Answer:
a) 95 % confidence interval for the mean estimate of the average income of all U.S. households based on the information
(58,808 ,71,192)
<u> </u>Step-by-step explanation
<u>Explanation</u>:-
Given data assume the 25 adults in the study can be considered an SRS from the population of all adult citizens of the United States.
Given sample size 'n' =25
The mean estimate x⁻ = $ 65 , 000
sample standard deviation 's' = $ 15 , 000
Degrees of freedom γ= n-1 =25-1=24
The tabulated value t = 2.0639 (check t- table blue mark) at 0.05 level of significance with 24 degrees of freedom.
<u>Step(ii):-</u>
<u>95 % confidence interval for the mean </u>
![(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} +t_{\alpha } \frac{S}{\sqrt{n} } )](https://tex.z-dn.net/?f=%28x%5E%7B-%7D%20-%20t_%7B%5Calpha%20%7D%20%5Cfrac%7BS%7D%7B%5Csqrt%7Bn%7D%20%7D%20%2Cx%5E%7B-%7D%20%2Bt_%7B%5Calpha%20%7D%20%5Cfrac%7BS%7D%7B%5Csqrt%7Bn%7D%20%7D%20%29)
![(65000- 2.0639 \frac{15000}{\sqrt{25} } ,65000 +2.0639 \frac{15000}{\sqrt{25} } )](https://tex.z-dn.net/?f=%2865000-%202.0639%20%5Cfrac%7B15000%7D%7B%5Csqrt%7B25%7D%20%7D%20%2C65000%20%2B2.0639%20%5Cfrac%7B15000%7D%7B%5Csqrt%7B25%7D%20%7D%20%29)
(65,000-6191.7 ,65,000+6191.7)
(58,808 ,71,192)
<u>Conclusion</u>:-
95 % confidence interval for the mean estimate of the average income of all U.S. households based on the information
(58,808 ,71,192)
<u> </u>