91Ó°ÊÓ

Construct a $95$% confidence interval for the mean number of hours slept for the population (assumed normal) from which you took the data. $8.2$; $9.1$; $7.7$; $8.6$; $6.9$; $11.2$; $10.1$; $9.9$; $8.9$; $9.2$; $7.5$;

If $\stackrel{-}{x}$and s are the mean and standard deviation of a random sample from a normal distribution with unknown variance ${\mathrm{σ}}^2$,100(1-$\mathrm{Î\pm }$) % confidence interval on is given by

$\stackrel{¯}{x}−t_{\frac{\mathrm{Î\pm }}{2},n−1}\frac{s}{\sqrt{n}}≤\mathrm{μ}≤\stackrel{¯}{x}+t_{\frac{\mathrm{Î\pm }}{2},n−1}\frac{s}{\sqrt{n}}$ (1)

$\text{where}{t}_{\frac{\mathrm{Î\pm }}{2},n−1}\text{is the upper}100\frac{\mathrm{Î\pm }}{2}\text{percentage point of the}t\text{distribution with}n−1\text{degrees of freedom.}$

The simple mean is

$\stackrel{¯}{x}=\frac{1}{n}\underset{i=1}{\overset{n}{∑}}{x}_i=\frac{8.2+9.1+⋯+10.5}{12}=8.98$

and standard deviation is

$s={\left(\frac{\underset{i=1}{\overset{n}{∑}}{(x_i−\stackrel{¯}{x})}^2}{n−1}\right)}^{\frac{1}{2}}={\left(\frac{\underset{i=1}{\overset{12}{∑}}{(x_i−8.98)}^2}{11}\right)}^{\frac{1}{2}}=1.29$

$\text{Now, we need to find a}95\mathrm{\%}\mathrm{Cl}\text{on the population mean, then}$

$\frac{\mathrm{Î\pm }}{2}=\frac{1−0.95}{2}=0.025⇒t_{\frac{\mathrm{Î\pm }}{2},n−1}=t_{0.025,11}=2.201.$ (2)

We used a probability table for the Student's t-distribution to find the value of t. The table gives t-scores that correspond to degrees of freedom (row) and the confidence level (column). The t-score is found where the row and column intersect in the table.

From Equation (1) and (2) we get

$8.98−2.2\frac{1.29}{\sqrt{12}}≤\mathrm{μ}≤8.98+2.2\frac{1.29}{\sqrt{12}}$

Therefore,a $95$% cl on the population mean is

$8.16≤\mathrm{μ}≤9.80$