Question

Suppose that the population of the scores of all high school seniors who took the SAT Math (SAT-M) test this year follows a Normal distribution with standard deviation σ = 100.

You read a report that says, "On the basis of a simple random sample of 100 high school seniors that took the SAT-M test this year, a confidence interval for μ is found to be 512.00 ± 25.76."

What was the confidence level used to calculate this confidence interval?

Answer 1

Confidence =
`1-\alpha=1-0.01=0.99`

And then the confidence level would be given by 99%

Explanation:

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Assuming the X follows a normal distribution

`X \sim N(\mu, \sigma)`

The distribution for the sample mean is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

`\bar x =512` represent the sample mean

`\sigma=100` represent the population standard deviation

n= 100 sample size selected.

The confidence interval is given by this formula:

`\bar X \pm z_(\alpha/2) (\sigma)/(√(n))` (1)

The marginof error for this case is given by Me=25.76. And we know that the formula for the margin of error is given by:

`Me=z_(\alpha/2) (\sigma)/(√(n))`

`25.76=z_(\alpha/2) (100)/(√(100))`

And we can find the critical value
`z_(\alpha/2)` like this:

`z_(\alpha/2)=(25.76(√(100)))/(100)=2.576`

And we know that on the right tail of the z score =2.576 we have
`\alpha/2` of the total area. We can find the area on the right of the z score using this excel code:

"=1-NORM.DIST(2.576,0,1,TRUE)" or using a table of the normal standard distribution, and we got 0.004998=
`\alpha/2`, so then
`\alpha=0.00498*2=0.009995 \approx 0.01`, and then we can find the confidence like this:

Confidence =
`1-\alpha=1-0.01=0.99`

And then the confidence level would be given by 99%