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

$CI=\bar {X}\pm t_{\frac{\alpha}{2},n-1}\frac{s}{\sqrt{n}}$

90%CI

$CI=\bar {X}\pm t_{0.05,n-1}\frac{s}{\sqrt{n}}$

$\bar{X}=71.5$

$s=12$

$n=16$

$t_{0.05,15}=1.753$

$CI=71.5\pm 1.753(\frac{12}{\sqrt{16}})$

=[66.242, 76.759]

95%CI

$CI=\bar {X}\pm t_{0.025,n-1}\frac{s}{\sqrt{n}}$

$t_{0.025,15}=2.131$

$CI=71.5\pm 2.131(\frac{12}{\sqrt{16}})$

$=[65.106, 77.894]$

99%CI

$CI=\bar {X}\pm t_{0.005,n-1}\frac{s}{\sqrt{n}}$

$t_{0.005,15}=2.947$

$CI=71.5\pm 2.947(\frac{12}{\sqrt{16}})$

$=[62.66, 80.34]$