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 5.4. Provide 90% and 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel.

90% confidence interval, to 2 decimals:

( ,  )

95% confidence interval, to 2 decimals:

( , )

Answer 1

Solution :

Given that,

a.

sample size = n = 65

Degrees of freedom = df = n - 1 = 65 - 1 = 64

t _/2,df = 1.669

Margin of error = E = t_/2,df * (s /n)

= 1.669 * (5.4 / 65)

= 1.12

The 90% confidence interval estimate of the population mean is,

- E < < + E

17.5 - 1.12 < < 17.5 + 1.12

16.38 < < 18.62

The 90% confidence interval: (16.38 , 18.62

b.

sample size = n = 65

Degrees of freedom = df = n - 1 = 65 - 1 = 64

t _/2,df = 1.998

Margin of error = E = t_/2,df * (s /n)

= 1.998 * (5.4 / 65)

= 1.34

The 95% confidence interval estimate of the population mean is,

- E < < + E

17.5 - 1.34 < < 17.5 + 1.34

16.16 < < 18.84

The 95% confidence interval: (16.16 , 18.84)