Question

Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 75 weekly reports showed a sample mean of 17.5 customer contacts per week. The sample standard deviation was 5.1. Provide 90% and 95 percent confidence intervals for the population mean number of weekly customer contacts for the sales personnel.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 17.5

sample standard deviation = s = 5.1

sample size = n = 75

Degrees of freedom = df = n - 1 = 75 - 1 = 74

a) At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z2 = Z0.05 = 1.645

t/2,df = 1.666

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

= 1.666* ( 5.1/ 75)

Margin of error = E = 0.98

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

  ± E

17.5  ± 0.98

(16.52 , 18.48)

b) At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

  t/2,df = 1.993

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

= 1.993* ( 5.1/ 75)

Margin of error = E = 0.17

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

  ± E

17.5  ± 0.17

(17.33 , 17.67)