Respuesta :
Answer:
a) The 95% confidence interval estimate for the mean weight fo the carry-on luggage
(17.38 , 22.61)
b) The 97% confidence interval estimate for the mean weight of the carry-on luggage
(17.11 , 22.89)
Step-by-step explanation:
Step1:-
a sample of 36 pieces of carry-on luggage was weighed
n=36
The average weight was 20 pounds
x⁻ = 20
given standard deviation of the population to be 8 pounds.
σ = 8
Confidence intervals :-
The values [tex](x^{-} - 1.96 \frac{S.D}{\sqrt{n} } ,x^{-} + 1.96 \frac{S.D}{\sqrt{n} } )[/tex] are called 95% of Confidence intervals for the mean of the population corresponding to the given sample.
The values [tex](x^{-} - 2.17 \frac{S.D}{\sqrt{n} } ,x^{-} + 2.17 \frac{S.D}{\sqrt{n} } )[/tex] are called 97% of Confidence intervals for the mean of the population corresponding to the given sample.
Step:-(1)
The 95% confidence interval estimate for the mean weight of the carry-on luggage
[tex](x^{-} - 1.96 \frac{S.D}{\sqrt{n} } ,x^{-} + 1.96 \frac{S.D}{\sqrt{n} } )[/tex]
[tex](20 - 1.96 \frac{8}{\sqrt{36} } ,20 + 1.96 \frac{8}{\sqrt{36} } )[/tex]
on calculation , we get
(20 -2.613 ,20 +2.613)
(17.3866 , 22.613)
The 95% confidence interval estimate for the mean weight fo the carry-on luggage
(17.3866 , 22.613)
Step2:-
The 97% confidence interval estimate for the mean weight of the carry-on luggage
[tex](x^{-} - 2.17 \frac{S.D}{\sqrt{n} } ,x^{-} + 2.17 \frac{S.D}{\sqrt{n} }[/tex]
[tex](20 - 2.17 \frac{8}{\sqrt{36} } ,20 + 2.17 \frac{8}{\sqrt{36} } )[/tex]
on calculation , we get
(20-2.89 , 20+2.89)
(17.11 , 22.89)
The 97% confidence interval estimate for the mean weight of the carry-on luggage
(17.11 , 22.89)
