Answer:
We need to sample at least 37 weeks of data.
Explanation:
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of
.
So it is z with a pvalue of
, so

Now, find the margin of error M as such

We want 98% confidence that the sample mean is within $500 of the population mean, and the population standard deviation is known to be $1300
This is at least n weeks, in which n is found when

So






Rounding up
We need to sample at least 37 weeks of data.