Answer:
The 95 % confidence interval of the population mean for the weights of adult elephants
(11,629.375 , 11,664.625) up to three decimals
Nearest whole number
(11,629 , 11,665)
Step-by-step explanation:
Explanation:-
A sample of 7 adult elephants had an average weight of 11,647 pounds
size of the sample 'n' =7
The mean of the sample x⁻ = 11,647 pounds
The standard deviation of the sample 'S' = 24 pounds
Step(ii):-
The 95 % confidence interval of the population mean for the weights of adult elephants
[tex](x^{-} - t_{0.95} \frac{S}{\sqrt{n} } , x^{-} + t_{0.95} \frac{S}{\sqrt{n} } )[/tex]
The degrees of freedom γ =n-1 = 7-1=6
t₆ = 1.943
[tex](11,647 - 1.943 \frac{24}{\sqrt{7} } , 11,647 + 1.943 \frac{24}{\sqrt{7} } )[/tex]
(11,647-17.625 , 11,647+17.625)
(11,629.375 , 11,664.625)
Conclusion:-
The 95 % confidence interval of the population mean for the weights of adult elephants
(11,629 , 11,665)
