91Ó°ÊÓ

Null hypotheses is,

${\mathrm{H}}_0:\mathrm{μ}=4$

Alternative hypotheses is,

${\mathrm{H}}_0:\mathrm{μ}≠4$

The random variable $\stackrel{¯}{\mathrm{X}}$ is the average$\mathrm{I}.\mathrm{Q}.$

$\mathrm{n}-1=11$

Sample mean is,

$\stackrel{¯}{\mathrm{x}}=\frac{5+4+⋯+8+5}{12}$

$=4.92$

Sample standard deviation is,

$\mathrm{s}=\underset{\mathrm{j}=1}{\overset{\mathrm{n}}{∑}}\frac{{\left({\mathrm{x}}_{\mathrm{j}}-\stackrel{¯}{\mathrm{x}}\right)}^2}{\mathrm{n}-1}$

$=\underset{\mathrm{j}=1}{\overset{12}{∑}}\frac{{\left({\mathrm{x}}_{\mathrm{j}}-4.92\right)}^2}{11}$

$=1.62$

Test is,

${\mathrm{t}}_{11}=\frac{4.92-4}{1.62/\sqrt{12}}$

$=1.958$

We do not reject the null hypothesis when the $\mathrm{p}$value is bigger than the established alpha value.

$\mathrm{p}\text{-value}=0.076>0.05$

$=\mathrm{Î\pm }$

Mean and variance of random is,

$\stackrel{¯}{\mathrm{x}}-{\mathrm{t}}_{\frac{\mathrm{Î\pm }}{2},\mathrm{n}-1}\frac{\mathrm{s}}{\sqrt{\mathrm{n}}}≤\mathrm{μ}≤\stackrel{¯}{\mathrm{x}}+{\mathrm{t}}_{\frac{\mathrm{Î\pm }}{2},\mathrm{n}-1}\frac{\mathrm{s}}{\sqrt{\mathrm{n}}}......1$

where ${\mathrm{t}}_{\frac{\mathrm{Î\pm }}{2},\mathrm{n}-1}$is the upper

$100\frac{\mathrm{Î\pm }}{2}$percentage point of the distribution with $\mathrm{n}-1$degrees of freedom.

$\frac{\mathrm{Î\pm }}{2}=0.025$

$⇒{\mathrm{t}}_{\frac{\mathrm{a}}{2},\mathrm{n}-1}={\mathrm{t}}_{0.025,11}$

$=2.2.......2$

From $1$and$2$,

$4.92-2.2\frac{1.62}{\sqrt{12}}≤\mathrm{μ}≤4.92+2.2\frac{1.62}{\sqrt{12}}$

$4.92-2.2×0.468≤\mathrm{μ}≤4.92+2.2×0.468$

Population proportion is,

$3.886≤\mathrm{μ}≤5.947$

Therefore, Mean I.Q is not four $\left(4\right).$