Confidence Intervals

Question

If n=22, (x-bar)=46, and s=8, 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

Find a 98% CI for n = 22, mean =66, s= 8 (the sample standard deviation), normally distributed

Since the population standard deviation is not know, we will use a t-distribution.

Where E = t_c (s/√n).  At the end the CI is (mean – E, mean + E)

Degrees of Freedom = n-1 = 22-1 = 21. Therefore, from a t-Distribution chart, t_c = 2.518

E = 2.518(8/√22) = 4.2947

The Confidence Interval is:

(66 - E, 66 + E )

(60 - 4.2947, 66 + 4.2947)

(61.7053, 70.2947)

(61.7, 70.3)

Answer 2

I see you have submitted 3 questions. It would help if you tell us how you solve them in your class. Do you use a calculator (what kind), a table, or computer software (what type)? Do you have an idea about what formula you should use in each of these questions?

For example, to do this question with a TI-83 calculator (how I would do it), you press STAT and highlight Tests. Then look for the appropriate choice which will have the word "interval" since it is a confidence interval question. Since you have one group, and an s (sample standard deviation), you would use Tinterval. Simply choose that and type in the numbers from the problem (make sure it is on Stats not Data). Leave pooled as No and choose calculate at the very bottom.