The recommended retail price of a brand of designer jeans is known. The price of the jeans in a sample of 16 retailers is on average $141 with a sample standard deviation of 4. If this is a ‘random’ sample and the prices can be assumed to be normally distributed, construct a 95% confidence interval for the average sale price.

The critical value for $\alpha = 0.05$ and $df = n-1 = 15$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} =2.131449.$

The corresponding confidence interval is computed as shown below:

$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$

$=(141-2.131449\times\dfrac{4}{\sqrt{16}}, 141+2.131449\times\dfrac{4}{\sqrt{16}})$

$=(138.87, 143.13)$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $138.87 < \mu < 143.137,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(138.87, 143.13).$