Let’s return to the slicing tomato exercise in Try It. 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 percent, 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.

The null and alternative hypotheses are as follows:

H₀: μ₁ = μ₂ = μ₃ = μ₄ = μ₅

H_a: μ_i ≠ μ_j for some i ≠ j

The one-way ANOVA results are shown in Table 13.5

Distribution for the test: F_4,10

df(num) = 5 – 1 = 4

df(denom) = 15 – 5 = 10

Test statistic: F = 4.4810

Figure 13.4

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 reject H₀.

Conclusion: At the 5 percent 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 the mulches led to different mean yields.

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

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.

H₀: μ₁ = μ₂ = μ₃ = μ₄

H_a: Not all of the means μ₁, μ₂, μ₃, μ₄ 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 13.6

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

Compare α and the p-value: α = 0.01

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.

Does it appear that the three media in which the bean plants were grown produce the same mean height? Test at a 3 percent level of significance.

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

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

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

then MS_between = $n{s}_{\overline{x}}{}^2$ = (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 $F=\frac{M{S}_{\text{between}}}{M{S}_{\text{within}}}=\frac{n{s}_{\overline{x}}{}^2}{s^2{}_{pooled}}=\frac{(5)(0.413)}{15.433}=0.134\text{.}$

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, do not reject H₀. Why?

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