As per given , we have
sample size : n= 65
degree of freedom : df=n-1=64
sample mean : [tex]\overline{x}=19.5[/tex]
sample standard deviation : s= 5.2
Since , 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)
