Here is the question to answer:

Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 20 of its subscribers. It found that the annual health insurance premi-ums for a family with coverage through an employer averaged $10,979. The standard deviation of the sample was $1,000. a. Based on this sample information, develop a 90% confidence interval for the popula-tion mean yearly premium. b. How large a sample is needed to find the population mean within $250 at 99% confidence?

Here is what I have from getting help. However, i still don't understand how to calculate t and look it up. I'm not finding this on the standard table. Can you help me understand how to use it?

mean + - t(a/2) * stdev / n

19 df 0.05

1.73 t

10979 + - 1.73 * 1000/20

386.8397601

10979 + - 387

10592 11366

Confidence interval from $10,592 to $11,366

B.

0.005

250 = 2.576 * 1000/ n

250 = 2576 /n

62500 = 6635776 /n

106.172416 = n

106 is number needed for 250 with 99% confidence