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JusSomeRandomGuy
24.03.2020 •
Mathematics
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
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Ответ:
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} }](/tpl/images/0561/8880/86958.png)
where,
= significance level = 1 - 0.95 = 0.05 and
= 0.025.
n = sample size
Also, at 0.025 significance level the z table gives critical value of 1.96.
So, margin of error is ;
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.
Ответ:
at answer C, that's supposed to be x= -4 , 1 and that's the answer.