Answer:
99% confidence interval of the true mean weight of the minivans
(3954.192 ,4345.808)
Step-by-step explanation:
step(i):-
Given sample size 'n' = 40
mean of the sample x⁻ = 4150 pounds
Standard deviation of the Population 'σ' = 480 pounds
Critical value:-
[tex]Z_{\frac{\alpha }{2} } = Z_{\frac{0.01}{2} } = Z_{0.005} = 2.58[/tex]
step(ii):-
99% confidence interval of the true mean weight of the minivans
[tex](x^{-} - Z_{\frac{\alpha }{2} } \frac{S.D}{\sqrt{n} } , x^{-} - Z_{\frac{\alpha }{2} } \frac{S.D}{\sqrt{n} })[/tex]
[tex](4150 - 2.58 \frac{480}{\sqrt{40} } , 4150 - 2.58 \frac{480}{\sqrt{40} })[/tex]
(4150 - 195.808 , 4150 + 195.808)
(3954.192 ,4345.808)
Conclusion:-
99% confidence interval of the true mean weight of the minivans
(3954.192 ,4345.808)
