Question

The annual camnings of 13 randomly selected computer soware engineers have a sample standard deviation of 53708. Assume the sample is from a normally distruled population Construct a confidence interval for the poor ance and the population standard deviation Use 5 vel of confidence internet the rest What is the confidence interval for the population variance C o und to the nearesteger as needed) are ther e the corect choice below and Round to the nearest Integer as needed.) in the box on your choice OA With 5% confidence, you can say that the population variance is greater than OB With 95% confidence, you can say that the population variance is less than inces between OCWith 95% confidence, you can say that the population and OD with 5% orice, you can say that the poinvances between and What is the confidence interval for the population sandard deviation ? Round to the nearest integer as needed interpret the results Select the correct choice below and fill in the answer boxes) locomplete your choice (Round to the nearest Integer as needed)

Answer 1

Given that, n=13,5 = 3708 Level of significance(a)=1-confidence = 1-0.95 = 0.05 The degree of freedom is (n-1)= 13-1 =12 By using the chi square table, the critical values at 5% level of significance are 7.05/2 = 23.337 , X1-0.05/2 = 4.404 2 (n-1) ) 95%CI=1" (13-1)37082 (13-1)37082 23.337 4 .404 = 7069939<o? <37463935 With 95% confidence you can say that the population variance is between 7069939 and 37463935

Given that, n=13,5 = 3708 Level of significance(a)=1-confidence = 1-0.95 = 0.05 The degree of freedom is (n-1)= 13-1 =12 By using the chi square table, the critical values at 5% level of significance are xsv2 =23.337 , 0.052 = 4.404 (n-1)s (n-1)32 95%CI = < 2 1 Kas y tich (13- 1)37082 (13-1)3708 V 23.337 V 4.404 = 7069939 <o< 37463935 = 2659<o<6121 With 95% confidence you can say that the population standard deviation is between 2659 and 6121