11.3 Facts about the f distribution

This time, we will perform the calculations that lead to the F' statistic. Notice that each group has the same number of plants so we will use the formula $\mathrm{F\text{'}}=\frac{n\cdot {s_{\stackrel{\_}{x}}}^2}{{{s^2}_{\text{pooled}}}^{}}$ .

First, calculate the sample mean and sample variance of each group.

Next, calculate the variance of the three group means (Calculate the variance of 24.2, 25.4, and 24.4). Variance of the group means = 0.413 = $s_{\overline{x}}{}^2$

Then ${\mathrm{MS}}_{\text{between}}=n{s}_{\overline{x}}{}^2=\left(5\right)\left(0.413\right)$ where $n=5$ is the sample size (number of plants each child grew).

Calculate the mean of the three sample variances (Calculate the mean of 11.7, 18.3, and 16.3). Mean of the sample variances = 15.433 $={s^2}_{\text{pooled}}{}^{}$

Then ${\mathrm{MS}}_{\text{within}}={s^2}_{\text{pooled}}{}^{}=15.433$ .

The $F$ statistic (or $F$ ratio) is $F=\frac{{\mathrm{MS}}_{\text{between}}}{{\mathrm{MS}}_{\text{within}}}=\frac{n\cdot {s_{\stackrel{\_}{x}}}^2}{{{s^2}_{\text{pooled}}}^{}}=\frac{\left(5\right)\cdot \left(0.413\right)}{15.433}=0.134$

The dfs for the numerator = $\text{the number of groups}-1=3-1=2$

The dfs for the denominator = $\text{the total number of samples}-\text{the number of groups}=15-3=12$

The distribution for the test is $F_{2,12}$ and the F statistic is $F=0.134$

The p-value is $P(F>0.134)=0.8759$ .

Decision: Since $\alpha =0.03$ and the $\text{p-value}=0.8759$ , do not reject $H_o$ . (Why?)

Conclusion: With a 3% the level of significance, from the sample data, the evidence is not sufficient to conclude that the mean heights of the bean plants are different.

(This experiment was actually done by three classmates of the son of one of the authors.)

• $F$ = 0.9496
• $\mathrm{p-value}$ = 0.4402

From the sample data, the evidence is not sufficient to conclude that the mean heights of the bean plants are different.