If n=20, ¯ x (x-bar)=36, and s=15, construct a confidence interval at a 98% confidence level. Assume the data came from a normally distributed population. Give your answers to one decimal place.

Question

If n=20, ¯x�¯(x-bar)=36, and s=15, construct a confidence interval at a 98% confidence level. Assume the data came from a normally distributed population.

Give your answers to one decimal place.

Answer 1

Since you don't have a population standard deviation that is given, this must be a t-interval

The sample is assumed to come from a normal distribution, so even though the sample size is small (n=20),

a t-interval is appropriate.

The formula for a t-interval is CI = x-bar +/- t-crit * s/sqrt(n)

The critical t value can be found for df=19 in a t-table or by using the invT function on a TI-84 calculator. Remember that the area in the left tail at 98% confidence is 0.01 and you'll get the negative critical value using

invT(0.01,19) the critical t at df=19 and 98% confidence is 2.539

so the CI is given as 36 +/- 2.539 * 15/sqrt(20) or 36 +/- 8.51 or (27.5,44.5)

If you have a Ti83/84 calculator you can also solve the problem using STAT--->TESTS---->TInterval

using summary statistics

x-bar = 36

s=15

n=20

conf=0.98

which gives (27.482,44.518) rounded to one decimal (27.5,44.5)