Use the t-distribution to find a confidence interval for a mean given the releva

Question

Use the t-distribution to find a confidence interval for a mean given the relevant sample results. Give the best point estimate for 1, the margin of error, and the confidence interval. Assume the results come from a random sample from a population that is approximately normally distributed A 99% confidence interval for using the sample results x = 87.8, s = 30.3, 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

Explanation / Answer

Solution:- X = 87.8 , s = 30.3 n = 15

=> df = n-1 = 15-1 = 14, t0.005 = 2.977

=> Point estmate : 87.8

=> margin of error : 2.977*(30.3/sqrt(15)) = 23.29

=> 99% confidence interval is : X +/- t*(sd/sqrt(n))

: 87.8 +/- 2.977*(30.3/sqrt(15))

: 87.8 +/- 23.29

: 64.51 ,  111.09 (rounded)

:

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