Question

A group of physical education majors was discussing the heights of female runners and whether female runners tended to be tall, on the average. They decided to estimate the mean height of female runners. A sample of 12 runners showed a sample mean height of 65.80 inches and a sample standard deviation of 1.95 inches. Assume the population is approximately normal. Find a 90% confidence interval for the mean µ of the population of female runners from which the 12 runners were selected.

Answer 1

Answer:

The 90% confidence interval for the mean µ of the population of female runners.

( 65.0328 , 66.5672)

Step-by-step explanation:

Step(i)

Given  A sample of 12 runners showed a sample mean height of 65.80 inches and a sample standard deviation of 1.95 inches.

Given sample size is n = 12 <30 so small sample

Given sample mean (x⁻) = 65.80 inches

sample standard deviation (S) = 1.95 inches.

Step(ii)

Assume the population is approximately normal.

The 90% confidence interval for the mean µ of the population of female runners.

$(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} +t_{\alpha } \frac{S}{\sqrt{n} } )$

substitute  all above interval

$(65.80 - t_{\alpha } \frac{1.95}{\sqrt{12} } ,65.80 +t_{\alpha } \frac{1.95}{\sqrt{12} } )$

The degrees of freedom γ=n-1 = 12-1=11

From t- table = 1.363 at 90 % 0r 0.10 level of significance

$(65.80 -1.363 \frac{1.95}{\sqrt{12} } ,65.80 +1.363 \frac{1.95}{\sqrt{12} } )$

on calculation , we get

(65.80 -0.7672 ,65.80 + 0.7672)

( 65.0328 , 66.5672)