T-distribution and normal distribution

Question

I couldn't differentiate between these 2 questions, why are we using t-distribution for question 1? The format looks the same as question 2?

1. The high price of medicines is a source of major expense for those seniors in the United States who have to pay for these medicines themselves. A random sample of 40 seniors who pay for their medicines showed that they spent an average of $4600 last year on medicines with a standard deviation of $800. Make a 98% confidence interval for the corresponding population mean. (t-distribution)

2. According to Moebs Services Inc., an individual checking account at major U.S. banks costs the banks between $350 and $450 per year (Time, November 21, 2011). A recent random sample of 600 such checking accounts produced a mean annual cost of $500 to major U.S. banks. Assume that the standard deviation of annual costs to major U.S. banks of all such checking accounts is $40. Make a 99% confidence interval for the current mean annual cost to major U.S. banks of all such checking accounts. (normal distribution)

Answer 1

You're right, the t distribution and the normal distribution are very similar. The key on when to use them depends on if you know the population parameters or not.

On question 1, you're told that the average and standard deviation of a sample of 40 seniors and we're asked to come up with a confidence interval for the population mean. In other words, we're being asked to draw conclusions about all seniors based on a smaller sample. This is when you use a t-test, when the parameters of the larger population are unknown.

Question 2 says that there is a sample of 600 checking accounts but that the standard deviation of ALL US banks is $40. We're told that the population standard deviation is $40 so in this case the population parameter is known so we would use a Z-test.

t-test is used when population mean/std dev are unknown

z-test is used when population mean/std dev are known (or given)

Hope this helps