The mature eucalyptus trees in a forest vary. It is known that the population of heights of all mature eucalyptuses in the forest is approximately normally distributed. An article in a conservation journal claims that the standard deviation of this population is 6.89 m. You are a researcher who wants to f this claim with a random sample of 44 mature eucalyptuses from the forest. Based on your sample, follow the steps below to construct a 90% confidence interval for the population standard deviation of all mature eucalyptus heights the forest. 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 from the random sample. (b) (c) Take Sample Point estimate of the population variance: 47.89 Sample size: 44 Left critical value: 0 Right critical value: 0 Compute 0.00 0.00 Number of mature eucalyptuses 44 To find the confidence interval for the population standard deviation, first find the confidence interval for the population variance. Enter the values of the point estimate of the population variance, the sample size, the left critical value, and the right critical value you need for you 90% confidence interval for the population variance. (Choose the correct critical values from the table of critical values provided.) When you are done, select "Compute". 2.00 Sample mean 90% confidence interval for the population variance: 90% confidence interval for the population standard deviation: 88.79 4.00 5.00 X 6.00 X0.995 90% confidence interval for the population standard deviation: Critical values Left Based on your sample, graph the 90% confidence interval for the population standard deviation of all mature eucalyptus heights in the forest. • Enter the values for the lower and upper limits on the graph to show your confidence interval. Round the values to two decimal places. . For the point (◆) enter the claim 6.89 from the article on your graph. Sample standard Right = 22.859 0.005=70.616 X0.975=26.785 0.025 = 62.99 X0.950 = 28.965 %0.050=59.304 deviation Does the 90% confidence interval you constructed contradict the article's claim? Choose the best answer from the choices below. 8.00 6.92 X 10.00 10.00 O No, the confidence interval does not contradict the claim. The claimed standard deviation 6.89 is inside the 90% confidence interval. O No, the confidence interval does not contradict the claim. The claimed standard deviation 6.89 is outside the 90% confidence interval. Sample variance O Yes, the confidence interval contradicts the claim. The claimed standard deviation 6.89 is inside the 90% confidence interval. 47.8864 O Yes, the confidence interval contradicts the claim. The claimed standard deviation 6.89 is outside the 90% confidence interval. x

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Question
The mature eucalyptus trees in a forest vary. It is known that the population of heights of all mature eucalyptuses in the forest is approximately
normally distributed. An article in a conservation journal claims that the standard deviation of this population is 6.89 m. You are a researcher who wants to test
this claim with a random sample of 44 mature eucalyptuses from the forest.
Based on your sample, follow the steps below to construct a 90% confidence interval for the population standard deviation of all mature eucalyptus heights in
the forest. 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 from the random sample.
(b)
(c)
Take Sample
Point estimate of the
population variance:
47.89
Sample size:
44
Left critical value:
0
Right critical value:
0
Compute
0.00
To find the confidence interval for the population standard deviation, first find the confidence interval for the population variance.
Enter the values of the point estimate of the population variance, the sample size, the left critical value, and the right critical value you need for your
90% confidence interval for the population variance. (Choose the correct critical values from the table of critical values provided.) When you are
done, select "Compute".
0.00
Number of mature
eucalyptuses
44
2.00
Sample mean
90% confidence interval for the
population variance:
90% confidence interval for the
population standard deviation:
4.00
88.79
5.00
X
90% confidence interval for the population standard deviation:
6.00
Based on your sample, graph the 90% confidence interval for the population standard deviation of all mature eucalyptus heights in the forest.
• Enter the values for the lower and upper limits on the graph to show your confidence interval. Round the values to two decimal places.
• For the point (◆) enter the claim 6.89 from the article on your graph.
Critical values
Sample standard
deviation
Left
Right
20.995 = 22.859
0.005 = 70.616
20.975=26.785
0.025 = 62.99
2
20.950=28.965 0.050 = 59.304
Does the 90% confidence interval you constructed contradict the article's claim?
Choose the best answer from the choices below.
6.92
8.00
X
10.00
10.00
O No, the confidence interval does not contradict the claim. The claimed standard deviation 6.89 is inside the 90%
confidence interval.
O No, the confidence interval does not contradict the claim. The claimed standard deviation 6.89 is outside the 90%
confidence interval.
Sample variance
O Yes, the confidence interval contradicts the claim. The claimed standard deviation 6.89 is inside the 90%
confidence interval.
47.8864
O Yes, the confidence interval contradicts the claim. The claimed standard deviation 6.89 is outside the 90%
confidence interval.
X
S
Transcribed Image Text:The mature eucalyptus trees in a forest vary. It is known that the population of heights of all mature eucalyptuses in the forest is approximately normally distributed. An article in a conservation journal claims that the standard deviation of this population is 6.89 m. You are a researcher who wants to test this claim with a random sample of 44 mature eucalyptuses from the forest. Based on your sample, follow the steps below to construct a 90% confidence interval for the population standard deviation of all mature eucalyptus heights in the forest. 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 from the random sample. (b) (c) Take Sample Point estimate of the population variance: 47.89 Sample size: 44 Left critical value: 0 Right critical value: 0 Compute 0.00 To find the confidence interval for the population standard deviation, first find the confidence interval for the population variance. Enter the values of the point estimate of the population variance, the sample size, the left critical value, and the right critical value you need for your 90% confidence interval for the population variance. (Choose the correct critical values from the table of critical values provided.) When you are done, select "Compute". 0.00 Number of mature eucalyptuses 44 2.00 Sample mean 90% confidence interval for the population variance: 90% confidence interval for the population standard deviation: 4.00 88.79 5.00 X 90% confidence interval for the population standard deviation: 6.00 Based on your sample, graph the 90% confidence interval for the population standard deviation of all mature eucalyptus heights in the forest. • Enter the values for the lower and upper limits on the graph to show your confidence interval. Round the values to two decimal places. • For the point (◆) enter the claim 6.89 from the article on your graph. Critical values Sample standard deviation Left Right 20.995 = 22.859 0.005 = 70.616 20.975=26.785 0.025 = 62.99 2 20.950=28.965 0.050 = 59.304 Does the 90% confidence interval you constructed contradict the article's claim? Choose the best answer from the choices below. 6.92 8.00 X 10.00 10.00 O No, the confidence interval does not contradict the claim. The claimed standard deviation 6.89 is inside the 90% confidence interval. O No, the confidence interval does not contradict the claim. The claimed standard deviation 6.89 is outside the 90% confidence interval. Sample variance O Yes, the confidence interval contradicts the claim. The claimed standard deviation 6.89 is inside the 90% confidence interval. 47.8864 O Yes, the confidence interval contradicts the claim. The claimed standard deviation 6.89 is outside the 90% confidence interval. X S
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