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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.

Note

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.

F -ratio or F Statistic

F = M S between M S within

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:

F -ratio formula when the groups are the same size

F = n s x ¯ 2 s 2 pooled

    Where ...

  • n = the sample size
  • df numerator = k – 1
  • df denominator = n k
  • s 2 pooled = the mean of the sample variances (pooled variance)
  • s x ¯ 2 = the variance of the sample means

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.

S S ( b e t w e e n ) = [ ( s j ) 2 n j ] ( s j ) 2 n  
=   s 1 2 4 + s 2 2 3 + s 3 2 3 ( s 1 + s 2 + s 3 ) 2 10

where n 1 = 4, n 2 = 3, n 3 = 3 and n = n 1 + n 2 + n 3 = 10

    = ( 16.5 ) 2 4 + ( 15 ) 2 3 + ( 5.5 ) 2 3 ( 16.5 + 15 + 15.5 ) 2 10
S S ( b e t w e e n ) = 2.2458
S ( t o t a l ) = x 2 ( x ) 2 n
  = ( 5 2 + 4.5 2 + 4 2 + 3 2 + 3.5 2 + 7 2 + 4.5 2 + 8 2 + 4 2 + 3.5 2 )
( 5 + 4.5 + 4 + 3 + 3.5 + 7 + 4.5 + 8 + 4 + 3.5 ) 2 10
= 244 47 2 10 = 244 220.9
S S ( t o t a l ) = 23.1
S S ( w i t h i n ) = S S ( t o t a l ) S S ( b e t w e e n )
=   23.1 2.2458
S S ( w i t h i n ) = 20.8542

One-Way 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
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Try it

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

  • bare soil
  • a commercial ground cover
  • black plastic
  • straw
  • compost

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 one-way 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.

One-Way 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 36 , 648 , 561 4 = 9 , 162 , 140 9 , 162 , 140 2 , 044 , 672.6 = 4.4810
Error (Within) 20,446,726 15 – 5 = 10 20 , 446 , 726 10 = 2 , 044 , 672.6
Total 57,095,287 15 – 1 = 14
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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 H 0 .

Notation

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 μ = d f ( n u m ) d f ( d e n o m ) 1

References

Tomato Data, Marist College School of Science (unpublished student research)

Chapter review

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.

Formula review

  S S between = [ ( s j ) 2 n j ] ( s j ) 2 n  

S S total = x 2 ( x ) 2 n

S S within = S S total S S between

df between = df ( num ) = k – 1

df within = df(denom) = n k

MS between = S S between d f between

MS within = S S within d f within

F = M S between M S within

F ratio when the groups are the same size: F = n s x ¯ 2 s 2 p o o l e d

Mean of the F distribution: µ = d f ( n u m ) d f ( d e n o m ) 1

where:

  • k = the number of groups
  • n j = the size of the j th group
  • s j = the sum of the values in the j th group
  • n = the total number of all values (observations) combined
  • x = one value (one observation) from the data
  • s x ¯ 2 = the variance of the sample means
  • s 2 p o o l e d = the mean of the sample variances (pooled variance)

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 one-way 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 Factor?

4,939.2

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What is the Sum of Squares Error?

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What is the df for the numerator?

2

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What is the df for the denominator?

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What is the Mean Square Factor?

2,469.6

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What is the Mean Square Error?

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What is the F statistic?

3.7416

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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 one-way 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 the df for the numerator?

3

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What is SS within ?

13.2

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What is the df for the denominator?

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What is MS within ?

0.825

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What is the F statistic?

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Judging by the F statistic, do you think it is likely or unlikely that you will reject the null hypothesis?

Because a one-way ANOVA test is always right-tailed, 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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Source:  OpenStax, Introductory statistics. OpenStax CNX. May 06, 2016 Download for free at http://legacy.cnx.org/content/col11562/1.18
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