Question

You are planning a survey of starting salaries for recent computer science majors. In a recent survey by the National Association of Colleges and Employers, the average starting salary was reported to be $61,467. Grade: 5 4 3 2 1 Communication strong ←− − − − − − −→ weak ×1 If the standard deviation is known to be $12,000, what sample size do you need to have a margin of error equal to $1000, with 95% confidence

Answer 1

Answer:

A sample size of 554 is needed.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$.

That is z with a pvalue of $1 - 0.025 = 0.975$, so Z = 1.96.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

Standard deviation is known to be $12,000

This means that $\sigma = 12000$

What sample size do you need to have a margin of error equal to $1000, with 95% confidence?

This is n for which M = 1000. So

$M = z\frac{\sigma}{\sqrt{n}}$

$1000 = 1.96\frac{12000}{\sqrt{n}}$

$1000\sqrt{n} = 1.96*12$

Dividing both sides by 1000:

$\sqrt{n} = 1.96*12$

$(\sqrt{n})^2 = (1.96*12)^2$

$n = 553.2$

Rounding up:

A sample size of 554 is needed.