Respuesta :
Answer:
The 95% confidence interval for the population mean mileage is
(27.917, 29.593)
The 98% confidence interval for the population mean mileage
(27.737, 29.762)
Step-by-step explanation:
Step(i):-
Given that the random sample size 'n' = 64
Given that the mean of sample x⁻ = 28.75miles
Given that the standard deviation of the sample (S) = 3.4 miles
Degrees of freedom = n-1 = 64-1 =63
t₀.₀₅ = 1.9983
Step(ii):-
95% confidence interval for the population mean mileage is determined by
[tex](x^{-} - t_{0.05} \frac{S}{\sqrt{n} } , x^{-} + t_{0.05} \frac{S}{\sqrt{n} })[/tex]
[tex](28.75-1.99\frac{3.4}{\sqrt{64} },28.75 + 1.99 \frac{3.4}{\sqrt{64} } )[/tex]
(28.75 - 0.845 , 28.75 + 0.845)
(27.917 , 29.593)
Step(iii):-
98% confidence interval for the population mean mileage is determined by
[tex](x^{-} - t_{0.02} \frac{S}{\sqrt{n} } , x^{-} + t_{0.02} \frac{S}{\sqrt{n} })[/tex]
[tex](28.75-2.3824\frac{3.4}{\sqrt{64} },28.75 + 2.3824 \frac{3.4}{\sqrt{64} } )[/tex]
(28.75 - 1.01252 , 28.75 + 1.01252)
(27.737 , 29.762)
Final answer:-
98% confidence interval for the population mean mileage is
(27.737 , 29.762)
