Alexander Holmes; Barbara Illowsky; Susan Dean

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 σ² 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 σ² 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: ∑n_j)
x = one value: ∑x = ∑s_j
Sum of squares of all values from every group combined: ∑x²
Between group variability: SS_total = ∑x² – $\frac{(\sum x^{2})}{n}$
Total sum of squares: ∑x² – $\frac{{(\sum x)}^{2}}{n}$
Explained variation: sum of squares representing variation among the different samples:
SS_between = $\sum [\frac{{(s_{j})}^{2}}{n_{j}}] - \frac{{(\sum s_{j})}^{2}}{n}$
Unexplained variation: sum of squares representing variation within samples due to chance: $S S_{within} = S S_{total} - S S_{between}$
df's for different groups (df's for the numerator): df = k – 1
Equation for errors within samples (df's for the denominator): df_within = n – k
Mean square (variance estimate) explained by the different groups: MS_between = $\frac{S S_{between}}{d f_{between}}$
Mean square (variance estimate) that is due to chance (unexplained): MS_within = $\frac{S S_{within}}{d f_{within}}$

MS_between and MS_within can be written as follows:

$M S_{between} = \frac{S S_{between}}{d f_{between}} = \frac{S S_{between}}{k - 1}$
$M S_{w i t h i n} = \frac{S S_{w i t h i n}}{d f_{w i t h i n}} = \frac{S S_{w i t h i n}}{n - k}$

The one-way 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.

F-Ratio or F Statistic $F = \frac{M S_{between}}{M S_{within}}$

If MS_between and MS_within estimate the same value (following the belief that H₀ is true), then the F-ratio 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 F-ratio 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 F-ratio can be written as:

F-Ratio Formula when the groups are the same size $F = \frac{n \cdot s_{\bar{x}}^{2}}{s^{2}_{pooled}}$

where ...

n = the sample size
df_numerator = k – 1
df_denominator = n – k
s² pooled = the mean of the sample variances (pooled variance)
$s_{\bar{x}}^{2}$ = the variance of the sample means

Data are typically put into a table for easy viewing. One-Way 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 – 1	MS(Factor) = SS(Factor)/(k – 1)	F = MS(Factor)/MS(Error)
Error (Within)	SS(Error)	n – k	MS(Error) = SS(Error)/(n – k)
Total	SS(Total)	n – 1

Table 12.3

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 one-way ANOVA results are shown in Table 12.4.

Plan 1: n₁ = 4	Plan 2: n₂ = 3	Plan 3: n₃ = 3
5	3.5	8
4.5	7	4
4		3.5
3	4.5

Table 12.4

s₁ = 16.5, s₂ =15, s₃ = 15.5

Following are the calculations needed to fill in the one-way ANOVA table. The table is used to conduct a hypothesis test.

S S (b e t w e e n) = \sum [\frac{{(s_{j})}^{2}}{n_{j}}] - \frac{{(\sum s_{j})}^{2}}{n}

12.1

= \frac{s_{1}^{2}}{4} + \frac{s_{2}^{2}}{3} + \frac{s_{3}^{2}}{3} - \frac{{(s_{1} + s_{2} + s_{3})}^{2}}{10}

12.2

where n₁ = 4, n₂ = 3, n₃ = 3 and n = n₁ + n₂ + n₃ = 10

= \frac{{(16.5)}^{2}}{4} + \frac{{(15)}^{2}}{3} + \frac{{(15.5)}^{2}}{3} - \frac{{(16.5 + 15 + 15.5)}^{2}}{10}

S S (b e t w e e n) = 2.2458

12.3

S (t o t a l) = \sum x^{2} - \frac{{(\sum x)}^{2}}{n}

12.4

= (5^{2} + {4.5}^{2} + 4^{2} + 3^{2} + {3.5}^{2} + 7^{2} + {4.5}^{2} + 8^{2} + 4^{2} + {3.5}^{2})

12.5

- \frac{{(5 + 4.5 + 4 + 3 + 3.5 + 7 + 4.5 + 8 + 4 + 3.5)}^{2}}{10}

12.6

= 244 - \frac{47^{2}}{10} = 244 - 220.9

12.7

S S (t o t a l) = 23.1

12.8

S S (w i t h i n) = S S (t o t a l) - S S (b e t w e e n)

