Question

A factor in determining the usefulness of an examination as a measure of demonstrated ability is the amount of spread that occurs in the grades. If the spread or variation of examination scores is very small, it usually means that the examination was either too hard or too easy. However, if the variance of scores is moderately large, then there is a definite difference in scores between "better," "average," and "poorer" students. A group of attorneys in a Midwest state has been given the task of making up this year's bar examination for the state. The examination has 500 total possible points, and from the history of past examinations, it is known that a standard deviation of around 60 points is desirable. Of course, too large or too small a standard deviation is not good. The attorneys want to test their examination to see how good it is. A preliminary version of the examination (with slight modifications to protect the integrity of the real examination) is given to a random sample of 24 newly graduated law students. Their scores give a sample standard deviation of 63points. Using a 0.01 level of significance, test the claim that the population standard deviation for the new examination is 60 against the claim that the population standard deviation is different from 60.

(a) Find a 99% confidence interval for the population variance. (Round your answers to two decimal places.)

lower limit
upper limit

(b) Find a 99% confidence interval for the population standard deviation. (Round your answers to two decimal places.)

lower limit	points
upper limit	points

Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of x was approximately σ² = 5.1. Suppose a recent study of age at first marriage for a random sample of 31 women in rural Quebec gave a sample variance s² = 2.9. Use a 5% level of significance to test the claim that the current variance is less than 5.1. Find a 90% confidence interval for the population variance.

(a) Find the requested confidence interval for the population variance. (Round your answers to two decimal places.)

lower limit
upper limit

Answer 1

(a)

Confidence interval of variance is,

where s is the sample standard deviation,

n is the sample size

is chi-square statistic at the significance level and degree of freedom n-1

99% confidence interval is equivalent to 0.01 significance level.

Degree of freedom = 24 - 1 = 23

99% Confidence interval of variance is,

Lower limit = 2066.192

Upper limit = 9857.755

(b)

99% Confidence interval of population standard deviation is,

Lower limit = 45.4554

Upper limit = 99.2862

As, the 99% confidence interval limit contains the value of 60, we fail to reject the null hypothesis and there is no signifcant evidence that the population standard deviation is different from 60.

(a)

90% confidence interval is equivalent to 0.1 significance level.

Degree of freedom = 31 - 1 = 30

99% Confidence interval of variance is,

Lower limit = 1.988

Upper limit = 4.705

As, the 95% confidence interval limit is under the variance of 5.2, there is signifcant evidence that current variance is less than 5.1.