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 5.2. 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 that,

=19.5

s =5.2

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    ( using student t table)

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

= 1.663 * (5.2 / 85)

E = 0.94

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

- E < < + E

19.5 - 0.94< <19.5 + 0.94

18.56 < < 20.44

(18.56 ,20.44 )

(B)

Given that,

=19.5

s =5.2

n = 85

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

a ) 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 ( using student t table)

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

= 1.989 * (5.2 / 85)

E=1.12

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

- E < < + E

19.5 -1.12 < < 19.5+1.12

18.38 < < 20.62

( 18.38 , 20.62 )