Question

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Chapter 6, Section 2-CI, Exercise 086 Use the t-distribution to find a confidence interval for a mean y given the relevant sample results. Give the best point estimate for y, the margin of error, and the confidence interval. Assume the results come from random sample from a population that is approximately normally distributed. A 99% confidence interval for u using the sample results I = 85.9, s = 30.0, and n = 15 Round your answer for the point estimate to one decimal place, and your answers for the margin of error and the confidence interval to two decimal places. point estimate = margin of error = The 99% confidence interval is to Click if you would like to Show Work for this question: Open Show Work

Answer 1

Point estimate=$\bar{X}$=85.9

Now , df=degrees of freedom=n-1=15-1=14

The critical value is , ; From t-table

tdf,a/2 = t14,0.01 /2 = 2.977

The margin of error is ,

s 30.0 E = tdf,a/27 - 2.977 * = 23.06 n /15

The 99% confidence interval is ,

$\bar{X}\pm E$

$85.9\pm 23.06$