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 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

Solutions

Expert Solution

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 = t0.05,67 = 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 = t0.025,67 = 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


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