Question

At an academically challenging high school, the average GPA of a high school senior is known to be normally distributed with a variance of 0.25. A sample of 20 seniors is taken and their average GPA is found to be 2.71. Develop a 90% confidence interval for the population mean GPA.

Answer 1

The 90% confidence interval for the population mean GPA is between 2.53 and 2.89

Explanation:

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

`\alpha = (1-0.9)/(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(square root of the variance) of the population and n is the size of the sample.

`M = 1.645*(√(0.25))/(√(20)) = 0.18`

The lower end of the interval is the sample mean subtracted by M. So it is 2.71 - 0.18 = 2.53

The upper end of the interval is the sample mean added to M. So it is 2.71 + 0.18 = 2.89.

The 90% confidence interval for the population mean GPA is between 2.53 and 2.89