A union of restaurant and foodservice workers would like to estimate the mean hourly wage, μ, 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 $0.35 of μ? Suppose that the standard deviation of wages of foodservice workers in the U.S. is about $2.25.
Carry your intermediate computations to at least three decimal places.

Assuming this complete 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 95% confident that its estimate is within $0.35 of ? Suppose that the standard deviation of wages of foodservice workers in the U.S. is about $2.15 .

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X[/tex] represent the sample mean for the sample

[tex]\mu[/tex] population mean (variable of interest)

[tex]\sigma=2.25[/tex] represent the population standard deviation

n represent the sample size

Solution to the problem

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] The margin of error is given by this formula:

[tex] ME=z_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (a)

And on this case we have that ME =0.35 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got [tex]z_{\alpha/2}=1.960[/tex], replacing into formula (b) we got:

[tex]n=(\frac{1.960(2.25)}{0.35})^2 =158.76 \approx 159[/tex]

So the answer for this case would be n=159 rounded up to the nearest integer