Question

The monthly earnings of a group of business students are normally distributed with a standard deviation of 528 dollars. A researcher wants to estimate the mean monthly earnings of all business students. Find the sample size needed to have a confidence level of​ 95% and a margin of error of 126 dollars.

Answer 1

Answer:--- Date:---30/4/2020 Solution : Given that, o= 528 margin of error = E = 126 At 95% confidence level the z is, alpha = 1 - 95% = 1 - 0.95 = 0.05 alpha / 2 = 0.05 / 2 = 0.025 Zalpha/2 = 20.025 = 1.96 Sample size = n =((Zalpha/2 *0) /E)? =(1.96 * 528)/1262 = (1034.88 / 126) = 67.4588 Sample size = 67.46