Answer:
99% confidence interval for the mean of college students
A) 112.48 < μ < 117.52
Step-by-step explanation:
step(i):-
Given sample size 'n' =150
mean of the sample = 115
Standard deviation of the sample = 10
99% confidence interval for the mean of college students are determined by
[tex](x^{-} -t_{0.01} \frac{S}{\sqrt{n} } , x^{-} + t_{0.01} \frac{S}{\sqrt{n} } )[/tex]
Step(ii):-
Degrees of freedom
ν = n-1 = 150-1 =149
t₁₄₉,₀.₀₁ = 2.8494
99% confidence interval for the mean of college students are determined by
[tex](115 -2.8494 \frac{10}{\sqrt{150} } , 115 + 2.8494\frac{10}{\sqrt{150} } )[/tex]
on calculation , we get
(115 - 2.326 , 115 +2.326 )
(112.67 , 117.326)