Question

The weekly salaries of sociologists in the United States are normally distributed and have a known population standard deviation of 425 dollars and an unknown population mean. A random sample of 22 sociologists is taken and gives a sample mean of 1520 dollars.
Find the margin of error for the confidence interval for the population mean with a 98% confidence level.
z0.10 z0.05 z0.025 z0.01 z0.005
1.282 1.645 1.960 2.326 2.576
You may use a calculator or the common z values above. Round the final answer to two decimal places.

Answer 1

Answer:

$ME =z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

The confidence level is 0.98 and the significance is $\alpha=1-0.98 =0.02$ and $\alpha/2 =0.01$ and the critical value using the table is:

$z_{\alpha/2}= 2.326$

And replacing we got:

$ME=2.326 \frac{425}{\sqrt{22}}= 210.760\ approx 210.76$

Step-by-step explanation:

For this case we have the following info given:

$\sigma = 425$ represent the population deviation

$n =22$ the sample size

$\bar X =1520$ represent the sample mean

We want to find the margin of error for the confidence interval for the population mean and we know that is given by:

$ME =z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

The confidence level is 0.98 and the significance is $\alpha=1-0.98 =0.02$ and $\alpha/2 =0.01$ and the critical value using the table is:

$z_{\alpha/2}= 2.326$

And replacing we got:

$ME=2.326 \frac{425}{\sqrt{22}}= 210.760\ approx 210.76$