In: Statistics and Probability
A website advertises job openings on its website, but job seekers have to pay to access the list of job openings. The website recently completed a survey to estimate the number of days it takes to find a new job using its service. It took the last 30 customers an average of 60 days to find a job. Assume the population standard deviation is 10 days. Construct a 90% confidence interval of the population mean number of days it takes to find a job.
[56.9966, 63.0034]
[56.4216, 63.5784]
[40.4000, 79.6000]
[55.9085, 64.0915]
Solution :
Given that,
Point estimate = sample mean =
= 60
Population standard deviation =
= 10
Sample size = n =30
At 90% confidence level the z is ,
= 1 - 90% = 1 - 0.90 = 0.1
/ 2 = 0.1 / 2 = 0.05
Z/2 = Z0.05 = 1.645 ( Using z table )
Margin of error = E = Z/2
* (
/n)
=1.645 * ( 10/
30)
= 3.0033
At 90% confidence interval estimate of the population mean
is,
- E <
<
+ E
60 - 3.0033 <
< 60+3.0033
56.996 <
< 63.0034
(56.996 ,63.0034)