ws ws ws ws ws
absent moredays than people from the East Side Plant..
ws sw ws ws ws ws ws ws ws
3. The mayor of a city would like to know if there is a difference in the systolic bl
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
ood pressure of those who live in her city compared to those who live in the rural
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws w
sarea outside the city. So, 77 city dwellers are selected and it is found that their m
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
ean systolic blood pressure is 142 with a standard deviation of 10.7. Also, 65 peopl
ws ws ws ws ws ws ws ws ws ws ws ws ws ws
e are selected fromthe surrounding rural area and it is found that their mean systol
ws ws ws w
s ws ws ws ws ws ws ws ws ws ws ws
ic blood pressure is 129 with a standard deviation of 8.6.
ws ws ws ws ws ws ws ws ws ws
a) Find the 98% confidence interval for estimating the difference in the population me
ws ws ws ws ws ws ws ws ws ws ws ws ws
ans(µ - µ ).w
s 1wswsws ws 2
Solution. When we look back at table 6.1, we see that 98% confidence corresponds to z
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
=2.33.If we say that the city residents corresponds to population 1 and the rural correspon
sw ws ws ws ws ws ws ws ws ws ws ws ws ws ws
ds to population 2, then:
ws ws ws ws
n =77, n =65, s =10.7, s =8.6, x
1 ws = 142, x
2 = 129 ws 1 ws 2 ws 1wswsws ws ws 2wswsws ws
a) We will use eqn. 8.1:ws ws ws ws
b) Notice that the entire 98% confidence interval is positive (it is never negative or zero).
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
Therefore, we can say that we are 98% confident that there is a difference in the two pop
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
ulationmeans. w
s
c) Since the entire confidence interval is positive, we can be 98% confident that (µ1 -
ws ws ws ws ws ws ws ws ws ws ws ws ws ws
µ2) is positive. This means that on average, people from the city have higher systolic bl
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
ood pressurethan those from the rural area.
ws sw ws ws ws ws ws
Problem Set 8.2 Solutions ws ws ws
1. Suppose we have independent random samples of size n
ws ws = 780 and n
ws = 7 ws ws ws ws ws ws 1wswsws ws ws ws 2wswsws ws
00. The number of successes in the two samples were x = 538 and x
ws ws = 434. Fi
ws ws ws ws ws ws ws ws 1 ws ws ws 2wswsws ws ws
nd the 95% confidence interval for the difference in the two population proportion
ws ws ws ws ws ws ws ws ws ws ws ws
s. Solution. Fromtable 6.1, we see that 95% confidence corresponds to z=1.96.
ws ws sw ws ws ws ws ws ws ws ws ws
Recall p = x /n = 538/780= .6897 and p
ws 1wswsws = x /n = 434/700= .62.
ws 1 1wswsws ws ws ws ws 2wswsws ws 2 2wswsws ws ws
Notice that the sample sizes are each greater than 30, so we may use eqn. 8.2:
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
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1. . Independent random samples were selected from population 1 and population 2.
ws ws ws ws ws ws ws ws ws ws ws
Thefollowing information was obtained from these samples:
ws w
s ws ws ws ws ws ws
a) Find the 95% confidence interval for estimating the difference in the population me
ws ws ws ws ws ws ws ws ws ws ws ws ws
ans(µ - µ ).w
s 1wswsws ws 2
Solution. When we look back at table 6.1, we see that 95% confidence corresponds to z=1
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
.96.
a) Notice that the sample sizes are each greater than 30, so we may use eqn. 8.1:
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
b) Notice that the 95% confidence interval covers both positive and negative values. Ther
ws ws ws ws ws ws ws ws ws ws ws ws
efore,we cannot be 95% confident that there is a difference in the two population means.
