91Ó°ÊÓ

$\begin{array}{c}\text{Pointestimatefor}\mathrm{μ},\\ \mathrm{μ}=\frac{\mathrm{Ex}}{\mathrm{n}}\\ =\frac{150+12+367+251+38+63+12+8+11+13+134+107+150+12}{12}\\ =97.1667\end{array}$

$\begin{array}{l}\mathrm{CI}=\stackrel{¯}{\mathrm{x}}Â\pm {\mathrm{t}}_{2/2},\mathrm{df}\frac{\mathrm{S}}{\sqrt{\mathrm{n}}}\\ \text{Fromsample,}\stackrel{¯}{\mathrm{x}}=\frac{\mathrm{Ex}}{\mathrm{n}}=97.16\end{array}$

$\mathrm{S}=\sqrt{\frac{\mathrm{E}{({\mathrm{x}}_{\mathrm{i}}−\stackrel{¯}{\mathrm{x}})}^2}{\mathrm{n}−1}}=113.7595$

Confidence Level = 99% = 0.99

$\begin{array}{c}\text{³¢=1–C}\\ \text{=1}−\text{0.99}\\ \text{=0.01}\end{array}$

${\mathrm{t}}_{2/2},\mathrm{df}=\frac{{\mathrm{t}}_{0.01}}{2},12−1=3.1058$

Now, the Confidence interval for 99% is:

$\begin{array}{c}\mathrm{CI}=\stackrel{¯}{\mathrm{x}}Â\pm {\mathrm{t}}_{2/2},\mathrm{df}\frac{\mathrm{S}}{\sqrt{\mathrm{n}}}\\ =97.16Â\pm 3.1058×\frac{113.75}{\sqrt{12}}\\ =(−4.8245,199.1445)\\ \mathrm{CI}=(−4.8245,199.1445)\end{array}$

Every sample's data value must be independent of one another. The measured values in the two samples ought to be independent of one another. The size of the sample should be at least 30 or almost typical in form. The two distinct samples must be randomly selected as well as a representative of the general population.

Yes, the conditions for creating a confidence interval are fulfilled. The 90% confidence interval is$(−4.8245,199.1445)$ .