An important problem in industry is shipment damage. A pottery producing company ships its product by truck and determines that it cannot meet its profit expectations if, on average, the number of damaged items per truckload is greater than 11. A random sample of 15 departing truckloads is selected at the delivery point and the average number of damaged items per truckload is calculated to be 11.3 with a calculated sample of variance of 0.64. Select a 90% confidence interval for the true mean of damaged items.a) [10.68, 11.92]b) [10.64, 11.36]c) [53.67, -33.86]d) [-0.3635, 0.3635]e) [10.94, 11.66]

Answer:e) (10.94, 11.66)Step-by-step explanation:We have a small sample size of n = 15 departing truck loads (we suppose that the sample comes from a normal distribution), the average number of damaged items per truck load is calculated to be [tex]\bar{x} = 11.3[/tex] with a calculated sample variance [tex]s^2[/tex] = 0.64. The 90% confidence interval for the true mean of damaged items is given by [tex]\bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}})[/tex] where [tex]t_{\alpha/2}[/tex] is the [tex]\alpha/2[/tex]th quantile of the t distribution with n - 1 = 14 degrees of freedom.  As we want the 90% confidence interval, we have that [tex]\alpha = 0.1[/tex] and the confidence interval is [tex]11.3\pm t_{0.05}(\frac{0.8}{\sqrt{15}})[/tex] where [tex]t_{0.05} = -1.7613[/tex]. Then, we have [tex]11.3\pm (-1.7613)(\frac{0.8}{\sqrt{15}})[/tex] and the 90% confidence interval is given by (10.94, 11.66)