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Ch 13: F Distribution and ANOVA

Module by: Susan Dean, Barbara Illowsky, Ph.D.. E-mail the authors

Summary: This module is the complementary teacher's guide for the "F Distribution and ANOVA" chapter of the Collaborative Statistics collection (col10522) by Barbara Illowsky and Susan Dean.

History:

The FF size 12{F} {} distribution is named after Ronald Fisher. Fisher is one of the most respected statisticians of all time. He did a lot of statistical work in biology and genetics and became chair of genetics at Cambridge University in England in 1949. In 1952, he was awarded knighthood.

This section is a very brief overview of the FF size 12{F} {} distribution and two of its applications - One Way Analysis of Variance (ANOVA) and test of two variances. There are college courses which deal exclusively with these topics. ANOVA, particularly, is used regularly in industry.

Explanation of Sum of Squares, Mean Square, and the F ratio for ANOVA

  • kk size 12{k} {} = the number of different groups
  • njnj size 12{n rSub { size 8{j} } } {} = the size of the jthjth size 12{ ital "jth"} {} group
  • sjsj size 12{s rSub { size 8{j} } } {}= the sum of the values in the jthjth size 12{ ital "jth"} {} group
  • NN size 12{N} {} = the total number of all the values combined
  • Total sample size: ∑nj∑nj size 12{ Sum {n rSub { size 8{j} } } } {}
  • xx = one value: ∑x=∑sj∑x=∑sj size 12{ Sum {x} = Sum {s rSub { size 8{j} } } } {}
  • Sum of squares of all values from every group combined: ∑x2∑x2 size 12{ Sum {x rSup { size 8{2} } } ={}} {}
  • Between group variability: SS total = ∑ x 2 − ∑ x 2 N SS total = ∑ x 2 − ∑ x 2 N size 12{ ital "SS" rSub { size 8{ ital "total"} } = Sum {x rSup { size 8{2} } } - { { left ( Sum {x} right ) rSup { size 8{2} } } over {N} } } {}
  • Total sum of squares: ∑ x 2 − ( ∑ x ) 2 N ∑ x 2 − ( ∑ x ) 2 N size 12{ Sum {x rSup { size 8{2} } } - { { \( Sum {x} \) rSup { size 8{2} } } over {N} } } {}
  • Explained variation- sum of squares representing variation among the different samples SS between = ∑ [ ( sj ) 2 n j ] − ( ∑ s j ) 2 N SS between = ∑ [ ( sj ) 2 n j ] − ( ∑ s j ) 2 N size 12{ ital "SS" rSub { size 8{ ital "between"} } = Sum { \[ { { \( ital "sj" \) rSup { size 8{2} } } over {n rSub { size 8{j} } } } \] } - { { \( Sum {s rSub { size 8{j} } \) rSup { size 8{2} } } } over {N} } } {}
  • Unexplained variation- sum of squares representing variation within samples due to chance: SS within = SS total − SS between SS within = SS total − SS between size 12{ ital "SS" rSub { size 8{ ital "within"} } = ital "SS" rSub { size 8{ ital "total"} } - ital "SS" rSub { size 8{ ital "between"} } } {}
  • df's for different groups (df's for the numerator): df between = k − 1 df between = k − 1 size 12{ ital "df" rSub { size 8{ ital "between"} } =k - 1} {}
  • Equation for errors within samples (df's for the denominator): df within = N − k df within = N − k size 12{ ital "df" rSub { size 8{ ital "within"} } =N - k} {}
  • Mean square (variance estimate) explained by the different groups: MS between = SS between df between MS between = SS between df between size 12{ ital "MS" rSub { size 8{ ital "between"} } = { { ital "SS" rSub { size 8{ ital "between"} } } over { ital "df" rSub { size 8{ ital "between"} } } } } {}
  • Mean square (variance estimate) that is due to chance (unexplained): MS within = SS within df within MS within = SS within df within size 12{ ital "MS" rSub { size 8{ ital "within"} } = { { ital "SS" rSub { size 8{ ital "within"} } } over { ital "df" rSub { size 8{ ital "within"} } } } } {}
  • F ratio or F statistic of two estimates of variance: F = MS between MS within F = MS between MS within size 12{F= { { ital "MS" rSub { size 8{ ital "between"} } } over { ital "MS" rSub { size 8{ ital "within"} } } } } {}

Note:

The above 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

F = n ⋅ ( s − x ) 2 ( s pooled ) 2 F = n ⋅ ( s − x ) 2 ( s pooled ) 2 size 12{F= { {n cdot \( s rSub { size 8{ - x} } \) rSup { size 8{2} } } over { \( s rSub { size 8{ ital "pooled"} } \) rSup { size 8{2} } } } } {}
(1)

where...

