The confidence interval: 5.06 < sigma2 < 23.33 is for the population variance based on the following sample statistics:
n = 25, x-bar = 41.2, and s = 3.1

What is the degree of confidence? Use only integers, no % sign and no decimal places.
(Points: 3)

The confidence interval for a population variance is calculated as:

(n - 1)s2 &#8804; &#963;2 &#8804; (n - 1)s2
X2&#945;/2, n - 1 X21- &#945;/2, n - 1

The X2&#945;/2, n - 1 and the X21- &#945;/2, n - 1 are critical values from the chi-squared distribution. We'll find them in a minute.

From the problem, we know that the left side of the inequality is equal to 5.06 and the right side is equal to 23.33. So, we have two equations:

(n - 1)s2 = 5.06
X2&#945;/2, n - 1

(25 - 1)(3.1)2 = 5.06
X2&#945;/2, 24

230.64 = 5.06
X2&#945;/2, 24

X2&#945;/2, 24 = 45.581

(n - 1)s2 = 23.33
X21- &#945;/2, n - 1

(25 - 1)(3.1)2 = 23.33
X21- &#945;/2, 24

230.64 = 23.33
X21- &#945;/2, 24

X21- &#945;/2, 24 = 9.886

If you look at a chi-squared distribution table with 24 degrees of freedom, you will see that a X2 value of 45.581 corresponds to a p-value of 0.005 and a X2 value of 9.886 corresponds to a p-value of 0.995.

That means that &#945;/2 = 0.005 (i.e. &#945; = 0.01), and 1 - &#945;/2 = 0.995 (again, &#945; = 0.01).

Therefore, this is a 99% confidence interval.

14.
Find the margin of error. 95% confidence interval; n = 91 ; x-bar = 55, s = 5.4
Round to the nearest two decimal places.
(Points: 3)

First, calculate the 95% confidence interval (I assume we're calculating it for the mean). It is calculated as:

x-bar - t&#945;/2, n - 1(s/&#8730;n) &#8804; &#956; &#8804; x-bar + t&#945;/2, n - 1(s/&#8730;n)

You use the t-distribution because the population standard deviation is unknown. A 95% confidence interval corresponds to a &#945; of 0.05, and &#945;/2 of 0.025. t0.025, 90 (the critical value of t with a p-value of 0.025 and 90 degrees of freedom) is 1.987.

x-bar - t&#945;/2, n - 1(s/&#8730;n) &#8804; &#956; &#8804; x-bar + t&#945;/2, n - 1(s/&#8730;n)

55 - t0.025, 90(5.4/&#8730;91) &#8804; &#956; &#8804; 55 + t0.025, 90(9.4/&#8730;91)

55 - (1.987)(5.4/&#8730;91) &#8804; &#956; &#8804; 55 + (1.987)(9.4/&#8730;91)

55 - 1.1248 &#8804; &#956; &#8804; 55 + 1.1248

53.8752 &#8804; &#956; &#8804; 56.1248

The margin of error is the distance from the estimation of the mean to the limits of the confidence interval. You can calculate it by finding the width of the confidence interval and dividing by 2 ((56.1248 - 53.8752)/2 = 1.1248), or by looking at the value that is subtracted from and added to the sample mean in the second to last line of the calculations above.

Either way, the margin of error is 1.13.