12.3 The F Distribution and the F-Ratio - Introductory Business Statistics 2e | OpenStax (2024)

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.

1. 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.
2. 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 somecases, may be weighted. We used sum of squares to calculate the sample variance andthe sample standard deviation in 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 (totalsamplesize: ∑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 – $\frac{\left({\displaystyle \sum x^2}\right)}{n}$
• Total sum of squares: ∑x² – $\frac{{\left(\sum x\right)}^2}{n}$
• Explained variation: sum of squares representing variation among the different samples:
  SS_between = $\sum \left[\frac{{(s_j)}^2}{n_j}\right]-\frac{{(\sum s_j)}^2}{n}$
• Unexplained variation: sum of squares representing variation within samples due to chance: $S{S}_{\text{within}}=S{S}_{\text{total}}–S{S}_{\text{between}}$
• 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 = $\frac{S{S}_{\text{between}}}{d{f}_{\text{between}}}$
• 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:

• $M{S}_{\text{between}}=\frac{S{S}_{\text{between}}}{d{f}_{\text{between}}}=\frac{S{S}_{\text{between}}}{k-1}$
• $M{S}_{within}=\frac{S{S}_{within}}{d{f}_{within}}=\frac{S{S}_{within}}{n-k}$

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.

F-Ratio 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 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 $F=\frac{n\cdot s_{\bar{x}}{}^2}{s^2{}_{\text{pooled}}}$

Data are typically put into a table for easy viewing. One-Way ANOVA results are often displayed in this manner by computer software.

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	$\frac{\text{36,648,561}}{\text{4}}\text{=9,162,140}$	$$\frac{\text{9,162,140}}{\text{2,044,672}\text{.6}}\text{=4}\text{.4810}$$
Error (Within)	20,446,726	15 – 5 = 10	$$\frac{\text{20,446,726}}{\text{10}}\text{=2,044,672}\text{.6}$$
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

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.

See Also

5.11: The F Distribution

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.

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 thesororities?

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

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:

Figure 12.5

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.