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 21 newly graduated law students. Their scores give a sample standard deviation of 68 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?
(b) Find the value of the chi-square statistic for the sample. (Use 2 decimal places.) What are the degrees of freedom?

Answer 1

Answer:

Step-by-step explanation:

From the information given :

The null hypothesis $H_o : \sigma = 60$

The alternative hypothesis  $Ha : \sigma \neq 60$

Given that:

the sample size n = 21

sample standard deviation s = 68

population standard deviation σ = 60

a) The level of significance $\alpha =0.01$

b) The chi-square statistics can be calculated as:

$\bar X = \dfrac{(n-1)s^2}{\sigma^2}$

$\bar X = \dfrac{(21-1)68^2}{60^2}$

$\bar X = \dfrac{(20)4624}{3600}$

$\bar X = 25.69$

The degree of freedom is :

= n - 1

= 21 - 1

= 20