Question 1182978
I would use a t-test since the sd of the sample is used as a variability estimator. It is true that n is > 30, so the difference between z and t is < 1%. Furthermore, given the ease of calculators today, dealing with a t-table is essentially the same as dealing with a z-table. 

The critical z-value (0.975) is 1.96; the t-value (0.975, df=168) is 1.97. My answer would be a, but some would say use z. The difference is small.


---
the half-interval for the interval is t*s/sqrt(n)
=1.97*7.3/13