Respuesta :
Answer:
yes it is little different from the confidence interval (30.4 ≤μ≤ 32.8) changes statistics
90% confidence interval estimate of the mean is
(30.1048 , 33.0952)
Step-by-step explanation:
Step(I):-
Given sample size 'n' = 153
Given mean of the sample x⁻ = 31.5
Sample standard deviation 'S' = 7.1 h g
90% confidence interval estimate of 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]
Degrees of freedom
ν = n-1 = 153-1 =152
t₀.₀₅ =1.9757
Step(ii)
90% confidence interval estimate of the mean is
[tex](x^{-} - t_{\frac{\alpha }{2} } \frac{S}{\sqrt{n} } , x^{-} + t_{\frac{\alpha }{2} } \frac{S}{\sqrt{n} } )[/tex]
[tex](31.5 - 1.9757 } \frac{7.1}{\sqrt{153} } , 31.5 + 1.9757 \frac{7.1}{\sqrt{153} } )[/tex]
( 31.5 - 1.1340 , 31.5 + 1.1340)
(30.366 , 32.634)
90% confidence interval estimate of the mean is
(30.4 , 32.6)
b)
Given sample size 'n' = 15
Given mean of the sample x⁻ = 31.6
Sample standard deviation 'S' = 2.7 h g
90% confidence interval estimate of 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]
Degrees of freedom
ν = n-1 = 15-1 =14
t₀.₀₅ =2.1448
90% confidence interval estimate of the mean is
[tex](x^{-} - t_{\frac{\alpha }{2} } \frac{S}{\sqrt{n} } , x^{-} + t_{\frac{\alpha }{2} } \frac{S}{\sqrt{n} } )[/tex]
[tex](31.6 - 2.1448 } \frac{2.7}{\sqrt{15} } , 31.6 + 2.1448 \frac{2.7}{\sqrt{15} } )[/tex]
( 31.6 - 1.4952 , 31.6 + 1.4952)
(30.1048 , 33.0952)
Conclusion:-
yes it is little different from the confidence interval (30.4 ≤μ≤32.8)