  • (s−x)2=(s−x)2= size 12{ \( s rSub { size 8{ - x} } \) rSup { size 8{2} } ={}} {}the variance of the sample means
  • n=n= size 12{n={}} {}the sample size of each group
  • (spooled)2=(spooled)2= size 12{ \( s rSub { size 8{ ital "pooled"} } \) rSup { size 8{2} } ={}} {}the mean of the sample variances (pooled variance)
  • df numerator = k − 1 df numerator = k − 1 size 12{ ital "df" rSub { size 8{ ital "numerator"} } =k - 1} {}
  • df denominator = k ( n − 1 ) = N − k df denominator = k ( n − 1 ) = N − k size 12{ ital "df" rSub { size 8{ ital "denominator"={}} } k \( n - 1 \) =N - k} {}

These calculations are easily done with a graphing calculator or a computer program. We present the information in the chapter assuming some kind of technology will be used.

For ANOVA, the samples must come from normally distributed populations with the same variance, and the samples must be independent. The ANOVA test is right-tailed.

In a test of two variances, the samples must come from normal populations and must be independent of each other.

Exercise 1

(One-Way ANOVA)

Three different diet plans are to be tested for average weight loss. For each diet plan, 4 dieters are selected and their weight loss (in pounds) in one month's time is recorded.

Table 1
Plan 1 Plan 2 Plan 3
5 3.5 8
4.5 7 4
4 6 3.5
3 4 4.5

Is the average weight loss the same for each plan? Conduct an ANOVA test with a 1% level of significance.

Solution

Let μ1μ1 size 12{μ rSub { size 8{1} } } {}, μ2μ2 size 12{μ rSub { size 8{2} } } {}, and μ3μ3 size 12{μ rSub { size 8{3} } } {} be the population means for the three diet plans.

  • H o : μ 1 = μ 2 = μ 3 H o : μ 1 = μ 2 = μ 3 size 12{H rSub { size 8{o} } :μ rSub { size 8{1} } =μ rSub { size 8{2} } =μ rSub { size 8{3} } } {}
  • Ha:Ha: size 12{H rSub { size 8{a} } :} {} Not all pairs of means are equal.
  • df numerator = 3 − 1 = 2 df numerator = 3 − 1 = 2 size 12{ ital "df" rSub { size 8{ ital "numerator"} } =3 - 1=2} {}
  • df deno min ator = 12 − 3 = 9 df deno min ator = 12 − 3 = 9 size 12{ ital "df" rSub { size 8{ ital "deno""min" ital "ator"} } ="12" - 3=9} {}

The distribution for the test is F2,9F2,9 size 12{F rSub { size 8{2,9} } } {}

Using a calculator or computer, the test statistic is F=0.47F=0.47 size 12{F=0 "." "47"} {}. The notation used for the FF size 12{F} {} statistic may also be F'F' size 12{F'} {} or F2,9F2,9 size 12{F rSub { size 8{2,9} } } {} (like the distribution). The TI-83/84 series has the function ANOVA in STAT TESTS. Enter the lists of data separated by commas.

If you use the formulas for groups of the same size, the calculations are as follows:

Sample means are 4.13, 5.13, and 5, respectively. Sample standard deviations are 0.8539, 1.6250, and 2.0412, respectively.

