Explanation:
Mean of the test of programming ability was 160 in the past.
Again now, twenty-five job applicants are randomly selected from one university and they produce a mean score and standard deviation of 183 and 12.
So in this case, the null hypothesis will be,

and alternate hypothesis will be,

Mean of the sample =
= 183
Standard deviation of the sample =
= 12
sample size = n = 25
Using t distribution,



P-value =

As the obtained value is less than 0.05 or 5% level of significance, so we have to reject the null hypothesis.