In: Statistics and Probability
Exhibit F
In order to estimate the average time that students spend on the
internet per day, data were collected for a sample of 36 business
students. Assume the population standard deviation is 2 hours.
Refer to Exhibit F. If the sample mean is 6 hours, then the 95% confidence interval for the population mean,μ, is
Group of answer choices
4.35 to 7.65
5.67 to 6.33
5.35 to 6.65
6 to 6.65
Solution :
Given that,
Point estimate = sample mean = =6
Population standard deviation =
= 2
Sample size n =36
At 95% confidence level the z is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
Z/2 = Z0.025 = 1.96 ( Using z table )
Margin of error = E = Z/2
* (
/n)
= 1.96* (2 / 36 )
E= 0.65
At 95% confidence interval estimate of the population mean
is,
- E <
<
+ E
6 - 0.65 <
<6 + 0.65
5.35<
< 6.65
( 5.35 to 6.65)