91Ó°ÊÓ

A random sample of $28$pizza delivery restaurants is taken and has a sample mean delivery time of $36$minutes.

Find a $90$% confidence interval estimate for the population mean delivery time.

If $\stackrel{-}{x}$is the sample mean of a random sample of size n from a normal population with unknown variance $\mathrm{σ}$^$2$ , a $100$($1$-$\mathrm{Î\pm }$) % cl on $\mathrm{μ}$is given by

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

where Z_{$\frac{\mathrm{Î\pm }}{2}$} is the upper $100$$\frac{\mathrm{Î\pm }}{2}$ percentage point of the standard normal distribution. We know that standard deviation is $\mathrm{σ}$=$6$minutes, a random sample n=$28$and a sample mean $\stackrel{-}{x}$=$36$minutes.

We need find a $90$% confidence interval estimate for the population mean delivery time. Therefore,

$\frac{\mathrm{Î\pm }}{2}=\frac{1−0.90}{2}=0.05⇒z_{\frac{a}{2}}=z_{0.05}=1.645.$ (2)

The previous implication was obtained on a probability table for the standard normal distribution.

From (1) and (2) we get

$36−z_{0.05}\frac{6}{\sqrt{28}}≤\mathrm{μ}≤36+z_{0.05}\frac{6}{\sqrt{28}}$

$36−1.645×1.13≤\mathrm{μ}≤36+1.645×1.13$

Therefore, $90\%Cl$for $\mathrm{μ}$is

$34.13≤\mathrm{μ}≤37.86$

We estimate with $90\%$confidence that the true population mean delivery time is between $34.13and37.86.3$

Ninety percent of all confidence intervals constructed in this way contain the true mean statistics exam score. For example, if we constructed $100$ of these confidence intervals, we would expect $90$of them to contain the true population mean.