Question

Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 68 weekly reports showed a sample mean of 17.5 customer contacts per week. The sample standard deviation was 5.2.

Provide a 90% confidence interval for the population mean number of weekly customer contacts for the sales personnel. (Round your answers to two decimal places.)

  contacts/week to  contacts/week

Provide 95% confidence interval for the population mean number of weekly customer contacts for the sales personnel. (Round your answers to two decimal places.)

  contacts/week to   contacts/week

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 17.5

sample standard deviation = s = 5.2

sample size = n = 68

Degrees of freedom = df = n - 1 = 68-1 = 67

At 90% confidence level

= 1-0.90% =1-0.9 =0.10

/2 =0.10/ 2= 0.05

t/2,df = t_{0.05,67 = 1.67}

t _/2,df = 1.67

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

= 1.67 * (5.2 / 68)

Margin of error = E = 1.052

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

- E < <  + E

17.5-1.052 < < 17.5+ 1.052

16.45 < < 18.55

(16.45 , 18.55)

Answer =16.45 and 18.55

At 95% confidence level

= 1-0.95% =1-0.95 =0.05

/2 =0.05/ 2= 0.025

t/2,df = t_{0.025,67 = 2.00}

t _/2,df = 2.00

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

= 2.00 * (5.2 / 68)

Margin of error = E = 1.259

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

- E < <  + E

17.5-1.259 < < 17.5+ 1.259

16.24 < < 18.76

(16.24, 18.76)

Answer = 16.24 and 18.76