Let μ 1 μ 1 , μ 2 μ 2 , μ 3 μ 3 , μ 4 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 size 5.

H o : μ 1 = μ 2 = μ 3 = μ 4 H o : μ 1 = μ 2 = μ 3 = μ 4

H a H a : Not all of the means μ 1 , μ 2 , μ 3 , μ 4 μ 1 , μ 2 , μ 3 , μ 4 are equal.

Distribution for the test: F 3 , 16 F 3 , 16

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

df(num) = k - 1 = 4 - 1 = 3 df(num)=k-1=4-1=3

df(denom) = n - k = 20 - 4 = 16 df(denom)=n-k=20-4=16

Calculate the test statistic: F = 2.23 F=2.23

Graph:

Figure 2

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

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

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

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

TI-83+ or TI 84: Put the data into lists L1, L2, L3, and L4. Press STAT and arrow over to TESTS. Arrow down to F:ANOVA. Press ENTER and Enter (L1,L2,L3,L4). The F statistic is 2.2303 and the p-value p-value is 0.1241. df(numerator) = 3df(numerator) = 3 (under "Factor") and df(denominator) = 16df(denominator) = 16 (under Error).

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' = n ⋅ s x_ 2 s 2 pooled F' = n ⋅ s x_ 2 s 2 pooled .

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

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 x¯ 2s x 2

Then MS between = n s x¯ 2 = ( 5 ) ( 0.413 ) MS between =ns x 2=(5)(0.413) where n=5n=5 is the sample size (number of plants each child grew).

Calculate the average of the three sample variances (Calculate the average of 11.7, 18.3, and 16.3). Average of the sample variances = 15.433 = s 2 pooled = s 2 pooled size 12{ \( s rSub { size 8{ ital "pooled"} } \) rSup { size 8{2} } ={}} {}

Then MS within = s 2 pooled = 15.433 MS within = s 2 pooled size 12{ \( s rSub { size 8{ ital "pooled"} } \) rSup { size 8{2} } ={}} {}=15.433.

The F F statistic (or F F ratio) is F = MS between MS within = n ⋅ s x_ 2 s 2 pooled = ( 5 ) ⋅ ( 0.413 ) 15.433 = 0.134 F= MS between MS within = n ⋅ s x_ 2 s 2 pooled = ( 5 ) ⋅ ( 0.413 ) 15.433 =0.134

The dfs for the numerator = the number of groups - 1 = 3 - 1 = 2 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 total number of samples-the number of groups=15-3=12

The distribution for the test is F 2 , 12 F 2 , 12 and the F statistic is F = 0.134 F=0.134

The p-value is P ( F > 0.134 ) = 0.8759 P(F>0.134)=0.8759.

Decision: Since α = 0.03 α=0.03 and the p-value = 0.8759 p-value=0.8759, do not reject H o H o . (Why?)

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

(This experiment was actually done by three classmates of the son of one of the authors.)

From the sample data, the evidence is not sufficient to conclude that the average heights of the bean plants are different.