Respuesta :
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 = [tex]Z_\frac{\alpha}{2} \times \frac{\sigma}{\sqrt{n} }[/tex]
where, [tex]\alpha[/tex] = significance level = 1 - 0.95 = 0.05 and [tex](\frac{\alpha}{2})[/tex] = 0.025.
[tex]\sigma[/tex] = 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 ;
[tex]500=1.96 \times \frac{3,850}{\sqrt{n} }[/tex]
[tex]\sqrt{n} = \frac{1.96 \times 3,850}{500}[/tex]
[tex]\sqrt{n}[/tex] = 15.092
Squaring both sides we get,
n = [tex]15.092^{2}[/tex] = 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.
