It’s A Math, Math World (ANOVA Part 2)

Today’s example is taken from the book, General Statistics, (2000) by Chase and Brown.

In previous weeks, we learned how to test a single hypothesis of the difference between two population means: i.e., test whether two means u1 and u2 are equal. What if we have more than two populations are we want to see if the means are equal? We want to compare more than 2 population means at the same time. This process is called Analysis of Variance (ANOVA).

Note we could conduct multiple pair-wise tests of the equality of means, but this would multiply the error rate considerably. In the ANOVA case, we test the following hypothesis:

H0: u1 = u2 = u3 = … = uk

HA: not all the means are equal

This week, we are NOT assuming equal sample sizes.

Swimmers Left Ventricular End Diastolic Volume (n1 =15)

  x1 (x1 – x1_bar) (x1 – x1_bar)2
  140 -41.733 1741.643
  140 -41.733 1741.643
  140 -41.733 1741.643
  148 -33.733 1137.915
  148 -33.733 1137.915
  175 -6.733 45.333
  185 3.267 10.673
  194 12.267 150.479
  194 12.267 150.479
  203 21.267 452.285
  203 21.267 452.285
  214 32.267 1041.159
  214 32.267 1041.159
  214 32.267 1041.159
  214 32.267 1041.159
SUM 2726   12926.929
x1­_bar 181.733    

 

Wrestlers Left Ventricular End Diastolic Volume (n2 =12)

  x2 (x2 – x2_Bar) (x2 – x2_Bar)2
  83 -27.167 738.046
  91 -19.167 367.374
  97 -13.167 173.370
  97 -13.167 173.370
  108 -2.167 4.696
  111 0.833 0.694
  111 0.833 0.694
  117 6.833 46.690
  117 6.833 46.690
  125 14.833 220.018
  125 14.833 220.018
  140 29.833 890.008
SUM 1322   2881.668
x2_Bar 110.167    

 

Example:

The 3 tables listed here contain measurements of heart size for 3 populations: swimmers, wrestlers and a control group of non-athletes. Left Ventricular End Diastolic Volume is the volume of the lower left heart chamber when it if filled with blood. Test that the mean size is the same for the 3 populations at 1% level of significance.

H0: u1 = u2 = u3

HA: not all of the population means are equal.

F-stat = (Estimate 1)/(Estimate 2)

Estimate #1 = Mean Squares Within (MSW) = Sum of Squares Within (SSW)/(n-k)

Estimate #2 = Mean Squares Between (MSB) = Sum of Squares Between (SSB)/(k-1)

Compare to F (0.01) with df = (k-1, n-k)

SSW = SUM(x1 – x1_bar)2 + SUM(x2 – x2_bar)2 + SUM (x3 – x3_bar)2

                = 12926.929 + 2881.668 + 437.756 = 20146.353

df = (n-k) = 43-3 = 40

MSW = 20146.353/40 = 503.659

Grand mean = ( 2726 + 1322 + 1614) /43 = 5662/43 = 131.674

 Left Ventricular End Diastolic Volume of Controls (n3 =16)

  x3 (x3 – x3_bar) (x3 – x3_bar)2
  64 -36.875 1359.766
  83 -17.875 319.516
  83 -17.875 319.516
  85 -15.875 252.016
  91 -9.875 97.516
  97 -3.875 15.016
  97 -3.875 15.016
  97 -3.875 15.016
  103 2.125 4.516
  108 7.125 50.766
  111 10.125 102.516
  111 10.125 102.516
  117 16.125 260.016
  117 16.125 260.016
  125 24.125 582.016
  125 24.125 582.016
SUM 1614   4337.756
X3_bar 100.875    

 

SSB = n1(x1_bar – grand_mean)2 + n2(x2_bar – grand_mean)2 + n3(x3_bar –grand_Mean)2

= 15(181.733 – 131.974)2 + 12(110.167 – 131.674)2 + 16(100.875 – 131.674)2

= 37588.552 + 5550.613 + 15177.254 = 58316.419

Df = (k-1) = 3-1 = 2

MSB = 58316.419/2 = 29158.210

Therefore, F-stat = 29158.210/503.659 = 57.89

Compare to F(0.01) with df = (2, 40)

From the table of the F-distribution, we have an F value of 5.18. Since 57.89 > 5.18, we reject the null hypothesis and conclude that not all of the 3 population means are equal.

Note: we have to have certain assumptions for this ANOVA calculation to be valid.

1) The sampled populations must be approximately normal.

2) There must be equal population standard deviations ; a property called homoscedasticity.

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