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?

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

[tex]X \sim N(\mu, \sigma)[/tex]

The distribution for the sample mean is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

[tex]\bar x =512[/tex] represent the sample mean

[tex]\sigma=100 [/tex] represent the population standard deviation

n= 100 sample size selected.

The confidence interval is given by this formula:

[tex]\bar X \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}[/tex] (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:

[tex]Me=z_{\alpha/2} \frac{\sigma}{\sqrt{n}}[/tex]

[tex]25.76=z_{\alpha/2} \frac{100}{\sqrt{100}}[/tex]

And we can find the critical value [tex]z_{\alpha/2}[/tex] like this:

[tex]z_{\alpha/2}=\frac{25.76(\sqrt{100})}{100}=2.576[/tex]

And we know that on the right tail of the z score =2.576 we have [tex]\alpha/2[/tex] 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=[tex]\alpha/2[/tex], so then [tex]\alpha=0.00498*2=0.009995 \approx 0.01[/tex], and then we can find the confidence like this:

Confidence =[tex]1-\alpha=1-0.01=0.99[/tex]

And then the confidence level would be given by 99%