A union of restaurant and foodservice workers would like to estimate the mean ho
ID: 3179156 • Letter: A
Question
A union of restaurant and foodservice workers would like to estimate the mean hourly wage, , of foodservice workers in the U.S. The union will choose a random sample of wages and then estimate using the mean of the sample. What is the minimum sample size needed in order for the union to be confident that its estimate is within of ? Suppose that the standard deviation of wages of foodservice workers in the U.S. is about . Carry your intermediate computations to at least three decimal places. Write your answer as a whole number (and make sure that it is the minimum whole number that satisfies the requirements).
Explanation / Answer
The question is incomplete as the data points are missing , hence i am going to rewrite the question as
A union of restaurant and foodservice workers would like to estimate the mean hourly wage, , of foodservice workers in the U.S. The union will choose a random sample of wages and then estimate using the mean of the sample. What is the minimum sample size needed in order for the union to be 95% confident that its estimate is within $0.40 of ? Suppose that the standard deviation of wages of foodservice workers in the U.S. is about $2.00
we know that the the sample size is given as
n = [zs/E]^2, where z is the z muliplier which would be 1.96 for 95% CI , from the Z tables, S is standard deviation , E is the margin of error
n = [1.96*2.00/0.4]^2 = 96.04 = approx 96
A union of restaurant and foodservice workers would like to estimate the mean hourly wage, , of foodservice workers in the U.S. The union will choose a random sample of wages and then estimate using the mean of the sample. What is the minimum sample size needed in order for the union to be 95% confident that its estimate is within $0.40 of ? Suppose that the standard deviation of wages of foodservice workers in the U.S. is about $2.00
we know that the the sample size is given as
n = [zs/E]^2, where z is the z muliplier which would be 1.96 for 95% CI , from the Z tables, S is standard deviation , E is the margin of error
n = [1.96*2.00/0.4]^2 = 96.04 = approx 96
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