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 17.5 customer contacts per week. The sample standard deviation was 4.2.
Required:
Provide 90%90% and 95%95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel.

Answer 1

Answer:

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$, and the critical value would be $t_{\alpha/2}=1.669$

And replacing we got

$17.5-1.669\frac{4.2}{\sqrt{65}}=16.63$    

$17.5+1.669\frac{4.2}{\sqrt{65}}=18.37$    

For the 95% confidence the critical value is $t_{\alpha/2}=1.998$

$17.5-1.998\frac{4.2}{\sqrt{65}}=16.46$    

$17.5+1.998\frac{4.2}{\sqrt{65}}=18.54$    

Step-by-step explanation:

Information given

$\bar X¿ 17.5$ represent the sample mean

$\mu$ population mean (variable of interest)

s¿4.2 represent the sample standard deviation

n¿65 represent the sample size  

Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

The degrees of freedom are given by:

$df=n-1=65-1=64$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$, and the critical value would be $t_{\alpha/2}=1.669$

And replacing we got

$17.5-1.669\frac{4.2}{\sqrt{65}}=16.63$    

$17.5+1.669\frac{4.2}{\sqrt{65}}=18.37$    

For the 95% confidence the critical value is $t_{\alpha/2}=1.998$

$17.5-1.998\frac{4.2}{\sqrt{65}}=16.46$    

$17.5+1.998\frac{4.2}{\sqrt{65}}=18.54$