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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.
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.
If MS between and MS within estimate the same value (following the belief that H 0 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:
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 |
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 [link] .
| 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 = 15.7
Following are the calculations needed to fill in the one-way 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 – 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 |
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