Suppose that the population of heights of all fully grown male West African giraffes is approximately normally distributed. A recent article published in the Zoology Now journal claims that the mean of this population is 2.12 m. You want to test the claim made in the article, so you select a random sample of 10 fully grown male West African giraffes and record the height of each. Follow the steps below to construct a 90% confidence interval for the population mean of all heights of fully grown male West African giraffes. Then state whether the confidence interval you construct contradicts the article's claim. (If necessary, consult a list of formulas.) (a) Click on "Take Sample" to see the results for your random sample. Take Sample Sample size: Point estimate: Sample standard deviation: Critical value: Number of giraffes Compute 10 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". Sample mean 5.223 Standard error: Margin of error: Sample standard deviation 90% confidence interval: 1.514 Critical values 10.00s= 3.250 ¹0.010 2.821 ¹0.025=2.262 10.050 1.833 0.100 1.383

MATLAB: An Introduction with Applications
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(b)
(c)
Based on your sample, graph the 90% confidence interval for the population mean of all the heights of fully grown male West African giraffes.
• Enter the values for the lower and upper limits on the graph to show your confidence interval.
For the point (), enter the claim 2.12 from the article.
0.000
0.000
2.000
90% confidence interval:
4.000
5.000
6.000
8.000
Does the 90% confidence interval you constructed contradict the claim made in the article?
Choose the best answer from the choices below.
X
10.000
10.000
S
No, the confidence interval does not contradict the claim. The mean of 2.12 m from the article is inside the 90%
confidence interval.
O No, the confidence interval does not contradict the claim. The mean of 2.12 m from the article is outside the 90%
confidence interval.
Yes, the confidence interval contradicts the claim. The mean of 2.12 m from the article is inside the 90%
confidence interval.
Yes, the confidence interval contradicts the claim. The mean of 2.12 m from the article is outside the 90%
confidence interval.
Transcribed Image Text:(b) (c) Based on your sample, graph the 90% confidence interval for the population mean of all the heights of fully grown male West African giraffes. • Enter the values for the lower and upper limits on the graph to show your confidence interval. For the point (), enter the claim 2.12 from the article. 0.000 0.000 2.000 90% confidence interval: 4.000 5.000 6.000 8.000 Does the 90% confidence interval you constructed contradict the claim made in the article? Choose the best answer from the choices below. X 10.000 10.000 S No, the confidence interval does not contradict the claim. The mean of 2.12 m from the article is inside the 90% confidence interval. O No, the confidence interval does not contradict the claim. The mean of 2.12 m from the article is outside the 90% confidence interval. Yes, the confidence interval contradicts the claim. The mean of 2.12 m from the article is inside the 90% confidence interval. Yes, the confidence interval contradicts the claim. The mean of 2.12 m from the article is outside the 90% confidence interval.
Suppose that the population of heights of all fully grown male West African giraffes is approximately normally distributed. A recent article published in the
Zoology Now journal claims that the mean of this population is 2.12 m. You want to test the claim made in the article, so you select a random sample of 10
fully grown male West African giraffes and record the height of each.
Follow the steps below to construct a 90% confidence interval for the population mean of all heights of fully grown male West African giraffes. Then state
whether the confidence interval you construct contradicts the article's claim. (If necessary, consult a list of formulas.)
(a) Click on "Take Sample" to see the results for your random sample.
Take Sample
Sample size:
0
Point estimate:
0
Sample standard deviation:
0
Critical value:
П
Number of giraffes
Compute
10
Sample mean
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".
5.223
Standard error:
Margin of error:
Sample standard
90% confidence interval:
deviation
1.514
Critical values
0.005 3.250
0.010 2.821
0.025 2.262
0.050
1.833
=
0.100 1.383
=
Transcribed Image Text:Suppose that the population of heights of all fully grown male West African giraffes is approximately normally distributed. A recent article published in the Zoology Now journal claims that the mean of this population is 2.12 m. You want to test the claim made in the article, so you select a random sample of 10 fully grown male West African giraffes and record the height of each. Follow the steps below to construct a 90% confidence interval for the population mean of all heights of fully grown male West African giraffes. Then state whether the confidence interval you construct contradicts the article's claim. (If necessary, consult a list of formulas.) (a) Click on "Take Sample" to see the results for your random sample. Take Sample Sample size: 0 Point estimate: 0 Sample standard deviation: 0 Critical value: П Number of giraffes Compute 10 Sample mean 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". 5.223 Standard error: Margin of error: Sample standard 90% confidence interval: deviation 1.514 Critical values 0.005 3.250 0.010 2.821 0.025 2.262 0.050 1.833 = 0.100 1.383 =
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