The distribution used for the hypothesis test is a new one. It is called the F distribution, invented by George Snedecor but named in honor of Sir Ronald Fisher, an English statistician. The F statistic is a ratio (a fraction). There are two sets of degrees of freedom; one for the numerator and one for the denominator.
For example, if F follows an F distribution and the number of degrees of freedom for the numerator is four, and the number of degrees of freedom for the denominator is ten, then F ~ F_{4,10}.
To calculate the F ratio, two estimates of the variance are made.
 Variance between samples: An estimate of Ïƒ^{2} that is the variance of the sample means multiplied by n (when the sample sizes are the same.). If the samples are different sizes, the variance between samples is weighted to account for the different sample sizes. The variance is also called variation due to treatment or explained variation.
 Variance within samples: An estimate of Ïƒ^{2} that is the average of the sample variances (also known as a pooled variance). When the sample sizes are different, the variance within samples is weighted. The variance is also called the variation due to error or unexplained variation.
 SS_{between} = the sum of squares that represents the variation among the different samples
 SS_{within} = the sum of squares that represents the variation within samples that is due to chance.
To find a "sum of squares" means to add together squared quantities that, in some cases, may be weighted. We used sum of squares to calculate the sample variance and the sample standard deviation in Chapter 2 Descriptive Statistics.
MS means "mean square." MS_{between} is the variance between groups, and MS_{within} is the variance within groups.
Calculation of Sum of Squares and Mean Square
 k = the number of different groups
 n_{j} = the size of the j^{th} group
 s_{j} = the sum of the values in the j^{th} group
 n = total number of all the values combined (total sample size: âˆ‘n_{j})
 x = one value: âˆ‘x = âˆ‘s_{j}
 Sum of squares of all values from every group combined: âˆ‘x^{2}
 Between group variability: SS_{total} = âˆ‘x^{2} â€“ $\frac{\left({\displaystyle \xe2\u02c6\u2018{x}^{2}}\right)}{n}$
 Total sum of squares: âˆ‘x^{2} â€“ $\frac{{\left(\xe2\u02c6\u2018x\right)}^{2}}{n}$
 Explained variation: sum of squares representing variation among the different samples:
SS_{between} = $\xe2\u02c6\u2018\left[\frac{{({s}_{j})}^{2}}{{n}_{j}}\right]\xe2\u02c6\u2019\frac{{(\xe2\u02c6\u2018{s}_{j})}^{2}}{n}$  Unexplained variation: sum of squares representing variation within samples due to chance: $S{S}_{\text{within}}=S{S}_{\text{total}}\xe2\u20ac\u201cS{S}_{\text{between}}$
 df's for different groups (df's for the numerator): df = k â€“ 1
 Equation for errors within samples (df's for the denominator): df_{within} = n â€“ k
 Mean square (variance estimate) explained by the different groups: MS_{between} = $\frac{S{S}_{\text{between}}}{d{f}_{\text{between}}}$
 Mean square (variance estimate) that is due to chance (unexplained): MS_{within} = $\frac{S{S}_{\text{within}}}{d{f}_{\text{within}}}$
MS_{between} and MS_{within} can be written as follows:
 $M{S}_{\text{between}}=\frac{S{S}_{\text{between}}}{d{f}_{\text{between}}}=\frac{S{S}_{\text{between}}}{k\xe2\u02c6\u20191}$
 $M{S}_{within}=\frac{S{S}_{within}}{d{f}_{within}}=\frac{S{S}_{within}}{n\xe2\u02c6\u2019k}$
The oneway ANOVA test depends on the fact that MS_{between} can be influenced by population differences among means of the several groups. Since MS_{within} compares values of each group to its own group mean, the fact that group means might be different does not affect MS_{within}.
The null hypothesis says that all groups are samples from populations having the same normal distribution. The alternate hypothesis says that at least two of the sample groups come from populations with different normal distributions. If the null hypothesis is true, MS_{between} and MS_{within} should both estimate the same value.
Note
The null hypothesis says that all the group population means are equal. The hypothesis of equal means implies that the populations have the same normal distribution, because it is assumed that the populations are normal and that they have equal variances.
FRatio or F Statistic $F=\frac{M{S}_{\text{between}}}{M{S}_{\text{within}}}$
If MS_{between} and MS_{within} estimate the same value (following the belief that H_{0} is true), then the Fratio should be approximately equal to one. Mostly, just sampling errors would contribute to variations away from one. As it turns out, MS_{between} consists of the population variance plus a variance produced from the differences between the samples. MS_{within} is an estimate of the population variance. Since variances are always positive, if the null hypothesis is false, MS_{between} will generally be larger than MS_{within}.Then the Fratio will be larger than one. However, if the population effect is small, it is not unlikely that MS_{within} will be larger in a given sample.
The foregoing calculations were done with groups of different sizes. If the groups are the same size, the calculations simplify somewhat and the Fratio can be written as:
FRatio Formula when the groups are the same size $F=\frac{n\xe2\u2039\dots {s}_{\stackrel{\xe2\u20ac\u201c}{x}}{}^{2}}{{s}^{2}{}_{\text{pooled}}}$
 n = the sample size
 df_{numerator} = k â€“ 1
 df_{denominator} = n â€“ k
 s^{2} pooled = the mean of the sample variances (pooled variance)
 ${s}_{\stackrel{\xe2\u20ac\u201c}{x}}{}^{2}$ = the variance of the sample means
Data are typically put into a table for easy viewing. OneWay ANOVA results are often displayed in this manner by computer software.
Source of variation  Sum of squares (SS)  Degrees of freedom (df)  Mean square (MS)  F 

