So I am working on a specific stats question. There is a part a and b to it.
A random sample of professional wrestlers was obtained, and the annual salary(in dollars) for each was recorded. The summary statistics were x = 47, 500 and n = 18. Assume the distribution of annual salary is normal with σ = 8, 500.
a. Find a 90% confidence interval for the true mean annual salary for all professional wrestlers.
b. How large a sample is necessary in order for the bound on the error of estimation to be 3000?
Now with Part a) I used the Confidence Formula to find the 90% Interval. It turned to be (44204.29314, 50795.70686).
But I am stuck on Part b) When they "Error of Estimation to be 3000 now does that mean that I subtract 3000 from the original x (mean) of 47,500. Or It it telling me to do something else? Can someone please hep me figure out this part.
Thank you.
As you know, a 90% confidence interval is of the form
$$\bar X \pm 1.645\frac{\sigma}{\sqrt{n}}.$$
The margin of error is $M = 1.645\frac{\sigma}{\sqrt{n}}.$ For (b) set $M = 3000$ and solve for $n.$