91Ó°ÊÓ

The null hypothesis of a statistical hypothesis test always predicts that there is no impact or association between variables.

Let us first draw the graph representing the p- values:

Null hypotheses is $H_0:\mathrm{μ}=10$

Alternative hypotheses is$H_0:\mathrm{μ}≠10$

Sample mean is,

$$\begin{gathered} \stackrel{¯}{\mathrm{x}}=\frac{12+4+⋯+6+8}{8} \\ \overline{x}=8.376 \end{gathered}$$

Sample standard deviation is,

$$\begin{gathered} \mathrm{σ}=\underset{\mathrm{j}=1}{\overset{8}{∑}}\frac{{\left({\mathrm{x}}_{\mathrm{j}}-\stackrel{¯}{\mathrm{x}}\right)}^2}{\mathrm{n}-1} \\ \mathrm{σ}=\underset{\mathrm{j}=1}{\overset{8}{∑}}\frac{{\left({\mathrm{x}}_{\mathrm{j}}-8.376\right)}^2}{7} \\ \mathrm{σ}=4.1 \end{gathered}$$

A Student's t-distribution with $n-1=7$degrees of freedom will be used for the test.

The test static of test is ,

$$\begin{gathered} {\mathrm{t}}_7=\frac{8.376-10}{4.1/\sqrt{8}} \\ {t}_7=-1.12 \end{gathered}$$

Level of confidence is $\mathrm{Î\pm }=5\%=0.05$

We do not reject the null hypothesis when the p-value is bigger than the established alpha value.

localid="1650384170896" $$\begin{gathered} pvalue=0.3>0.05 \\ =\mathrm{Î\pm } \end{gathered}$$

Do not reject null hypotheses.

A random sample from a normal distribution with unknown variance has a mean and standard deviation of,

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

where, $t_{\frac{\mathrm{Î\pm }}{2},n-1}$is the upper limit.

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

$\frac{\mathrm{Î\pm }}{2}=0.025$

$$\begin{gathered} ⇒{\mathrm{t}}_{\frac{\mathrm{Î\pm }}{2},\mathrm{n}-1}={\mathrm{t}}_{0.025,7} \\ =2.36.....2 \end{gathered}$$

From equation 1and 2

$$\begin{gathered} =8.375-2.36\frac{4.1}{\sqrt{8}}≤\mathrm{μ}≤8.375+2.36\frac{4.1}{\sqrt{8}} \\ =8.375-2.36×1.45≤\mathrm{μ}≤8.375+2.36×1.45 \end{gathered}$$

So, population mean is,

$4.944≤\mathrm{μ}≤11.806$