Factor (Between) 
SS(Factor)  k â€“ 1  MS(Factor) = SS(Factor)/(k â€“ 1)  F = MS(Factor)/MS(Error) 
Error (Within) 
SS(Error)  n â€“ k  MS(Error) = SS(Error)/(n â€“ k)  
Total  SS(Total)  n â€“ 1 
Example 12.2
Three different diet plans are to be tested for mean weight loss. The entries in the table are the weight losses for the different plans. The oneway ANOVA results are shown in Table 12.4.
Plan 1: n_{1} = 4  Plan 2: n_{2} = 3  Plan 3: n_{3} = 3 

5  3.5  8 
4.5  7  4 
4  3.5  
3  4.5 
s_{1} = 16.5, s_{2} =15, s_{3} = 15.5
Following are the calculations needed to fill in the oneway ANOVA table. The table is used to conduct a hypothesis test.
where n_{1} = 4, n_{2} = 3, n_{3} = 3 and n = n_{1} + n_{2} + n_{3} = 10
Source of variation  Sum of squares (SS)  Degrees of freedom (df)  Mean square (MS)  F 

Factor (Between) 
SS(Factor) = SS(Between) = 2.2458 
k â€“ 1 = 3 groups â€“ 1 = 2 
MS(Factor) = SS(Factor)/(k â€“ 1) = 2.2458/2 = 1.1229 
F = MS(Factor)/MS(Error) = 1.1229/2.9792 = 0.3769 
Error (Within) 
SS(Error) = SS(Within) = 20.8542 
n â€“ k = 10 total data â€“ 3 groups = 7 
MS(Error) = SS(Error)/(n â€“ k) = 20.8542/7 = 2.9792 

Total  SS(Total) = 2.2458 + 20.8542 = 23.1 
n â€“ 1 = 10 total data â€“ 1 = 9 
Try It 12.2
As part of an experiment to see how different types of soil cover would affect slicing tomato production, Marist College students grew tomato plants under different soil cover conditions. Groups of three plants each had one of the following treatments
 bare soil
 a commercial ground cover
 black plastic
 straw
 compost
All plants grew under the same conditions and were the same variety. Students recorded the weight (in grams) of tomatoes produced by each of the n = 15 plants:
Bare: n_{1} = 3  Ground Cover: n_{2} = 3  Plastic: n_{3} = 3  Straw: n_{4} = 3  Compost: n_{5} = 3 

2,625  5,348  6,583  7,285  6,277 
2,997  5,682  8,560  6,897  7,818 
4,915  5,482  3,830  9,230  8,677 
Create the oneway ANOVA table.
The oneway ANOVA hypothesis test is always righttailed because larger Fvalues are way out in the right tail of the Fdistribution curve and tend to make us reject H_{0}.
Example 12.3
Problem
Letâ€™s return to the slicing tomato exercise in Try It 12.2. The means of the tomato yields under the five mulching conditions are represented by Î¼_{1}, Î¼_{2}, Î¼_{3}, Î¼_{4}, Î¼_{5}. We will conduct a hypothesis test to determine if all means are the same or at least one is different. Using a significance level of 5%, test the null hypothesis that there is no difference in mean yields among the five groups against the alternative hypothesis that at least one mean is different from the rest.
Solution
The null and alternative hypotheses are:
H_{0}: Î¼_{1} = Î¼_{2} = Î¼_{3} = Î¼_{4} = Î¼_{5}
H_{a}: Î¼_{i} â‰ Î¼_{j} some i â‰ j
The oneway ANOVA results are shown in Table 12.7
Source of variation  Sum of squares (SS)  Degrees of freedom (df)  Mean square (MS)  F 

