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;
<u>Margin of error</u> =
where, = significance level = 1 - 0.95 = 0.05 and
= 0.025.
= standard deviation = $3,850
n = sample size
<em>Also, at 0.025 significance level the z table gives critical value of 1.96.</em>
So, margin of error is ;
= 15.092
Squaring both sides we get,
n = = 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.