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

`ME = z_(\alpha/2) (\sigma)/(√(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 (0.73)/(√(60))= 0.185`

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) (\sigma)/(√(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 (0.73)/(√(60))= 0.185`