Question

A factor in determining the usefulness of a homework as a measure of demonstrated ability is the amount of spread that occurs in the grades. If the spread or variation of homework scores is very small, it usually means that the homework 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 homework for the state. The homework has 500 total possible points, and from the history of past homework, 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 homework to see how good it is. A preliminary version of the homework (with slight modifications to protect the integrity of the real homework) is given to a random sample of 27 newly graduated law students. Their scores give a sample standard deviation of 73 points. Using a 0.01 level of significance, test the claim that the population standard deviation for the new homework 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.
Ho: σ = 60; H1: σ ≠ 60
Ho: σ = 60; H1: σ < 60
Ho: σ = 60; H1: σ > 60
Ho: σ > 60; H1: σ = 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 uniform population distribution.
We assume a exponential population distribution.
We assume a normal population distribution.
We assume a binomial population distribution.

(c) Find or estimate the P-value of the sample test statistic.
P-value > 0.100
0.050 < P-value < 0.100
0.025 < P-value < 0.050
0.010 < P-value < 0.025
0.005 < P-value < 0.010
P-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?
Since the P-value > α, we fail to reject the null hypothesis.
Since the P-value > α, we reject the null hypothesis.
Since the P-value ≤ α, we reject the null hypothesis.
Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.
At the 1% level of significance, there is insufficient evidence to conclude that the standard deviation of scores on the preliminary exam is different from 60.
At the 1% level of significance, there is sufficient evidence to conclude that the standard deviation of scores on the preliminary exam 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

Answer 1

Answer a)

The level of significance 0.01

The null and alternate hypotheses are:

Ho: σ = 60; H1: σ ≠ 60

Answer b)

The value of the chi-square statistic for the sample is 38.49

Degrees of freedom = n-1 = 27-1 = 26

We assume a normal population distribution.

Answer c)

P-value for corresponding to χ2 = 38.487 and df = 26 for a two tailed test is 2*P(χ2 > 38.487)

P-value = 2*P(χ2 > 38.487) = 2*0.05453 = 0.1091

Thus, correct answer is P-value > 0.100

Answer d)

Since, P-value = 0.1091 > α = 0.01, we fail to eject null hypothesis.

Since the P-value > α, we fail to reject the null hypothesis.

Answer e)

At the 1% level of significance, there is insufficient evidence to conclude that the standard deviation of scores on the preliminary "exam" is different from 60.

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