Question

The weights of a certain brand of cereal boxes are normally distributed. The mean weight of a SAMPLE of 13 boxes was 14.91 ounces with a sample standard deviation of .22 ounces. The margin of error on a 90% confidence interval would be smaller than the margin of error on a 95% confidence interval. True or false?

Answer 1

Given that,

= 14.91

s =0.22

n = 13

Degrees of freedom = df = n - 1 = 13- 1 = 12

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t /2,df = t0.05,12 = 1.782    ( using student t table)

Margin of error = E = t/2,df * (s /n)

= 1.782 * (0.22 / 13)

E= 0.1087

Margin of error = E =0.1087

(B)

Given that,

= 14.91

s =0.22

n = 13

Degrees of freedom = df = n - 1 = 13- 1 = 12

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.025

t /2,df = t0.025,12 = 2.179 ( using student t table)

Margin of error = E = t/2,df * (s /n)

= 2.179 * (0.22 / 13)

E= 0.1330

Margin of error = E =0.1330

(C) yes 90% confidence interval would be smaller than the margin of error on a 95% confidence interval.