Question

A sample of 16 ten-year-old girls had a mean weight of 71.5 and a standard deviation of 

12 pounds, respectively. Assuming normality, find the 90, 95, and 99 percent confidence 

intervals forthe mean of the population.

Answer 1

$M=71.5 \\ \sigma=12 \\ n=16$

Two-sided confidence interval:

$CI = (M - \frac{Z_c \times \sigma}{\sqrt{n}}, M + \frac{Z_c \times \sigma}{\sqrt{n}})$

For 90 % confidence interval

$Z_c=1.645 \\ CI = (71.5 - \frac{1.645 \times 12}{\sqrt{16}}, 71.5 + \frac{1.645 \times 12}{\sqrt{16}}) \\ = (71.5 -4.935, 71.5+4.935) \\ = (66.565, 76.435)$

For 95 % confidence interval

$Z_c= 1.96 \\ CI = (71.5 - \frac{1.96 \times 12}{\sqrt{16}}, 71.5 + \frac{1.96 \times 12}{\sqrt{16}}) \\ = (71.5 -5.88, 71.5+5.88) \\ = (65.62, 77.38)$

For 99 % confidence interval

$Z_c= 2.576 \\ CI = (71.5 - \frac{2.576 \times 12}{\sqrt{16}}, 71.5 + \frac{2.576 \times 12}{\sqrt{16}}) \\ = (71.5 -7.728, 71.5+7.728) \\ = (63.772, 79.228)$