sw ws ws ws ws ws ws ws ws ws ws ws ws ws ws
2. 2. A company would like to determine if there is a difference in the number of da
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
ys thatemployees are absent from the East Side Plant compared to the West Side
ws w
s ws ws ws ws ws ws ws ws ws ws ws ws ws
Plant. So, the company takes a sample of 54 employees from the East Side Plant a
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
nd finds that these people missed an average of 5.3 days last year with a standard
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
deviation of 1.3 days. A sample of 41 employees from the West Side plant revealed
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
that these people were absentan average of 6.8 days last year with a standard devi
ws ws ws ws ws w
s ws ws ws ws ws ws ws ws ws ws
ation of 1.8 days. ws ws ws
a) Find the 96% confidence interval for estimating the difference in the population me
ws ws ws ws ws ws ws ws ws ws ws ws ws
ans(µ - µ ).w
s 1wswsws ws 2
Solution. When we look back at table 6.1, we see that 96% confidence corresponds to z=
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
2.05.If we say that the East Side Plant corresponds to population 1 and the West Side Pl
sw ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
ant corresponds to population 2, then:
ws ws ws ws ws
n =54, n =41, s =1.3, s =1.8, x ぁ= 5.3, x あ= 6.8,
1 ws 2 ws 1 ws 2 ws ws ws ws ws ws s
w ws
a) We will use eqn. 8.1: ws ws ws ws
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, b) Notice that the entire 96% confidence interval is negative (it is never positive or zero).
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
Therefore, we can say that we are 96% confident that there is a difference in the two pop
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
ulationmeans. w
s
c) Since the entire confidence interval is negative, we can be 96% confident that (µ1 -
ws ws ws ws ws ws ws ws ws ws ws ws ws ws
µ2) isnegative. This means that on average, people from the West Side Plant will be
ws ws sw ws ws ws ws ws ws ws ws ws ws ws ws ws ws
absent moredays than people from the East Side Plant..
ws sw ws ws ws ws ws ws ws
3. The mayor of a city would like to know if there is a difference in the systolic bl
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
ood pressure of those who live in her city compared to those who live in the rural
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws w
sarea outside the city. So, 77 city dwellers are selected and it is found that their m
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
ean systolic blood pressure is 142 with a standard deviation of 10.7. Also, 65 peopl
ws ws ws ws ws ws ws ws ws ws ws ws ws ws
e are selected fromthe surrounding rural area and it is found that their mean systol
ws ws ws w
s ws ws ws ws ws ws ws ws ws ws ws
ic blood pressure is 129 with a standard deviation of 8.6.
ws ws ws ws ws ws ws ws ws ws
a) Find the 98% confidence interval for estimating the difference in the population me
ws ws ws ws ws ws ws ws ws ws ws ws ws
ans(µ - µ ).w
s 1wswsws ws 2
Solution. When we look back at table 6.1, we see that 98% confidence corresponds to z
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
=2.33.If we say that the city residents corresponds to population 1 and the rural correspon
sw ws ws ws ws ws ws ws ws ws ws ws ws ws ws
ds to population 2, then:
ws ws ws ws
n =77, n =65, s =10.7, s =8.6, x
1 ws = 142, x
2 = 129 ws 1 ws 2 ws 1wswsws ws ws 2wswsws ws
a) We will use eqn. 8.1:ws ws ws ws
b) Notice that the entire 98% confidence interval is positive (it is never negative or zero).
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
Therefore, we can say that we are 98% confident that there is a difference in the two pop
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
ulationmeans. w
s
c) Since the entire confidence interval is positive, we can be 98% confident that (µ1 -
ws ws ws ws ws ws ws ws ws ws ws ws ws ws
µ2) is positive. This means that on average, people from the city have higher systolic bl
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
ood pressurethan those from the rural area.
ws sw ws ws ws ws ws
Problem Set 8.2 Solutions ws ws ws
1. Suppose we have independent random samples of size n
ws ws = 780 and n
ws = 7 ws ws ws ws ws ws 1wswsws ws ws ws 2wswsws ws
00. The number of successes in the two samples were x = 538 and x
ws ws = 434. Fi
ws ws ws ws ws ws ws ws 1 ws ws ws 2wswsws ws ws
nd the 95% confidence interval for the difference in the two population proportion
ws ws ws ws ws ws ws ws ws ws ws ws
s. Solution. Fromtable 6.1, we see that 95% confidence corresponds to z=1.96.
ws ws sw ws ws ws ws ws ws ws ws ws
Recall p = x /n = 538/780= .6897 and p
ws 1wswsws = x /n = 434/700= .62.
ws 1 1wswsws ws ws ws ws 2wswsws ws 2 2wswsws ws ws
Notice that the sample sizes are each greater than 30, so we may use eqn. 8.2:
ws ws ws ws ws ws ws ws ws ws ws ws ws ws ws
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