In: Statistics and Probability
Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 67 weekly reports showed a sample mean of 19.5 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
Solution :
(a)
t
/2,df = 1.668
Margin of error = E = t/2,df
* (s /
n)
= 1.668 * (5.2 /
67)
Margin of error = E = 1.06
The 90% confidence interval estimate of the population mean is,
- E <
<
+ E
19.5 - 1.06 <
< 19.5 + 1.06
18.44 <
< 20.56
(18.44 , 20.56)
(b)
t
/2,df = 1.997
Margin of error = E = t/2,df
* (s /
n)
= 1.997 * (5.2 /
67)
Margin of error = E = 1.27
The 95% confidence interval estimate of the population mean is,
- E <
<
+ E
19.5 - 1.27 <
< 19.5 + 1.27
18.23 <
< 20.77
(18.23 , 20.77)