Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 65 weekly reports showed a sample mean of 19.5 customer contacts per week. The sample standard deviation was 5.2. Provide 90% and 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel.

Answer with explanation:As per given , we havesample size : n= 65degree of freedom : df=n-1=64sample mean : [tex]\overline{x}=19.5[/tex]sample standard deviation : s= 5.2Since , the population standard deviation is not given  , so we apply t-test.Significance level  for 90% confidence : [tex]\alpha=1-0.90=0.10[/tex]t-critical value for significance level 0.10 and df = 64 would be :[tex]t_c=t_{\alpha/2, df}=t_{0.05,\ 64}=1.6690[/tex]Formula for Confidence interval : [tex]\overline{x}\pm t_c\dfrac{s}{\sqrt{n}}[/tex]Then , 90%  confidence intervals for the population mean number of weekly customer contacts for the sales personnel would be : [tex]19.5\pm(1.669)\dfrac{5.2}{\sqrt{65}}[/tex] [tex]=19.5\pm1.076[/tex][tex](19.5-1.076,\ 19.5+1.076)=(18.424,\ 20.576)[/tex]∴ 90% confidence intervals for the population mean number of weekly customer contacts for the sales personnel : (18.424, 20.576)Significance level  for 95% confidence : [tex]\alpha=1-0.95=0.05[/tex]t-critical value for significance level 0.05 and df = 64 would be :[tex]t_c=t_{\alpha/2, df}=t_{0.05,\ 64}=1.9977[/tex]Then , 95%  confidence intervals for the population mean number of weekly customer contacts for the sales personnel would be : [tex]19.5\pm(1.9977)\dfrac{5.2}{\sqrt{65}}[/tex] [tex]=19.5\pm1.288[/tex][tex](19.5-1.288,\ 19.5+1.288)=(18.212,\ 20.788)[/tex]∴ 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel : (18.212, 20.788)