Answer to Question #210973 in Statistics and Probability for mafi

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.

The sample standard deviation was 5 contacts.

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

1. Based on the provided information (n = 61, α = 0,05) the critical z-value is Zc = 1.96

M=22.4

$\sigma = 5$

Two-sided confidence interval:

$CI = (M - \frac{Z_c \times \sigma}{\sqrt{n}}, M + \frac{Z_c \times \sigma}{\sqrt{n}}) \\ CI = (22.4 - \frac{1.96 \times 5}{\sqrt{61}}, 22.4 + \frac{1.96 \times 5}{\sqrt{61}}) \\ CI = (22.4 - 1.25, 22.4 + 1.25) \\ CI = (21.15, 23.65)$

2. n=61, P=0.95

t=2.00 (from table)

$CI = (M - \frac{t \times \sigma}{\sqrt{n}}, M + \frac{t \times \sigma}{\sqrt{n}}) \\ CI = (22.4 - \frac{2.00 \times 5}{\sqrt{61}}, 22.4 + \frac{2.00 \times 5}{\sqrt{61}}) \\ CI = (22.4 - 1.28, 22.4 + 1.28) \\ CI = (21.12, 23.68)$

3. CI from (a) is less than CI from (b).