Question

8.38 Suppose, for a random sample selected from a normally dis-tributed population, x = 68.50 and s = 8.9.

a. Construct a 95% confidence interval for μ assuming n = 16.

b. Construct a 90% confidence interval for μ assuming n = 16. Is the width of the 90% confidence interval smaller than the width of the 95% confidence interval calculated in part a? If yes, explain why.

Answer 1

Solution:-

x = 68.50, s = 8.9

a) 95% confidence interval for μ assuming n = 16 is C.I = ( 64.139, 72.861).

C.I = 68.50 + 1.96 × 2.225

C.I = 68.50 + 4.361

C.I = ( 64.139, 72.861)

b) Yes the width of the 90% confidence interval is smaller than the width of the 95% confidence interval calculated in part a, because confidence coefficient for 90% confidence interval is smaller that of the 95% confidence interval.

90% confidence interval for μ assuming n = 16 is C.I = ( 64.839, 72.160).

C.I = 68.50 + 1.645 × 2.225

C.I = 68.50 + 3.6601

C.I = ( 64.839, 72.160)

Yes the width of the 90% confidence interval is smaller than the width of the 95% confidence interval calculated in part a, because confidence coefficient for 90% confidence interval is smaller that of the 95% confidence interval.