Table 2
( s − x ) 2 = 0 . 2956 ( s − x ) 2 = 0 . 2956 size 12{ \( s rSub { size 8{ - x} } \) rSup { size 8{2} } =0 "." "2956"} {} The variance of the sample means
( s pooled ) 2 = 2 . 5416 ( s pooled ) 2 = 2 . 5416 size 12{ \( s rSub { size 8{ ital "pooled"} } \) rSup { size 8{2} } =2 "." "5416"} {} The mean of the sample variances
n = 4 n = 4 size 12{n=4} {} The sample size of each group
F = 4 ⋅ 0 . 2956 2 . 5416 F = 4 ⋅ 0 . 2956 2 . 5416 size 12{F= { {4 cdot 0 "." "2956"} over {2 "." "5416"} } } {}
(2)

Probability Statement: P ( F > 0 . 47 ) = 0 . 6395 P ( F > 0 . 47 ) = 0 . 6395 size 12{P \( F>0 "." "47" \) =0 "." "6395"} {}

Figure 1: pp size 12{p} {}-value =0.6395=0.6395 size 12{ {}=0 "." "6395"} {}
Probability graph showing the all values greater than .47 shaded

Since α<pα<p size 12{a<p} {}-value, do not reject HoHo size 12{H rSub { size 8{o} } } {}.

There is not sufficient evidence to conclude that the three diet plans are different. It appears that the three diet plans work equally well. The average weight loss is the same for all three plans.

Exercise 2

(Test of Two Variances):

Machine A makes a box and machine B makes a lid. For the lid to fit the box correctly, the variances should be nearly the same. There is a suspicion that the variance of the box is greater than the variance of the lid. The following data was collected.

Table 3
  Machine A (Box) Machine B (Lid)
Number of Parts 9 11
Variance 150 45

Are the machines working properly? Test at a 5% level of significance.

Solution

Let σA2σA2 size 12{σ rSub { size 8{A rSup { size 6{2} } } } } {} and σB2σB2 size 12{σ rSub { size 8{B rSup { size 6{2} } } } } {}be the population variances for machine A and machine B, respectively

  • H o : σ A 2 = σ B 2 H o : σ A 2 = σ B 2 size 12{H rSub { size 8{o} } :σ rSub { size 8{A rSup { size 6{2} } } } =σ rSub {B rSup { size 6{2} } } } {}
  • H a : σ A 2 > σ B 2 H a : σ A 2 > σ B 2 size 12{H rSub { size 8{a} } :σ rSub { size 8{A rSup { size 6{2} } } } >σ rSub {B rSup { size 6{2} } } } {}
  • n A = 9 n A = 9 size 12{n rSub { size 8{A} } =9} {}
  • n B = 11 n B = 11 size 12{n rSub { size 8{B} } ="11"} {}
  • df numerator = 9 − 1 df numerator = 9 − 1 size 12{ ital "df" rSub { size 8{ ital "numerator"} } =9 - 1} {}
  • df deno min ator = 11 − 1 = 10 df deno min ator = 11 − 1 = 10 size 12{ ital "df" rSub { size 8{ ital "deno""min" ital "ator"} } ="11" - 1="10"} {}

The distribution for the hypothesis test is F8,10F8,10 size 12{F rSub { size 8{8,"10"} } } {}

If you are using the TI-83/84 calculators, use the function 2-SAMPFTest for the test.

Using the formulas,

The test statistic is F=[(sA)2(σA)2][(sB)2(σB)2]=F=[(sA)2(σA)2][(sB)2(σB)2]= size 12{F= { { \[ { { \( s rSub { size 8{A} } \) rSup { size 8{2} } } over { \( σ rSub { size 8{A} } \) rSup { size 8{2} } } } \] } over { \[ { { \( s rSub { size 8{B} } \) rSup { size 8{2} } } over { \( σ rSub { size 8{B} } \) rSup { size 8{2} } } } \] } } ={}} {}

( s A ) 2 ( s B ) 2 = 150 45 = 3 . 3 3 ¯ ( s A ) 2 ( s B ) 2 = 150 45 = 3 . 3 3 ¯ size 12{ { { \( s rSub { size 8{A} } \) rSup { size 8{2} } } over { \( s rSub { size 8{B} } \) rSup { size 8{2} } } } = { {"150"} over {"45"} } =3 "." 3 {overline {3}} } {}

Since α>pα>p size 12{a>p} {}-value, reject the null hypothesis.

There is sufficient evidence to conclude that the box and lid do not fit each other. The variance of the box is larger.

Assign Practice

Have the students work collaboratively to complete the Practice.

Assign Homework

Assign Homework. Suggested homework: 1, 3, 4, 5.

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