95% = mean ± 1.96 SEm
SEm = SD/√n
Variance = SD^2
99% = mean ± 2.575 SEm
Can someone guide me on how to solve this problem: A sample of 52 night students ages is obtained in order to estimate the mean age of night school students. sample mean = 25.9 years. the population variance is 23. Give a point estimate for the mean. Find the 95% confidence interval for the mean lower and upper and the 99% confidence interval
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