Question

A study of stress on the campus of your university reported a mean stress level of 78 (on a 0 to 100 scale with a higher score indicating more stress) with a margin of error of 5 for 95% confidence. The study was based on a random sample of 64 undergraduates. If you wanted 90% confidence for the same study, what would be the confidence interval

Answer 1

The 90% confidence interval would be between 73.80 and 82.20.

Explanation:

95% confidence interval:

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`.

So it is z with a pvalue of
`1-0.025 = 0.975`, so
`z = 1.96`

Now, find the margin of errorM as such

`M = z*(\sigma)/(√(n))`

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

We have that M = 5, n = 64. We have to find the standard deviation of the population. So

`M = z*(\sigma)/(√(n))`

`5 = 1.96*(\sigma)/(√(64))`

`1.96\sigma = 5*8`

`\sigma = (5*8)/(1.96)`

`\sigma = 20.41`

90% confidence interval.

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

`\alpha = (1-0.9)/(2) = 0.05`

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

So it is z with a pvalue of
`1-0.05 = 0.95`, so
`z = 1.645`

Now, find 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.

`M = 1.645*(20.41)/(√(64)) = 4.20`

The lower end of the interval is the sample mean subtracted by M. So it is 78 - 4.20 = 73.80.

The upper end of the interval is the sample mean added to M. So it is 78 + 4.20 = 82.20

The 90% confidence interval would be between 73.80 and 82.20.