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 19.5 customer contacts per week. The sample standard deviation was 5.5. Provide 90% and 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel.

a. 90% confidence interval, to 2 decimals:

b. 95% confidence interval, to 2 decimals:

Answer 1

Given that,

= 19.5

s =5.5

n = 85

Degrees of freedom = df = n - 1 = 85- 1 = 84

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t _/2,df = t_0.05,84=1.663

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

= 1.663* ( 5.5/ 85) = 0.9921

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

- E < < + E

19.5- 0.9921< < 19.5+ 0.9921

18.5079 < < 20.4921

(18.5079 ,20.4921)

(B)Given that,

= 19.5

s =5.5

n = 85

Degrees of freedom = df = n - 1 = 85- 1 = 84

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,84=1.989

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

= 1.989* ( 5.5/ 85) = 1.1866

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

- E < < + E

19.5- 1.1866< < 19.5+ 1.1866

18.3134< < 20.6866

(18.3134 , 20.6866)