In: Statistics and Probability
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:
( , )
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 = t0.05,84 = 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 = t0.025,84 = 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 )