Answer:
B) 92.03 < μ < 97.97
99% confidence interval for the mean score of all students.
92.03 < μ < 97.97
Step-by-step explanation:
Step(i):-
Given sample mean (x⁻) = 95
standard deviation of the sample (s) = 6.6
Random sample size 'n' = 30
99% confidence interval for the mean score of all students.
[tex]((x^{-} - Z_{0.01} \frac{S}{\sqrt{n} } , (x^{-} + Z_{0.01} \frac{S}{\sqrt{n} })[/tex]
step(ii):-
Degrees of freedom
ν = n-1 = 30-1 =29
[tex]t_{0.01} = 2.462[/tex]
99% confidence interval for the mean score of all students.
[tex]((95 - 2.462 \frac{6.6}{\sqrt{30} } , 95 + 2.462\frac{6.6}{\sqrt{30} } )[/tex]
( 95 - 2.966 , 95 + 2.966)
(92.03 , 97.97)
Final answer:-
99% confidence interval for the mean score of all students.
92.03 < μ < 97.97
