Question

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 1

Answer with explanation:

As per given , we have

sample size : n= 65

degree of freedom : df=n-1=64

sample mean : $\overline{x}=19.5$

sample standard deviation : s= 5.2

Since , the population standard deviation is not given  , so we apply t-test.

Significance level  for 90% confidence : $\alpha=1-0.90=0.10$

t-critical value for significance level 0.10 and df = 64 would be :

$t_c=t_{\alpha/2, df}=t_{0.05,\ 64}=1.6690$

Formula for Confidence interval :

$\overline{x}\pm t_c\dfrac{s}{\sqrt{n}}$

Then , 90%  confidence intervals for the population mean number of weekly customer contacts for the sales personnel would be :

$19.5\pm(1.669)\dfrac{5.2}{\sqrt{65}}$

$=19.5\pm1.076$

$(19.5-1.076,\ 19.5+1.076)=(18.424,\ 20.576)$

∴ 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 : $\alpha=1-0.95=0.05$

t-critical value for significance level 0.05 and df = 64 would be :

$t_c=t_{\alpha/2, df}=t_{0.05,\ 64}=1.9977$

Then , 95%  confidence intervals for the population mean number of weekly customer contacts for the sales personnel would be :

$19.5\pm(1.9977)\dfrac{5.2}{\sqrt{65}}$

$=19.5\pm1.288$

$(19.5-1.288,\ 19.5+1.288)=(18.212,\ 20.788)$

∴ 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel : (18.212, 20.788)