Question

8. Given the standard deviation of 16 for a population and a 95% confidence level, find the margin

Answer 1

Answer:

4.0486

Step-by-step explanation:

The margin of error is calculated using  $E=z_{\frac{\alpha}{2} }*\frac{\sigma}{\sqrt{n} }$

Where n=60 is the sample size and $\sigma=16$ is the population standard deviation.

The critical value corresponding to 95% confidence interval is $z_{\frac{\alpha}{2}}=1.96$

We substitute the values to obtain;

$E=1.96*\frac{16}{\sqrt{60} }$

$E=1.96*2.06559=4.0486$