Question

In: Statistics and Probability

Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week....

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 17.5 customer contacts per week. The sample standard deviation was 5.3. 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:

(__,___)

Solutions

Expert Solution

Solution :

Given that,

A.

t /2,df = 1.663

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

= 1.663 * (5.3 / 85)

Margin of error = E = 0.96

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

- E < < + E

17.5 - 0.96 < < 17.5 + 0.96

16.54 < < 18.46

90% Confidence interval is (16.54 , 18.46)

B.

t /2,df = 1.989

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

= 1.989 * (5.3 / 85)

Margin of error = E = 1.14

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

- E < < + E

17.5 - 1.14 < < 17.5 + 1.14

16.36 < < 18.61

95% Confidence interval is (16.36 , 18.61)

orchestra answered 3 years ago
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