Question

Since 2007, a psychological association has supported an annual nationwide survey to examine stress across the United States. One year, a total of 760 millennials (18- to 33-year-olds) were asked to indicate their average stress level (on a 10-point scale) during a month. The mean score was 5.5. Assume that the population standard deviation is 2.8. Give the margin of error for a 95% confidence interval. (Enter your answer to three decimal places.)

Answer 1

The margin of error for a 95% confidence interval is 0.199.

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1 - 0.95)/(2) = 0.025`

Now, we have to find z in the Ztable as such z has a pvalue of
`1 - \alpha`.

That is z with a pvalue of
`1 - 0.025 = 0.975`, so Z = 1.96.

Now, find the margin of error M as such

`M = z(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

Assume that the population standard deviation is 2.8.

This means that
`\sigma = 2.8`

760 millennials (18- to 33-year-olds)

This means that
`n = 760`

Give the margin of error for a 95% confidence interval.

`M = z(\sigma)/(√(n)) = 1.96(2.8)/(√(760)) = 0.199`

The margin of error for a 95% confidence interval is 0.199.