A factor in determining the usefulness of an examination as a measure of demonst

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 gi examinations, it is known that a standard deviation of around 78 points is desirable. Of course, too large or too small a standard deviation is not good. The atormeys want to test their examination to see how good it is. A preliminary version of the examination (with slight modification 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 50 points Find a 99% confidence interval for the population standard deviation. 36.8S to 80.50 b36.08 to 78.80 057.49 to 12557 56.28 to 12293 58.67 to 128.16

Explanation / Answer

CONFIDENCE INTERVAL FOR STANDARD DEVIATION
ci = (n-1) s^2 / ^2 right < ^2 < (n-1) s^2 / ^2 left
where,
s = standard deviation
^2 right = (1 - confidence level)/2
^2 left = 1 - ^2 right
n = sample size
since alpha =0.01
^2 right = (1 - confidence level)/2 = (1 - 0.99)/2 = 0.01/2 = 0.005
^2 left = 1 - ^2 right = 1 - 0.005 = 0.995
the two critical values ^2 left, ^2 right at 23 df are 44.1813 , 9.26
s.d( s^2 )=50
sample size(n)=24
confidence interval for ^2= [ 23 * 2500/44.1813 < ^2 < 23 * 2500/9.26 ]
= [ 57500/44.1813 < ^2 < 57500/9.2604 ]
[ 1301.4556 < ^2 < 6209.235 ]
and confidence interval for = sqrt(lower) < < sqrt(upper)
= [ sqrt (1301.4556) < < sqrt(6209.235), ]
= [ 36.0757 < < 78.7987 ]
= [ 36.08 to 78.80 ]

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