In anticipation of an upcoming election, officials in Grand Ridge County are looking at the distance from each voter's home to that voter's nearest polling station. Assume that the population of all such distances for voters in Grand Ridge County is approximately normally distributed. An article for the newspaper Politically Sound claimed that the mean of this population is 4.82 km. You want to test this claim, so you select a random sample of 22 Grand Ridge County voters, and for each you record the distance the voter lives from their nearest polling station. Follow the steps below to construct a 90% confidence interval for the population mean of all the distances voters in Grand Ridge County live from their nearest polling station. Then state whether the confidence interval you construct contradicts the reporter's claim. (If necessary, consult a list of formulas.) (a) Click on "Take Sample" to see the results for your random sample. Take Sample Number of people 22 Sample mean 4.971 Sample standard deviation 2.433 Enter the values of the sample size, the point estimate of the mean, the sample standard deviation, and the critical value you need for your 90% confidence interval. (Choose the correct critical value from the table of critical values provided.) When you are done, select "Compute".

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In anticipation of an upcoming election, officials in Grand Ridge County are looking at the distance from each voter's home to that voter's nearest polling station.
Assume that the population of all such distances for voters in Grand Ridge County is approximately normally distributed. An article for the newspaper Politically
Sound claimed that the mean of this population is 4.82 km. You want to test this claim, so you select a random sample of 22 Grand Ridge County voters, and for
each you record the distance the voter lives from their nearest polling station.
Follow the steps below to construct a 90% confidence interval for the population mean of all the distances voters in Grand Ridge County live from their nearest
polling station. Then state whether the confidence interval you construct contradicts the reporter's claim. (If necessary, consult a list of formulas.)
(a) Click on "Take Sample" to see the results for your random sample.
esc
Take Sample
Sample size:
Point estimate:
0
FI
Sample standard deviation:
0
Critical value:
0
Enter the values of the sample size, the point estimate of the mean, the sample standard deviation, and the critical value you need for your 90%
confidence interval. (Choose the correct critical value from the table of critical values provided.) When you are done, select "Compute".
Compute
Explanation
Check
Number of people
F2
22
000
OOD F4
Sample mean
Standard error:
Margin of error:
4.971
F5
90% confidence interval:
F6
AA
Sample standard
deviation
F7
2.433
X
Critical values
0.005 2.831
0.010-2.518
0.025 -2.080
0.050 1.721
+333
$
2022 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility
F8
▶
F9
F10
4)
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Transcribed Image Text:In anticipation of an upcoming election, officials in Grand Ridge County are looking at the distance from each voter's home to that voter's nearest polling station. Assume that the population of all such distances for voters in Grand Ridge County is approximately normally distributed. An article for the newspaper Politically Sound claimed that the mean of this population is 4.82 km. You want to test this claim, so you select a random sample of 22 Grand Ridge County voters, and for each you record the distance the voter lives from their nearest polling station. Follow the steps below to construct a 90% confidence interval for the population mean of all the distances voters in Grand Ridge County live from their nearest polling station. Then state whether the confidence interval you construct contradicts the reporter's claim. (If necessary, consult a list of formulas.) (a) Click on "Take Sample" to see the results for your random sample. esc Take Sample Sample size: Point estimate: 0 FI Sample standard deviation: 0 Critical value: 0 Enter the values of the sample size, the point estimate of the mean, the sample standard deviation, and the critical value you need for your 90% confidence interval. (Choose the correct critical value from the table of critical values provided.) When you are done, select "Compute". Compute Explanation Check Number of people F2 22 000 OOD F4 Sample mean Standard error: Margin of error: 4.971 F5 90% confidence interval: F6 AA Sample standard deviation F7 2.433 X Critical values 0.005 2.831 0.010-2.518 0.025 -2.080 0.050 1.721 +333 $ 2022 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility F8 ▶ F9 F10 4) F11 D 1 امه M
(b)
Based on your sample, graph the 90% confidence interval for the population mean of all the distances the voters in Grand Ridge County live from
their nearest polling station.
(c)
Enter the values for the lower and upper limits on the graph to show your confidence interval.
. For the point (), enter the claim 4.82 from the article.
0.000
0.000
2,000
90% confidence interval:
4.000
5.000
Explanation
6.000
Does the 90% confidence interval you constructed contradict the claim made in the article?
Choose the best answer from the choices below.
8.000
Check
S
O No, the confidence interval does not contradict the claim. The mean of 4.82 km from the article is inside the 90%
confidence interval.
Yes, the confidence interval contradicts the claim. The mean of 4.82 km from the article is inside the 90%
confidence interval.
O No, the confidence interval does not contradict the claim. The mean of 4.82 km from the article is outside the 90%
confidence interval.
Yes, the confidence interval contradicts the claim. The mean of 4.82 km from the article is outside the 90%
confidence interval.
10.000
10.000
2022 McGraw Hill LLC. All Rights Reserved. Terms of Use Privacy Center Accessibility
Transcribed Image Text:(b) Based on your sample, graph the 90% confidence interval for the population mean of all the distances the voters in Grand Ridge County live from their nearest polling station. (c) Enter the values for the lower and upper limits on the graph to show your confidence interval. . For the point (), enter the claim 4.82 from the article. 0.000 0.000 2,000 90% confidence interval: 4.000 5.000 Explanation 6.000 Does the 90% confidence interval you constructed contradict the claim made in the article? Choose the best answer from the choices below. 8.000 Check S O No, the confidence interval does not contradict the claim. The mean of 4.82 km from the article is inside the 90% confidence interval. Yes, the confidence interval contradicts the claim. The mean of 4.82 km from the article is inside the 90% confidence interval. O No, the confidence interval does not contradict the claim. The mean of 4.82 km from the article is outside the 90% confidence interval. Yes, the confidence interval contradicts the claim. The mean of 4.82 km from the article is outside the 90% confidence interval. 10.000 10.000 2022 McGraw Hill LLC. All Rights Reserved. Terms of Use Privacy Center Accessibility
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