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.
[tex](x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} +t_{\alpha } \frac{S}{\sqrt{n} } )[/tex]
substitute all above interval
[tex](65.80 - t_{\alpha } \frac{1.95}{\sqrt{12} } ,65.80 +t_{\alpha } \frac{1.95}{\sqrt{12} } )[/tex]
The degrees of freedom γ=n-1 = 12-1=11
From t- table = 1.363 at 90 % 0r 0.10 level of significance
[tex](65.80 -1.363 \frac{1.95}{\sqrt{12} } ,65.80 +1.363 \frac{1.95}{\sqrt{12} } )[/tex]
on calculation , we get
(65.80 -0.7672 ,65.80 + 0.7672)
( 65.0328 , 66.5672)
