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Question: Suppose we want to estimate the mean salary of all X-university students who graduated in Spring 2016 as business majors. In previous years, the standard deviation for these salaries was \$5000, i.e. $\sigma$ = 5000. How many business majors do we have to sample to estimate the mean salary to within \$1000 with 95% confidence?

Approach: So the data is: $\sigma = 5000,\; \alpha=0.05,\; \overline{x} = 1000$

Straight from the theory, I got that: $\overline{x}\pm Z_{\alpha/2}\left(\frac{\sigma}{\sqrt{n}}\right) = 1000\pm 1.96\left(\frac{5000}{\sqrt{n}}\right) $

Is there a way to find $n$ exactly or is the answers just: "we need to sample more than 30 students"?

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You know that $$\sigma =5000$$ $$ME=1000$$

The formula for finding the margin of error is given by $$\text{ME}=\frac{\sigma * Z}{\sqrt{n}}$$ You can solve algebraically for n now

Thus, $$n=\left(\frac{\sigma * Z}{\text{ME}}\right)^2$$

Now,

$$n=\left(\frac{1.96 * 5000}{1000}\right)^2$$

$$n=96.04$$

It might be safe to round this to 97.

So in conclusion, 97 business majors would have to sampled to estimate the mean salary to within $1000 with 95% confidence

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    $\begingroup$ I was a good idea to post your answer. Nice work. I was and I´m still confused. Therefore I to do some additional statistic work. $\endgroup$ – callculus Dec 2 '16 at 21:03

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