Question

Refer to Exercise $9.119$. Conduct a hypothesis test to see if your decision and conclusion would change if your belief were that the brown trout's mean I.Q is not four.

Short answer

The brown trout's mean I.Q is not four.

Step-by-step solution

Introduction

Null hypotheses is,

${\mathrm{H}}_0:\mathrm{\mu }=4$

Alternative hypotheses is,

${\mathrm{H}}_0:\mathrm{\mu }\ne 4$

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

$\mathrm{n}-1=11$

Sample mean is,

$\overline{\mathrm{x}}=\frac{5+4+\cdots +8+5}{12}$

$=4.92$

Sample standard deviation is,

$\mathrm{s}=\sum _{\mathrm{j}=1}^{\mathrm{n}}\frac{{\left({\mathrm{x}}_{\mathrm{j}}-\overline{\mathrm{x}}\right)}^2}{\mathrm{n}-1}$

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

$=1.62$

Test is,

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

$=1.958$

Explanation

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

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

$=\mathrm{\alpha }$

Mean and variance of random is,

$\overline{\mathrm{x}}-{\mathrm{t}}_{\frac{\mathrm{\alpha }}{2},\mathrm{n}-1}\frac{\mathrm{s}}{\sqrt{\mathrm{n}}}\le \mathrm{\mu }\le \overline{\mathrm{x}}+{\mathrm{t}}_{\frac{\mathrm{\alpha }}{2},\mathrm{n}-1}\frac{\mathrm{s}}{\sqrt{\mathrm{n}}}......1$

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

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

$\frac{\mathrm{\alpha }}{2}=0.025$

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

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

From $1$and$2$,

$4.92-2.2\frac{1.62}{\sqrt{12}}\le \mathrm{\mu }\le 4.92+2.2\frac{1.62}{\sqrt{12}}$

$4.92-2.2\times 0.468\le \mathrm{\mu }\le 4.92+2.2\times 0.468$

Population proportion is,

$3.886\le \mathrm{\mu }\le 5.947$

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