Question

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.

1. Develop a 95% confidence interval estimate for the mean number of weekly

customer contacts for the population of sales personnel.

2. Assume that the population of weekly contact data has a normal distribution.

Use the t distribution to develop a 95% confidence interval for the mean number

of weekly customer contacts.

3. Compare your answer for parts (a) and (b). What do you conclude from your

results?

Answer 1

Solution:

Given, $n=61,\mu=22.4$

Standard deviation is not given(or missing), so we assume any value, say, $\sigma=5.2$ and proceed with this value.

(1):

Z value for 95% confidence is Z=1.96

Then, confidence interval, CI$=\mu\pm1.96\dfrac{\sigma}{\sqrt n}$

$=22.4\pm1.96(\dfrac{5.2}{\sqrt {61}})=22.4\pm1.30495 \\=(21.09505,23.70495)$

(2):

Degrees of freedom, df=61-1=60 and $\alpha=(1-0.95)/2=0.025$

Then, from its table, $t_{\alpha/2}=2.000298$ for 95% CI.

Then, CI$=\mu\pm2.000298\dfrac{\sigma}{\sqrt n}$

$=22.4\pm2.000298(\dfrac{5.2}{\sqrt {61}})=22.4\pm1.33178 \\=(21.06822,23.73178)$

(3):

Conclusion: Both CIs are very close to each other. But CI in part (1) is closer to given mean, i.e., 22.4.