For a large sample size we would use the Z score (2.326) corresponding to a two-tail probability of 98%. This would give us an interval corresponding to 98% of all annual incomes of real estate agents (realtors). It would tell us where we would expect 98% of the annual incomes to be in the whole population of these agents’ annual incomes.
But the sample size is small and the number of degrees of freedom (dof) is 16-1=15, so we use the t-distribution for this dof to find out for this sample what the spread would be expected to be. The t value for 98% (2-tail) is 2.602. The extreme limits for this confidence interval is given by (X-52000)/7500=2.602, from which X-52000=19515, giving us the extremes $32485 and $71515, the interval [$32485,$71515], $52000±$19515, or $32,485<X<$71,515, where X represents the annual income.
(This compares to the same interval in the whole population of such annual incomes of $34,555<X<$69,445.)