Question

Two accounting professors decided to compare the variance of their grading procedures. To accomplish this, they each graded the same 10 exams, with the following results: Mean Grade Standard Deviation Professor 1 79.3 22.4 Professor 2 82.1 12.0 What is the alternate hypothesis?a. H1:σ21=σ22. b. H1:σ21≠σ22. c. H1:μ1=μ2 d. H1:μ1≠μ2

Answer 1

H1:
`\sigma^2_1 \\eq \sigma^2_2`

So then the correct option for this case would be:

H1:σ21≠σ22

Explanation:

Data given and notation

`n_1 = 10` represent the sampe size for professor 1

`n_2 =10` represent the sample size for professor 2

`\bar X_1 =79.3` represent the sample mean for professor 1

`\bar X_2 =82.1` represent the sample mean for professor 2

`s_1 = 22.4` represent the sample deviation for professor 1

`s^2_1 = 501.76` represent the sample variance for professor 1

`s_2 = 12` represent the sample deviation for professor 2

`s^2_2 = 144` represent the sample variance for professor 2

`\alpha` represent the significance level provided

F test is a statistical test that uses a F Statistic to compare two population variances, with the sample deviations s1 and s2. The F statistic is always positive number since the variance it's always higher than 0. The statistic is given by:

`F=(s^2_1)/(s^2_2)`

Solution to the problem

System of hypothesis

We want to test if the variation between their grading procedure, so the system of hypothesis are:

H0:
`\sigma^2_1 = \sigma^2_2`

H1:
`\sigma^2_1 \\eq \sigma^2_2`

So then the correct option for this case would be:

H1:σ21≠σ22

Calculate the statistic

Now we can calculate the statistic like this:

`F=(s^2_1)/(s^2_2)=(22.4^2)/(12^2)=3.484`

Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have
`n_1 -1 =10-1=9` and for the denominator we have
`n_2 -1 =10-1=9` and the F statistic have 9 degrees of freedom for the numerator and 9 for the denominator. And the P value is given by:

`p_v =P(F_(9,9)>3.484)=0.0385`

And we can use the following excel code to find the p value:"=1-F.DIST(3.484,9,9;TRUE)"

Conclusion

Assuming
`\alpha=0.05` and since the
`p_v < \alpha` we have enough evidence to reject the null hypothesis.