Question

On the basis of data provided by a salary survey, the variance in annual salaries for seniors in public accounting firms is approximately 2.1 and the variance in annual salaries for managers in public accounting firms is approximately 11.1. The salary data were provided in thousands of dollars. Assuming that the salary data were based on samples of 25 seniors and 26 managers, test to determine whether there is a significant difference between the variances of salaries for seniors and managers. At a 0.05 level of significance, what is your conclusion?

Answer 1

`F=(s^2_2)/(s^2_1)=(11.1)/(2.1)=5.286`

`p_v =2*P(F_(25,24)>5.286)=0.000116`

And we can use the following excel code to find the p value:"=2*(1-F.DIST(5.286,25,24,TRUE))"

Since the
`p_v < \alpha` we have enough evidence to reject the null hypothesis. And we can say that we have enough evidence to conclude that we have significant differences between the two variances at 5% of significance.

Explanation:

Data given and notation

`n_1 = 25` represent the sampe size for seniors

`n_2 =26` represent the sample size for managers

`s^2_1 = 2.1` represent the sample variance for seniors

`s^2_2 = 11.1` represent the sample variance for managers

`\alpha=0.05` represent the significance level provided

Confidence =0.95 or 95%

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_2)/(s^2_1)`

Solution to the problem

System of hypothesis

We want to test if we have a significant difference between the variances of salaries for seniors and managers, so the system of hypothesis are:

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

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

Calculate the statistic

Now we can calculate the statistic like this:

`F=(s^2_2)/(s^2_1)=(11.1)/(2.1)=5.286`

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_2 -1 =26-1=25` and for the denominator we have
`n_1 -1 =25-1=24` and the F statistic have 25 degrees of freedom for the numerator and 24 for the denominator. And the P value is given by:

P value

`p_v =2*P(F_(25,24)>5.286)=0.000116`

And we can use the following excel code to find the p value:"=2*(1-F.DIST(5.286,25,24,TRUE))"

Conclusion

Since the
`p_v < \alpha` we have enough evidence to reject the null hypothesis. And we can say that we have enough evidence to conclude that we have significant differences between the two variances at 5% of significance.