Fortynine items are randomly selected from a population of 5

Forty-nine items are randomly selected from a population of 500 items. The sample mean is 40 and the sample standard deviation 9.

Develop a 99% confidence interval for the population mean. (Round your answers to 3 decimal places.)

36.511 is the not answer. I have tried the answers everybody had put. They are incorrect

Solution

Sample mean = 40
Standard deviation = 9
Standard error of mean = s / ? n
Standard error of mean = 9 / ? 49
SE = 9/7
Standard error of mean 1.2857
The 99% critical t-value is 2.682

Confidence interval 40-(1.2857)(2.682)
and 40+(1.2857)(2.682)
(36.5517, 43.4483)

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Let n = sample size

n= [ z sd /error]^2

z=2.575 (from normal probability table)
standard deviation of proportion = sqrt[ (0.45)(1-0.45) ] sqrt[(0.45)(0.55)] =0.49749
error =0.10

n = [ (2.575)(0.49749) / 0.1]^2=164.1
sample size = 162