Question

A student wants to estimate the average annual starting salary of recent graduates with a bachelor's degree in statistics. He wants his estimate to be within $500 from the true population mean salary with 95% confidence level. Which of the following is the most appropriate sample size to achieve this goal? Use $2000 as the estimate for population standard deviation.

Answer 1

We will calculate the width of the confidence interval by 2 methods and then equate them to find the sample size.

According to the question, the student wants his estimate to be within 500 from the true population mean . So

Width = 2*500 = 1000 .......(i)

Now, if we calculate the 95% confidence interval around the true population mean then

$\small Width = 2*1.96\frac{\sigma}{\sqrt{n}}$

From the question we have the value of $\small \sigma$ which is 2000 and n is the sample size which we want to calculate

2000 = 2 * 1.96 * n

$\small =\frac{7840}{\sqrt{n}}$ .......(ii)

From equation (i) and (ii)

7840 1000 vn

$\small \sqrt{n}=\frac{7840}{1000}$

$\small \sqrt{n}=7.84$

$\small n=61.4656$

We can take the sample size of 62 or any value greater than this.

Note :

1. 95% confidence interval of $\small \mu$ is :

$\small (\mu-{\color{Blue} Z_\frac{\alpha }{2}*\frac{\sigma}{\sqrt{n}}} , \mu+{\color{Blue} Z_\frac{\alpha }{2}*\frac{\sigma}{\sqrt{n}}} )$

Here the blue coloured term represent the margin of error and twice the margin of error is width which we used.

2. Here 95% confidence level is used so by  $\small \alpha/2$ we used z value at 2.5% (5/2).