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

Answer:

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

Step-by-step explanation:

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

$\alpha = \frac{1 - 0.95}{2} = 0.025$

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

$M = z\frac{\sigma}{\sqrt{n}}$

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.

$M = z\frac{\sigma}{\sqrt{n}} = 1.96\frac{2.8}{\sqrt{760}} = 0.199$

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