Question

Suppose you are asked to take a sample of men and women in order to measure their emotional IQ. The emotional IQ scale that you administer has a minimum of 0 and a maximum of 100, with higher scores indicating higher emotional intelligence (the range of the scale is not pertinent to the question -- I simply provide it so that you have some context in which to provide your answer). You sample 25 men and 25 women. The mean for men is 70 with a standard deviation of 10 and the mean for women is 82 with a standard deviation of 15.
Required:
What is the 99% confidence interval for the scores of men and women?

Answer 1

Answer:

The 99% confidence interval for the scores of men is (64.406,75.594).

The 99% confidence interval for the scores of women is (73.609,90.391).

Step-by-step explanation:

We have the standard deviations for the sample, which means that the t-distribution will be used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 25 - 1 = 24

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 24 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$. So we have T = 2.797.

For men:

Standard deviation of 10.

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.797\frac{10}{\sqrt{25}} = 5.594$

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

The lower end of the interval is the sample mean subtracted by M. So it is 70 - 5.594 = 64.406

The upper end of the interval is the sample mean added to M. So it is 70 + 5.594 = 75.594

The 99% confidence interval for the scores of men is (64.406,75.594).

For women:

Standard deviation of 15.

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.797\frac{15}{\sqrt{25}} = 8.391$

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

The lower end of the interval is the sample mean subtracted by M. So it is 82 - 8.391 = 73.609

The upper end of the interval is the sample mean added to M. So it is 82 + 8.391 = 90.391

The 99% confidence interval for the scores of women is (73.609,90.391).