STAT 211 Topic 9

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Lecture 18 Notes

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Topic 8 covered comparison of 2 populations based on:

  • population mean
  • population proportion
  • population variable


ANOVA

ANalysis Of VAriance — extension of pooled t-test.

Suppose we have many samples from many populations. Each population has it's own mean… Are all population means equal?

Assumptions:

  1. All populations have normal distributions
  2. All samples are independent
  3. All population variances are equal

We use the F-test due to sums of squares (since t2=F)


Procedure

Set up the following hypothesis test:

  • H0: μ1=μ2==μI
  • Ha: H0 is not true. (at least one μ is different)


1. Sums of Squares

Total
SStot=i,j(xijx¯)2
Treatment
SStrt=i,j(xix¯)2
Error
SStrt=i,j(xijx¯i)2

Where:

  • xij represents data of j-th sample subject in i-th population (treatment).
  • x¯i=1Jj=1Jxij, J represents number of subjects in each sample (x¯i is the average of each population's sample)
  • x¯=1IJi,jxij, I represents number of treatments or populations


2. Mean Squares

We can show that SStot=SStrt+SSerr:

Treatment Mean Square
MStrt=SStrtDFtrt, where DFtrt is the treatment degrees of freedom: I1
Error Mean Square
MSerr=SSerrDFerr, where DFerr is the error degrees of freedom: I(J1)

Therefore, DFtot=DFtrt+DFerr=IJ1


3. Test Statistic

Think about the ratio:

f=MStrtMSerrFDFtrt, DFerr
(f follows F-distribution with df DFtrt and DFerr, respectively)

If H0 is true, then f ≈ 1.

4. ANOVA Table

Source DF SS MS f
Treatment I1 SStrt MStrt=SStrtI1 f=MStrtMSerr
Error I(J1) SSerr MSerr=SSerrI(J1)
Total N1 SStot  

Perform F-test and reject H0 if f>Fα,DFtrt,DFerr


Example

Study effects of diet pills (four different brands):

  1. Randomly assign 20 women to each of 5 different groups: one for each diet pill and one placebo
  2. Let women take pills for a month and record weight loss (in pounds):
Group 1 2 3 4 5 (placebo)
Average Loss (x¯) 14 12 10 8 6
Standard Deviation (s) 1.3 1.5 0.8 1.0 1.7

If we wanted to see the effectiveness of one brand, do a pooled t-test between that group and group 5 (placebo):

  • H0: μ1=μ2
  • Ha: μ1>μ2

ANOVA Table:

Source DF SS MS f
Treatment 4 800 200 118.0638
Error 95 160.93 1.694
Total 99 960.93  

If F0.05, 4, 95 = 2.47 and F0.01, 4, 95 = 3.52, P-value is almost 0 since 118.0638 > 2.47. Therefore, we find enough evidence to conclude that not all of the means are equal.


Multiple Comparison

ANOVA only tells us whether a population mean differs from the others. To find out which ones are different, we could perform multiple t-tests, but that would throw off our significance level.

Tukey's Procedure

Only works if all treatments contain same number of observations.

  1. Select α and find Qα,I,DFerr (Q-distribution)
  2. Determine w=Qα,I,DFerrMSerrJ, where J is the number of observations per treatment.
  3. List sample means in increasing order. Underline pairs that differ by less than w.
  4. Any pairs not underlined by same line are the ones that are significantly different