Question

Assume that a sample is used to estimate a population mean μ . Find the 99.9% confidence interval for a sample of size 24 with a mean of 32.5 and a standard deviation of 12.9. Enter your answer accurate to one decimal place. I am 99.9% confident that the mean μ is between ____and____

Answer 1

solution:

Given that,

mean = = 32.5

standard deviation = = 12.9

n = 24

= 32.5

= / n = 12.9 /24=2.63

middle 99.9% of score is

P(-z < Z < z) = 0.999

P(Z < z) - P(Z < -z) = 0.999

2 P(Z < z) - 1 = 0.999

2 P(Z < z) = 1 + 0.999 = 1.999

P(Z < z) =1.999 / 2 = 0.9995

P(Z <3.29) = 0.9995

z  ±3.29 using z table

Using z-score formula  

= z * +   

= ±3.29 *2.63+32.5

= 23.85 and 41.15