Question

The accounting department analyzes the variance of the weekly unit costs reported by two production departments. A sample of cost reports for each of the two departments shows cost variances of and , respectively. Is this sample sufficient to conclude that the two production departments differ in terms of unit cost variance?

Answer 1

`F=(s^2_2)/(s^2_1)=(5.4)/(2.3)=2.348`

`p_v =2*P(F_(15,15)>2.348)=0.109`

And we can use the following excel code to find the p value:"=2*(1-F.DIST(2.348,15,15,TRUE))"

Since the
`p_v > \alpha` we have enough evidence to FAIL to reject the null hypothesis. And we can say that we don't have enough evidence to conclude that the variation is different between the two groups at 10% of significance.

Explanation:

Assuming the following question :"The accounting department analyzes the variance of the weekly unit costs reported by two production departments. Asample of 16 cost reports for each of the two departments shows cost variances of 2.3 and 5.4, respectively. Is this sample sufficient to conclude that the two production departments differ in terms of unit cost variance? Use α= .10."

Data given and notation

`n_1 = 16` represent the sampe size for sample 1

`n_2 =16` represent the sample size for sample 2

`s^2_1 = 2.3` represent the sample variance for sample 1

`s^2_2 = 5.4` represent the sample variance for sample 2

`\alpha=0.1` represent the significance level provided

Confidence =0.9 or 9%

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 the variation is equal for both groups or no, 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)=(5.4)/(2.3)=2.348`

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 =16-1=15` and for the denominator we have
`n_1 -1 =16-1=15` and the F statistic have 15 degrees of freedom for the numerator and 15 for the denominator. And the P value is given by:

P value

Since is a bilateral test the p value is given by

`p_v =2*P(F_(15,15)>2.348)=0.109`

And we can use the following excel code to find the p value:"=2*(1-F.DIST(2.348,15,15,TRUE))"

Conclusion

Since the
`p_v > \alpha` we have enough evidence to FAIL to reject the null hypothesis. And we can say that we don't have enough evidence to conclude that the variation is different between the two groups at 10% of significance.