The mean number of sick days an employee takes per year is believed to be about ten. Members of a personnel department do not believe this figure. They randomly survey eight employees. The number of sick days they took for the past year are as follows: $12;4;15;3;11;8;6;8$. Let $x=$ the number of sick days they took for the past year. Should the personnel team believe that the mean number is ten?

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 $H_0:\mu =10$

Alternative hypotheses is$H_0:\mu \ne 10$

Sample mean is,

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

Sample standard deviation is,

$$\begin{gathered} \sigma =\sum _{\mathrm{j}=1}^8\frac{{\left({\mathrm{x}}_{\mathrm{j}}-\overline{\mathrm{x}}\right)}^2}{\mathrm{n}-1} \\ \sigma =\sum _{\mathrm{j}=1}^8\frac{{\left({\mathrm{x}}_{\mathrm{j}}-8.376\right)}^2}{7} \\ \sigma =4.1 \end{gathered}$$

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 $\alpha =5\%=0.05$

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

localid="1650384170896" $$\begin{gathered} pvalue=0.3>0.05 \\ =\alpha \end{gathered}$$

Do not reject null hypotheses.

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

$\overline{x}-t_{\frac{a}{2},n-1}\frac{s}{\sqrt{n}}\le \mu \le \overline{x}+t_{\frac{\alpha }{2},n-1}\frac{s}{\sqrt{n}}......1$

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

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

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

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

From equation 1and 2

$$\begin{gathered} =8.375-2.36\frac{4.1}{\sqrt{8}}\le \mu \le 8.375+2.36\frac{4.1}{\sqrt{8}} \\ =8.375-2.36\times 1.45\le \mathrm{\mu }\le 8.375+2.36\times 1.45 \end{gathered}$$

So, population mean is,

$4.944\le \mathrm{\mu }\le 11.806$