Answer:
The 95% confidence interval of the true mean.
(29.4261 ,36.9739)
Step-by-step explanation:
Step :- (i)
Given sample size 'n' =15
sample of the mean x⁻ = 33.2
The standard deviation of the sample 'S' = 8.3
95% of confidence intervals
[tex](x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} + t_{\alpha }\frac{S}{\sqrt{n} } )[/tex]
Step:-(ii)
The degrees of freedom γ=n-1 = 15-1=14
The tabulated value t = 1.761 at 0.05 level of significance.
now substitute all possible values, we get
[tex](33.2 - 1.761\frac{8.3}{\sqrt{15} } ,33.2+ 1.761\frac{8}{\sqrt{15} } )[/tex]
After calculation , we get
(33.2-3.7739 , 33.2+3.7739
(29.4261 ,36.9739)
Conclusion:-
the 95% confidence interval of the true mean.
(29.4261 ,36.9739)
