Answer :
Answer:
90% confidence intervals for the mean
( 4111.05 , 4288.95)
Step-by-step explanation:
Step(i):-
Given sample size 'n' = 12
Mean of the sample 'x⁻' = 4200
Standard deviation of the sample 's' = 140
Step(ii):-
90% confidence intervals for the mean is determined by
[tex](x^{-} - t_{\frac{\alpha }{2} } \frac{S}{\sqrt{n} } , x^{-} + t_{\frac{\alpha }{2} } \frac{S}{\sqrt{n} })[/tex]
[tex](x^{-} - t_{\frac{0.10}{2} } \frac{S}{\sqrt{n} } , x^{-} + t_{\frac{0.10}{2} } \frac{S}{\sqrt{n} })[/tex]
[tex](x^{-} - t_{0.05 } \frac{S}{\sqrt{n} } , x^{-} + t_{0.05} \frac{S}{\sqrt{n} })[/tex]
The degrees of freedom
ν = n-1 = 12-1 =11
t₀.₀₅ = 2.201
Step(iii):-
[tex](4200 - 2.201\frac{140}{\sqrt{12} } ,(4200 -+2.201\frac{140}{\sqrt{12} })[/tex]
( 4200 - 88.95 , 4200+88.95)
( 4111.05 , 4288.95)
Conclusion:-
90% confidence intervals for the mean
( 4111.05 , 4288.95)