A website advertises job openings on its website but job see

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. Calculate a 95% confidence interval of the population mean number of days it takes to find a job.

Solution

Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=60
Standard deviation( sd )=10
Sample Size(n)=30
Confidence Interval = [ 60 ± Z a/2 ( 10/ Sqrt ( 30) ) ]
= [ 60 - 1.96 * (1.8257) , 60 + 1.96 * (1.8257) ]
= [ 56.4215,63.5785 ]