Suppose that the population of lengths of all fully grown male killer whales is approximately normally distributed. A recent article published in the Zoology Now journal claims that the mean of this population is 6.99 m. You want to test the claim made in the article, so you select a random sample of 10 fully grown male killer whales and record the length of each. Follow the steps below to construct a 90% confidence interval for the population mean of all lengths of fully grown male killer whales. 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. Number of killer whales Sample standard Sample mean deviation Take Sample 10 7.241 1.436 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 size: X 5 ? Standard error: Point estimate: Critical values Sample standard deviation: Margin of error: to.005 = 3.250 to.010 = 2.821 Critical value: 0 0.025 = 2.262 90% confidence interval: 0.050 = 1.833 Compute 0.100 = 1.383 (b) Based on your sample, graph the 90% confidence interval for the population mean of all the lengths of fully grown male killer whales. . Enter the values for the lower and upper limits on the graph to show your confidence interval. . For the point (*), enter the claim 6.99 from the article.90% confidence interval: 0.000 10.000 5.000 0.000 2.000 4.000 6.000 8.000 10.000 X 5 ? (c) Does the 90% confidence interval you constructed contradict the claim made in the article? Choose the best answer from the choices below. O No, the confidence interval does not contradict the claim. The mean of 6.99 m from the article is inside the 90% confidence interval. O No, the confidence interval does not contradict the claim. The mean of 6.99 m from the article is outside the 90% confidence interval. O Yes, the confidence interval contradicts the claim. The mean of 6.99 m from the article is inside the 90% confidence interval. O Yes, the confidence interval contradicts the claim. The mean of 6.99 m from the article is outside the 90% confidence interval