Question

We have IQ test scores of 31 seventh-grade girls in a Midwest school district. We have calculated that sample mean is 105.84 and the standard deviation is 14.27.
Give a 99% confidence interval for the average score in the population. What is the margin of error?

Answer 1

Answer:

The 99% confidence interval for the average score in the population is between 99.24 and 112.44.

The margin of error is 6.60.

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:

$\alpha = \frac{1-0.99}{2} = 0.005$

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 $1-0.005 = 0.995$, so $z = 2.575$

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.

$M = 2.575*\frac{14.27}{\sqrt{31}} = 6.60$

The lower end of the interval is the mean subtracted by M. So it is 105.84 - 6.60 = 99.24

The upper end of the interval is the mean added to M. So it is 105.84 + 6.60 = 112.44

The 99% confidence interval for the average score in the population is between 99.24 and 112.44.