Answer:
The 99% confidence interval is (3.0493, 3.4907).
We are 99% sure that the true mean of the students Perry score is in the above interval.
Explanation:
Our sample size is 21.
The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So
.
Then, we need to subtract one by the confidence level
and divide by 2. So:

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 20 and 0.005 in the two-sided t-distribution table, we have

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

Now, we multiply T and s

Then
The lower end of the interval is the mean subtracted by M. So:

The upper end of the interval is the mean added to M. So:

The 99% confidence interval is (3.0493, 3.4907).
Interpretation:
We are 99% sure that the true mean of the students Perry score is in the above interval.