Factor (Between)  36,648,561  5 â€“ 1 = 4  $\frac{\text{36,648,561}}{\text{4}}\text{=9,162,140}$  $$\frac{\text{9,162,140}}{\text{2,044,672}\text{.6}}\text{=4}\text{.4810}$$ 
Error (Within)  20,446,726  15 â€“ 5 = 10  $$\frac{\text{20,446,726}}{\text{10}}\text{=2,044,672}\text{.6}$$  
Total  57,095,287  15 â€“ 1 = 14 
Distribution for the test: F_{4,10}
df(num) = 5 â€“ 1 = 4
df(denom) = 15 â€“ 5 = 10
Test statistic: F = 4.4810
Probability Statement: pvalue = P(F > 4.481) = 0.0248.
Compare Î± and the pvalue: Î± = 0.05, pvalue = 0.0248
Make a decision: Since Î± > pvalue, we cannot accept H_{0}.
Conclusion: At the 5% significance level, we have reasonably strong evidence that differences in mean yields for slicing tomato plants grown under different mulching conditions are unlikely to be due to chance alone. We may conclude that at least some of mulches led to different mean yields.
Try It 12.3
MRSA, or Staphylococcus aureus, can cause a serious bacterial infections in hospital patients. Table 12.8 shows various colony counts from different patients who may or may not have MRSA. The data from the table is plotted in Figure 12.5.
Conc = 0.6  Conc = 0.8  Conc = 1.0  Conc = 1.2  Conc = 1.4 

9  16  22  30  27 
66  93  147  199  168 
98  82  120  148  132 
Plot of the data for the different concentrations:
Test whether the mean number of colonies are the same or are different. Construct the ANOVA table, find the pvalue, and state your conclusion. Use a 5% significance level.
Example 12.4
Four sororities took a random sample of sisters regarding their grade means for the past term. The results are shown in Table 12.9.
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 
Problem
Using a significance level of 1%, is there a difference in mean grades among the sororities?
Solution
Let Î¼_{1}, Î¼_{2}, Î¼_{3}, Î¼_{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}: ${\mathrm{\xce\xbc}}_{1}={\mathrm{\xce\xbc}}_{2}={\mathrm{\xce\xbc}}_{3}={\mathrm{\xce\xbc}}_{4}$
H_{a}: Not all of the means ${\mathrm{\xce\xbc}}_{1},{\mathrm{\xce\xbc}}_{2},{\mathrm{\xce\xbc}}_{3},{\mathrm{\xce\xbc}}_{4}$are equal.
Distribution for the test: F_{3,16}
where k = 4 groups and n = 20 samples in total
df(num)= k â€“ 1 = 4 â€“ 1 = 3
df(denom) = n â€“ k = 20 â€“ 4 = 16
Calculate the test statistic: F = 2.23
Graph:
Probability statement: pvalue = P(F > 2.23) = 0.1241
Compare Î± and the pvalue: Î± = 0.01
pvalue = 0.1241
Î± < pvalue
Make a decision: Since Î± < pvalue, 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.
Try It 12.4
Four sports teams took a random sample of players regarding their GPAs for the last year. The results are shown in Table 12.10.
Basketball  Baseball  Hockey  Lacrosse 

3.6  2.1  4.0  2.0 
2.9  2.6  2.0  3.6 
2.5  3.9  2.6  3.9 
3.3  3.1  3.2  2.7 
3.8  3.4  3.2  2.5 
Use a significance level of 5%, and determine if there is a difference in GPA among the teams.
Example 12.5
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 Table 12.11.
Tommy's plants  Tara's plants  Nick's plants 

24  25  23 
21  31  27 
23  23  22 
30  20  30 
23  28  20 
Problem
Does it appear that the three media in which the bean plants were grown produce the same mean height? Test at a 3% level of significance.
Solution
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 F' = $\frac{n\xe2\u2039\dots {s}_{\stackrel{\xe2\u20ac\u201c}{x}}{}^{2}}{{s}^{2}{}_{\text{pooled}}}$ .
First, calculate the sample mean and sample variance of each group.
Tommy's plants  Tara's plants  Nick's plants  

Sample mean  24.2  25.4  24.4 
Sample variance  11.7  18.3  16.3 
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 = ${s}_{\stackrel{\xe2\u20ac\u201c}{x}}{}^{2}$
Then MS_{between} = $n{s}_{\stackrel{\xe2\u20ac\u201c}{x}}{}^{2}$ = (5)(0.413) 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} pooled
Then MS_{within} = s^{2}_{pooled} = 15.433.
The F statistic (or F ratio) is $F=\frac{M{S}_{\text{between}}}{M{S}_{\text{within}}}=\frac{n{s}_{\stackrel{\xe2\u20ac\u201c}{x}}{}^{2}}{{s}^{2}{}_{pooled}}=\frac{(5)(0.413)}{15.433}=0.134$
The dfs for the numerator = the number of groups â€“ 1 = 3 â€“ 1 = 2.
The dfs for the denominator = the total number of samples â€“ the number of groups = 15 â€“ 3 = 12
The distribution for the test is F_{2,12} and the F statistic is F = 0.134
The pvalue is P(F > 0.134) = 0.8759.
Decision: Since Î± = 0.03 and the pvalue = 0.8759, then you cannot reject H_{0}. (Why?)
Conclusion: With a 3% level of significance, from the sample data, the evidence is not sufficient to conclude that the mean heights of the bean plants are different.
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 $\mathrm{\xce\xbc}=\frac{df(num)}{df(denom)\xe2\u20ac\u201c2}$