Question

González Industries requires its sales personnel to keep track of their weekly contacts with customers. A sample of 16 reports showed a mean of 32.4 customer contacts per week for the sales personnel, and a sample standard deviation of 5.7 contacts. Assuming customer contacts is a normally distributed variable, generate a 95% confidence interval estimate of the true mean number of customer contacts per week at González Industries. Begin by stating whether this estimation problem should use the student t distribution or the normal (Z) distribution. Should the t distribution be used?

If you indicated the t distribution should be used, give the value that should be used here. If you said, "no", then indicate which Z value should be used.

What is the lower limit of the confidence interval?

What is the upper limit of the confidence interval?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 32.4

sample standard deviation = s = 5.7

sample size = n = 16

Degrees of freedom = df = n - 1 = 16 - 1 = 15

Using  t distribution

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t0.025,15 = 2.131

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

= 2.131 * ( 5.7/ 16)

Margin of error = E = 3.04

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

  ± E  

= 32.4   ± 3.04

= ( 29.36, 35.44 )

lower limit = 29.36

upper limit = 35.44