Suppose average pizza delivery times are normally distributed with an unknown population mean and a population standard deviation of six minutes. 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.

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 $\sigma$^$2$ , a $100$($1$-$\alpha$) % cl on $\mu$is given by

$\overline{x}-z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}\le \mu \le \overline{x}+z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}$ (1)

where Z_{$\frac{\alpha }{2}$} is the upper $100$$\frac{\alpha }{2}$ percentage point of the standard normal distribution. We know that standard deviation is $\sigma$=$6$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,

$\frac{\alpha }{2}=\frac{1-0.90}{2}=0.05\Rightarrow z_{\frac{a}{2}}=z_{0.05}=1.645.$ (2)

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

From (1) and (2) we get

$36-z_{0.05}\frac{6}{\sqrt{28}}\le \mu \le 36+z_{0.05}\frac{6}{\sqrt{28}}$

$36-1.645\times 1.13\le \mu \le 36+1.645\times 1.13$

Therefore, $90\%Cl$for $\mu$is

$34.13\le \mu \le 37.86$

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.