12.9

= 23.1 - 2.2458

12.10

S S (w i t h i n) = 20.8542

12.11

Source of variation	Sum of squares (SS)	Degrees of freedom (df)	Mean square (MS)	F
Factor (Between)	SS(Factor) = SS(Between) = 2.2458	k – 1 = 3 groups – 1 = 2	MS(Factor) = SS(Factor)/(k – 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 – k = 10 total data – 3 groups = 7	MS(Error) = SS(Error)/(n – k) = 20.8542/7 = 2.9792
Total	SS(Total) = 2.2458 + 20.8542 = 23.1	n – 1 = 10 total data – 1 = 9

Table 12.5

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: n₁ = 3	Ground Cover: n₂ = 3	Plastic: n₃ = 3	Straw: n₄ = 3	Compost: n₅ = 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

Table 12.6

Create the one-way ANOVA table.

The one-way ANOVA hypothesis test is always right-tailed because larger F-values are way out in the right tail of the F-distribution curve and tend to make us reject H₀.

Example 12.3

Problem

Let’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 μ₁, μ₂, μ₃, μ₄, μ₅. 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:

H₀: μ₁ = μ₂ = μ₃ = μ₄ = μ₅

H_a: μ_i ≠ μ_j some i ≠ j

The one-way 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 – 1 = 4	$\frac{36,648,561}{4} = 9,162,140$	$\frac{9,162,140}{2,044,672 .6} = 4 .4810$
Error (Within)	20,446,726	15 – 5 = 10	$\frac{20,446,726}{10} = 2,044,672 .6$
Total	57,095,287	15 – 1 = 14

Table 12.7

Distribution for the test: F_4,10

df(num) = 5 – 1 = 4

df(denom) = 15 – 5 = 10

Test statistic: F = 4.4810

This graph shows a nonsymmetrical F distribution curve. The horizontal axis extends from 0 - 5, and the vertical axis ranges from 0 - 0.7. The curve is strongly skewed to the right.

Figure 12.4

Probability Statement: p-value = P(F > 4.481) = 0.0248.

Compare α and the p-value: α = 0.05, p-value = 0.0248

Make a decision: Since α > p-value, we cannot accept H₀.

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

Table 12.8

Plot of the data for the different concentrations:

This graph is a scatterplot for the data provided. The horizontal axis is labeled 'Colony counts' and extends from 0 - 200. The vertical axis is labeled 'Tryptone concentrations' and extends from 0.6 - 1.4.

Figure 12.5

Test whether the mean number of colonies are the same or are different. Construct the ANOVA table, find the p-value, 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

Table 12.9 Mean grades for four sororities

Problem

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

Solution

Let μ₁, μ₂, μ₃, μ₄ 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.

H₀: $μ_{1} = μ_{2} = μ_{3} = μ_{4}$

H_a: Not all of the means $μ_{1}, μ_{2}, μ_{3}, μ_{4}$ 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:

Figure 12.6

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 H₀.

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

Table 12.10 GPAs for four sports teams

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

Table 12.11

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' = $\frac{n \cdot s_{\bar{x}}^{2}}{s^{2}_{pooled}}$ .

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

Table 12.12

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 = $s_{\bar{x}}^{2}$

Then MS_between = $n s_{\bar{x}}^{2}$ = (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 = s² pooled

Then MS_within = s²_pooled = 15.433.

The F statistic (or F ratio) is $F = \frac{M S_{between}}{M S_{within}} = \frac{n s_{\bar{x}}^{2}}{s^{2}_{p o o l e d}} = \frac{(5) (0.413)}{15.433} = 0.134$

The dfs for the numerator = the number of groups – 1 = 3 – 1 = 2.

The dfs for the denominator = the total number of samples – 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 α = 0.03 and the p-value = 0.8759, then you cannot reject H₀. (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 $μ = \frac{d f (n u m)}{d f (d e n o m) - 2}$

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

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

12.3 The F Distribution and the F-Ratio

Problem

Solution

Problem

Solution

Problem

Solution

Notation

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