Question

Exercise 2.3 Sales personnel for X Company are required to submit weekly reports

listing customer contacts made during the week. A sample of 61 weekly contact

reports showed a mean of 22.4 customer contacts per week for the sales personnel.

Answer 1

a)

The critical value for $\alpha=0.05$ and $df=n-1=61-1=60$ degrees of freedom is $t_c= 2.000298.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{ s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{ s}{\sqrt{n}})$$

$$=(22.4-2.000298\times\dfrac{ 5}{\sqrt{61}}, 22.4+2.000298\times\dfrac{ 5}{\sqrt{61}})$$

$$\approx(21.11944, 23.68056)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $21.11944<\mu<23.68056,$ which indicates that we are 95% confident that the true population mean $\mu$  is contained by the interval $(21.11944, 23.68056).$

b)

The critical value for $\alpha=0.10$ and $df=n-1=61-1=60$ degrees of freedom is $t_c= 1.670649.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{ s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{ s}{\sqrt{n}})$$

$$=(22.4-1.670649\times\dfrac{ 5}{\sqrt{61}}, 22.4+1.670649\times\dfrac{ 5}{\sqrt{61}})$$

$$\approx(21.33048, 23.46952)$$

Therefore, based on the data provided, the 90% confidence interval for the population mean is $21.33048<\mu<23.46952,$ which indicates that we are 90% confident that the true population mean $\mu$  is contained by the interval $(21,33048, 23.46952).$

Level of significance is a statistical term for how willing you are to be wrong. With a 95 percent confidence interval, you have a 5 percent chance of being wrong. With a 90 percent confidence interval, you have a 10 percent chance of being wrong. A 99 percent confidence interval would be wider than a 95 percent confidence interval. A 90 percent confidence interval would be narrower.