Question

A researcher works on a study that has a population standard deviation of 0.73a sample mean of 84.5 and a sample size of 60 .What is the margin of error with a confidence level of 95%

Answer 1

Answer:

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

The critical value can be founded in the normal standard distribution table using the value of $\alpha/2 =0.025$ and we got $z_{\alpha/2}=1.96$. Replacing the info given we got:

$ME = 1.96 \frac{0.73}{\sqrt{60}}= 0.185$

Step-by-step explanation:

For this case we have the following info given:

$\sigma = 0.73$ represent the population deviation

$\bar X = 84.5$ represent the sample mean

$n =60$ represent the sample size

And we want to find the margin of error for a confidence level of 95%. So then the significance level would be $\alpha=1-0.95 = 0.05$ and $\alpha/2 =0.025$. The margin of error is given by:

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

The critical value can be founded in the normal standard distribution table using the value of $\alpha/2 =0.025$ and we got $z_{\alpha/2}=1.96$. Replacing the info given we got:

$ME = 1.96 \frac{0.73}{\sqrt{60}}= 0.185$