Question

Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 85 weekly reports showed a sample mean of 18.5 customer contacts per week. The sample standard deviation was 5.6. 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

Given,

Sample size = n = 85

sample mean = μ_S = 18.5

population mean = μ_P = μ_S = 18.5

sample standard deviation = σ_S = 5.6

Population standard deviation σ_P = σ_S/ √n = 5.6/√ 85 = 0.607

For 90% confidence interval, z = 1.28. Let the weekly sales be x for 90% confidence interval

z = (x - μ) / σ_P

=> 1.28 = (x - 18.5)/0.607

=> x = 1.28*0.607 + 18.5 = 19.28

For 95% confidence interval, z = 1.64. Let the weekly sales be x for 95% confidence interval

z = (x - μ) / σ_P

=> 1.64 = (x - 18.5)/0.607

=> x = 1.64*0.607 + 18.5 = 19.50