SOLUTION: a survey of 144 retail stores revealed that a particular brand and model of a VCR retails for 375 with a standard deviation of 20. What is the 99% confidence interval to estimate

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Question 296799: a survey of 144 retail stores revealed that a particular brand and model of a VCR retails for 375 with a standard deviation of 20. What is the 99% confidence interval to estimate the true cost of the VCR? and what is the 95% confidence interval?
Answer by jrfrun(3) (Show Source):
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Assuming your data is representative and randomly selected, then since the sample is large (>30) and estimating averages, the CLT allows you to use normal approximation to estimate the population mean
confidence intervals for the population mean via normal approximations are: point estimate +/- Noise
where the point estimate is the sample average and Noise= Z*Standard Error
or basically Confidence interval for mean : Xbar +/-Z*S/Sqrt(n)
The sample of 144 stores indicates that Xbar = 375 and S=20
The Z value corresponding to 99% is the number of standard deviations which capture 99% of the normal distribution in this case +/-2.576
So. 99% confidence interval for the population mean = 375-2.576*20/Sqrt(144) and 375+2.576*20/Sqrt(144)
The 95% confidence interval for the population mean is done the same way but substituting the Z =1.96