Question

You are planning a survey of starting salaries for recent computer science majors. In the latest survey by the National Association of Colleges and Employers, the average starting salary was reported to be $61,238. If you assume that the standard deviation is $3850, what sample size do you need to have a margin of error equal to $500 with 95% confidence

Answer 1

Answer:

A sample size of at least 228 must be needed.

Step-by-step explanation:

We are given that in the latest survey by the National Association of Colleges and Employers, the average starting salary was reported to be $61,238. Assume that the standard deviation is $3850.

And we have to find that what sample size do we need to have a margin of error equal to $500 with 95% confidence.

As we know that the Margin of error formula is given by;

 Margin of error = $Z_\frac{\alpha}{2} \times \frac{\sigma}{\sqrt{n} }$

where, $\alpha$ = significance level = 1 - 0.95 = 0.05 and  $(\frac{\alpha}{2})$ = 0.025.

            $\sigma$ = standard deviation = $3,850

            n = sample size

Also, at 0.025 significance level the z table gives critical value of 1.96.

So, margin of error is ;

                        $500=1.96 \times \frac{3,850}{\sqrt{n} }$

                         $\sqrt{n} = \frac{1.96 \times 3,850}{500}$

                          $\sqrt{n}$ = 15.092

 Squaring both sides we get,

                            n = $15.092^{2}$ = 227.8 ≈ 228

So, we must need at least a sample size of 228 to have a margin of error equal to $500 with 95% confidence.