Question

g You have drawn a simple random sample of 68 college students, asking each student their grade point average (GPA). Each individual draw is normally distributed with a standard deviation of 0.2. You obtain a sample mean grade point average of 2.89. Calculate the upper confidence limit of a 92% confidence interval for the population mean of grade point average. If necessary, round your answer to four decimal places. HINT: For a confidence interval of [a,b] we refer to "a" as the "lower confidence limit" and we refer to "b" as the "upper confidence limit".

Answer 1

Answer:

The upper confidence limit of a 92% confidence interval for the population mean of grade point average is 2.93.

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:

$\alpha = \frac{1-0.92}{2} = 0.04$

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 $1-0.04 = 0.96$, so $z = 175$

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.

$M = 1.75*\frac{0.2}{\sqrt{68}} = 0.04$

The upper end of the interval is the sample mean added to M. So it is 2.89 + 0.04 = 2.93.

The upper confidence limit of a 92% confidence interval for the population mean of grade point average is 2.93.