factor in determining the usefulness of an examination as a measure of demonstrated ability is the amount of spread that occurs in the grades.

factor in determining the usefulness of an examination as a measure of demonstrated ability is the amount of spread that occurs in the grades.

asked 2021-01-31

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 20 newly graduated law students. Their scores give a sample standard deviation of 70 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.
(a) What is the level of significance?
State the null and alternate hypotheses.

\(H_{1}:\sigma\ <\ 60H_{0}:\sigma\ >\ 60\)


\(H_{1}:\sigma\ >\ 60H_{0}:\sigma=60\)

\(H_{1}:\sigma\ \neq\ 60\)
(b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)
What are the degrees of freedom?
What assumptions are you making about the original distribution?
We assume a binomial population distribution.We assume a exponential population distribution. We assume a normal population distribution.We assume a uniform population distribution.

Answers (1)

sample standard deviation \(= 70\)
confidence interval \(= 99\%\)
sample size \(= 20\)
degree of freedom \(= 19\)
a)For \(99\%\) confidence interval,
The upper critical value has right tail area \(=\frac{1\ -\ 0.99}{2}=0.005\) and for 19 degree of freedom the critical value is 38.582
The lower critical value has right tail area \(=\frac{1\ +\ 0.99}{2}=0.995\) and for 19 degree of freedom the critical value is 6.844
The confidence interval for variance is calculated as shown below
Lower limit \(=\frac{(n\ -\ 1)s^{2}}{X_{0.005}^{2}}=\frac{(20\ -\ 1)70^{2}}{38.582}=2413.04235\)
Upper limit \(=\frac{(n\ -\ 1)s^{2}}{X_{0.095}^{2}}=\frac{(20\ -\ 1)70^{2}}{6.844}=13603.156\)
The \(99\%\) confidence interval for variance is
Lower limit \(= 2413.04\) (rounded to 2 decimals)
Upper limit \(= 13603.16\) (rounded to 2 decimals)
b)The \(99\%\) confidence interval of standard deviation is calculated by applying square root to
\(99\%\) confidence interval of variance.
The \(99\%\) confidence interval for population standard deviation is given below
Lower limit \(=\sqrt{\frac{(n\ -\ 1)s^{2}}{X_{0.005}^{2}}}=\sqrt{\frac{(20\ -\ 1)70^{2}}{38.582}}=\sqrt{2413.04235}=49.1227\)
Upper limit \(=\sqrt{\frac{(n\ -\ 1)s^{2}}{X_{0.095}^{2}}}=\sqrt{\frac{(20\ -\ 1)70^{2}}{6.844}}=\sqrt{13603.156}=116.6326\)
The \(99\%\) confidence interval for standard deviation is
Lower limit \(= 49.12\) (rounded to 2 decimals)
Upper limit \(= 116.63\) (rounded to 2 decimals)

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