12.3 The f distribution and the f-ratio  (Page 4/7)

Four sororities took a random sample of sisters regarding their grade means for the past term. The results are shown in [link] .

Mean grades for four sororities

Sorority 1	Sorority 2	Sorority 3	Sorority 4
2.17	2.63	2.63	3.79
1.85	1.77	3.78	3.45
2.83	3.25	4.00	3.08
1.69	1.86	2.55	2.26
3.33	2.21	2.45	3.18

Using a significance level of 1%, is there a difference in mean grades among the sororities?

Let μ ₁ , μ ₂ , μ ₃ , μ ₄ be the population means of the sororities. Remember that the null hypothesis claims that the sorority groups are from the same normal distribution. The alternate hypothesis says that at least two of the sorority groups come from populations with different normal distributions. Notice that the four sample sizes are each five.

Note

This is an example of a balanced design , because each factor (i.e., sorority) has the same number of observations.

H ₀ : ${\mu }_1={\mu }_2={\mu }_3={\mu }_4$

H _a : Not all of the means ${\mu }_1,{\mu }_2,{\mu }_3,{\mu }_4$ are equal.

Distribution for the test: F _3,16

where k = 4 groups and n = 20 samples in total

df ( num )= k – 1 = 4 – 1 = 3

df ( denom ) = n – k = 20 – 4 = 16

Calculate the test statistic: F = 2.23

Graph:

Probability statement: p -value = P ( F >2.23) = 0.1241

Compare α and the p -value: α = 0.01
p -value = 0.1241
α < p -value

Make a decision: Since α < p -value, you cannot reject H ₀ .

Conclusion: There is not sufficient evidence to conclude that there is a difference among the mean grades for the sororities.

A fourth grade class is studying the environment. One of the assignments is to grow bean plants in different soils. Tommy chose to grow his bean plants in soil found outside his classroom mixed with dryer lint. Tara chose to grow her bean plants in potting soil bought at the local nursery. Nick chose to grow his bean plants in soil from his mother's garden. No chemicals were used on the plants, only water. They were grown inside the classroom next to a large window. Each child grew five plants. At the end of the growing period, each plant was measured, producing the data (in inches) in [link] .

Notation

The notation for the F distribution is F ~ F _{df ( num ), df ( denom )}

where df ( num ) = df _between and df ( denom ) = df _within

The mean for the F distribution is $\mu =\frac{df(num)}{df(denom)–1}$

References

Tomato Data, Marist College School of Science (unpublished student research)

Chapter review

Analysis of variance compares the means of a response variable for several groups. ANOVA compares the variation within each group to the variation of the mean of each group. The ratio of these two is the F statistic from an F distribution with (number of groups – 1) as the numerator degrees of freedom and (number of observations – number of groups) as the denominator degrees of freedom. These statistics are summarized in the ANOVA table.

Formula review

$S{S}_{\text{between}}={{\displaystyle \sum }}^{\text{}}\left[\frac{{(s_j)}^2}{n_j}\right]-\frac{{\left({{\displaystyle \sum }}^{\text{}}{s}_j\right)}^2}{n }$

$S{S}_{\text{total}}={{\displaystyle \sum }}^{\text{}}{x}^2-\frac{{\left({{\displaystyle \sum }}^{\text{}}x\right)}^2}{n}$

$S{S}_{\text{within}}=S{S}_{\text{total}}-S{S}_{\text{between}}$

df _between = df ( num ) = k – 1

df _within = df(denom) = n – k

MS _between = $\frac{S{S}_{\text{between}}}{d{f}_{\text{between}}}$

MS _within = $\frac{S{S}_{\text{within}}}{d{f}_{\text{within}}}$

F = $\frac{M{S}_{\text{between}}}{M{S}_{\text{within}}}$

• k = the number of groups
• n _j = the size of the j ^th group
• s _j = the sum of the values in the j ^th group
• n = the total number of all values (observations) combined
• x = one value (one observation) from the data
• $s_{\bar{x}}{}^2$ = the variance of the sample means
• $s^2{}_{pooled}$ = the mean of the sample variances (pooled variance)

Use the following information to answer the next eight exercises. Groups of men from three different areas of the country are to be tested for mean weight. The entries in the table are the weights for the different groups. The one-way ANOVA results are shown in [link] .

Use the following information to answer the next eight exercises. Girls from four different soccer teams are to be tested for mean goals scored per game. The entries in the table are the goals per game for the different teams. The one-way ANOVA results are shown in [link] .