Q3b) F test of the Ratio of Two Variances

ADMS 2320: FINAL EXAM PREP · Summer 2025 2022 Summer Final Q3b) F test of the Ratio of Two Variances

SOLUTION

Given

Variance for MFM: $s_{\text{MFM}}^2 = 0.0562$

Variance for GH: $s_{\text{GH}}^2 = 0.0330$

Degrees of Freedom: $df_{\text{MFM}} = (n_{\text{MFM}} - 1) = 27 - 1 = 26$ , $df_{\text{GH}} = (n_{\text{GH}} - 1) = 19 - 1 = 18$

Hypotheses

Null Hypothesis:

$H_0: \frac{\sigma_{\text{MFM}}^2}{\sigma_{\text{GH}}^2} = 1$

Alternative Hypothesis:

$H_1: \frac{\sigma_{\text{MFM}}^2}{\sigma_{\text{GH}}^2} < 1$

Rejection Region

Critical F-value for one-tailed (lower tail) test at 5% significance level:

$F_{\text{critical}} = \frac{1}{F_{\alpha, df_2, df_1}} = \frac{1}{F_{0.05, 16, 26}} = \frac{1}{2.02} = 0.495$

F-Statistic Calculation

$F = \frac{s_{\text{MFM}}^2}{s_{\text{GH}}^2} = \frac{0.0562}{0.0330}$

$F = 1.7030$

Conclusion

Since $F = 1.7030$ is greater than $F_{\text{critical}} = 0.495$ , we fail to reject $H_0$ . There is not enough evidence at the 5% significance level to conclude that MFM’s variance is significantly less than GH’s.

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