A study of 141 in-home caregivers revealed the average number of hours
worked per week to be 106.9 with a sample standard deviation of 68.2
hours.  Form a confidence interval at the 90% level for the population
mean number of caregiver hours.

Naturally, you should use the t distribution to solve this question.
However as the sample size is 141, there is very little difference if
you were to use the z distribution, which I am going to use here.
A 90% confidence interval for the population mean is 
= 106.9 +/- z(0.05) x s /[sq root (141)]
= 106.9 +/- 1.645   x 68.2 /[sq root (141)]
= (97.45, 116.35)