Question

5. A simple random sample of size n is drawn from a population that is known to be normally distributed. The sample variance, s, is determined to be 9.1. Construct and interpret a 90% confidence interval for o if the sample size, n, is 14. Show formula and final answer to two decimal places. O (6.94, 13.52) O (2.30, 3.68) O (48.15, 182.71) O (5.29.20.08)

Answer 1

2 = 9.1 n = 14 The critical values for a = 0.1 and df = 13 degrees of freedom are xź xį-a/2n-1 = 5.8919 and xi = xả/2.n-1 = 22.362. The corresponding confidence interval is computed as shown below: CI(Variance) = (n-1)s (n - 1) Xa/2, 0-1 Xi-a/2,1-1 7 = (14 – 1) x 9.1 (14 – 1) ~ 9.1 22.362 5.8919 (5.2902, 20.0785) Now that we have the limits for the confidence interval, the limits for the 90% confidence interval for the population standard deviation are obtained by simply taking the squared root of the limits of the confidence interval for the variance, so then: CI(Standard Deviation) 5.2902, 20.0785) = (2.3, 4.4809) Therefore, based on the data provided, the 90% confidence interval for the population variance is 5.2902 <o? < 20.0785, and the 90% confidence interval for the population standard deviation is 2.3 <o< 4.4809.

Last option is correct.

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