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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. OneWay 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 oneway 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 oneway 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
OneWay ANOVA Table: The formulas for SS (Total), SS (Factor) = SS (Between) and SS (Error) = SS (Within) as shown previously. The same information is provided by the TI calculator hypothesis test function ANOVA in STAT TESTS (syntax is ANOVA(L1, L2, L3) where L1, L2, L3 have the data from Plan 1, Plan 2, Plan 3 respectively).
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 
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
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: n _{1} = 3  Ground Cover: n _{2} = 3  Plastic: n _{3} = 3  Straw: n _{4} = 3  Compost: n _{5} = 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 
Create the oneway ANOVA table.
Enter the data into lists L1, L2, L3, L4 and L5. Press STAT and arrow over to TESTS. Arrow down to ANOVA. Press ENTER and enter L1, L2, L3, L4, L5). Press ENTER. The table was filled in with the results from the calculator.
OneWay ANOVA table:
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 
The oneway ANOVA hypothesis test is always righttailed because larger F values are way out in the right tail of the F distribution curve and tend to make us reject H _{0} .
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 $\mu =\frac{df(num)}{df(denom)\u20131}$
Tomato Data, Marist College School of Science (unpublished student research)
Analysis of variance compares the means of a response variable for several groups. ANOVA compares the variation within each group to the variation of the mean of each group. The ratio of these two is the F statistic from an F distribution with (number of groups – 1) as the numerator degrees of freedom and (number of observations – number of groups) as the denominator degrees of freedom. These statistics are summarized in the ANOVA table.
$S{S}_{\text{between}}={{\displaystyle \sum}}^{\text{}}\left[\frac{{({s}_{j})}^{2}}{{n}_{j}}\right]\frac{{\left({{\displaystyle \sum}}^{\text{}}{s}_{j}\right)}^{2}}{n}$
$S{S}_{\text{total}}={{\displaystyle \sum}}^{\text{}}{x}^{2}\frac{{\left({{\displaystyle \sum}}^{\text{}}x\right)}^{2}}{n}$
$S{S}_{\text{within}}=S{S}_{\text{total}}S{S}_{\text{between}}$
df _{between} = df ( num ) = k – 1
df _{within} = df(denom) = n – k
MS _{between} = $\frac{S{S}_{\text{between}}}{d{f}_{\text{between}}}$
MS _{within} = $\frac{S{S}_{\text{within}}}{d{f}_{\text{within}}}$
F = $\frac{M{S}_{\text{between}}}{M{S}_{\text{within}}}$
F ratio when the groups are the same size: F = $\frac{n{s}_{\overline{x}}{}^{2}}{{s}^{\text{2}}{}_{pooled}}$
Mean of the F distribution: µ = $\frac{df(num)}{df(denom)1}$
where:
Use the following information to answer the next eight exercises. Groups of men from three different areas of the country are to be tested for mean weight. The entries in the table are the weights for the different groups. The oneway ANOVA results are shown in [link] .
Group 1  Group 2  Group 3 

216  202  170 
198  213  165 
240  284  182 
187  228  197 
176  210  201 
What is the Sum of Squares Error?
What is the df for the denominator?
What is the Mean Square Error?
Use the following information to answer the next eight exercises. Girls from four different soccer teams are to be tested for mean goals scored per game. The entries in the table are the goals per game for the different teams. The oneway ANOVA results are shown in
[link] .
Team 1  Team 2  Team 3  Team 4 

1  2  0  3 
2  3  1  4 
0  2  1  4 
3  4  0  3 
2  4  0  2 
What is SS _{between} ?
What is MS _{between} ?
What is the df for the denominator?
What is the F statistic?
Judging by the F statistic, do you think it is likely or unlikely that you will reject the null hypothesis?
Because a oneway ANOVA test is always righttailed, a high F statistic corresponds to a low p value, so it is likely that we will reject the null hypothesis.
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