Question

In: Statistics and Probability

A factor in determining the usefulness of an examination as a measure of demonstrated ability is...

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 29 newly graduated law students. Their scores give a sample standard deviation of 65 points. 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.

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

lower limit
upper limit    


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

lower limit points
upper limit     points

Solutions

Expert Solution

From the given data:s= 65, Therefore s2 = 4225, n = 29, and degrees of freedom = n - 1 = 28. = 0.01

We find critical values for /2 = 0.005 and 1- /2 = 1 – 0.995 = 0.975, df = 28

The lower critical value and the upper critical critical value

The confidence interval is given by:

Therefore:

Lower Limit = 2319.93

Upper Limit = 9493.62

___________________________________

(b) For the population standard deviation, we take the square root of the limits of the variance.

Therefore:

Lower Limit = Sqrt(2319.93) = 48.17

Upper Limit = Sqrt(9493.62) = 97.44

_____________________________________


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