Faculty at a local university complain that achieving tenure

Faculty at a local university complain that achieving tenure at their university...

Faculty at a local university complains that achieving tenure at their university takes much longer than at comparable universities. They randomly sample nine tenured faculties at their university and find that the average time of service prior to tenure was 13.55 years with a standard deviation of 1.98 years. Construct a 95% confidence interval on the mean number of years of service to achieve tenure at this university. (12.028.15.072) (12.057. 15.043) (12.256. 14.844) (8.984. 18.116)

Solution

CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=13.55
Standard deviation( sd )=1.98
Sample Size(n)=9
Confidence Interval = [ 13.55 ± t a/2 ( 1.98/ Sqrt ( 9) ) ]
= [ 13.55 - 2.306 * (0.66) , 13.55 + 2.306 * (0.66) ]
= [ 12.028,15.072 ]

[ANSWER]
A. [ 12.028,15.072 ]