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?

The confidence interval and margin of error for the given confidence level are;

CI = (99.24, 112.44)

E = 6.60

We are given;

Sample size; n = 31

sample mean; x' = 105.84

standard deviation; σ = 14.27

Now, formula for confidence interval is;

CI = x' ± z(σ/√n)

where z is the critical value at the given confidence level.

z at CL of 99% is 2.576

Thus;

CI = 105.84 ± 2.576(14.27/√31)

CI = 105.84 ± 2.576(2.56297)

CI = 105.84 ± 6.60

CI = (105.84 - 6.60), (105.84 + 6.60)

CI = (99.24, 112.44)

Formula for margin of error is;

E = z(σ/√n)

From what we have above;

E = 6.60

Read more about Confidence intervals at; link

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 level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find the margin of error M as such

In which is the standard deviation of the population and n is the size of the sample.

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.

alright well the answers to your question is

2xy/9 hope this have a nice day : )

step-by-step explanation: