ANOVA

SINGLE FACTOR

Fisher Test for Differences

SITUATION:

  • Sampling From 3 or more Populations Based on a Single Factor
  • No common link between the k-th observation from Population 1 and the k-th observation of Population 2, or 3, etc.
  • Have determined that there ARE differences in the means

QUESTION:

Which means differ from the others?

Example:

Can we conclude that there are differences inaverage battery life of 5 different types of notebook computer batteries? And if so, which ones differ from the others?

The Statistical F-Test:

H0:A = B = C = D= E

H1: At least one of these j's does not equal the others

So which means differ from the others?

We can use the FISHER approach.

The FISHER Approach

The Fisher approach states that two sample means ('s) that differs by "a lot" indicates that their corresponding population means ('s) differ.

The Fisher approach defines "a lot" as the LEAST SIGNIFICANT DIFFERENCE (LSD):

Here:

MSE= Mean Square Error (found on Excel printout)

n1= Number sampled from the first population

n2= Number sampled from the second population

NOTES:

1.SAME SAMPLE SIZES

LSD must be recalculated for each comparison since the ni’s can vary. If, as in the case illustrated here the sample sizes (ni’s) are all the same value, n, only one LSD need be calculated and can be used for all comparisons. The above formula reduces to:

2.WHAT ARE THE DEGREES OF FREEDOM?

The degrees of freedom to use in tα/2 is the degrees of freedom for ERROR (Within Groups)

3.WHAT IS THE VALUE OF α?

Recall that α = Probability of concluding there is a difference when there is not. If we are interested only in comparing the differences of two of the populations, choose α in the usual manner, say α = .05.

But if we are comparing k different groups (here k = 5), we will doing many hypothesis tests: µ1 to µ3, µ2 to µ4, µ4 to µ5, etc. The total number of tests that will be done is labeled C. C can be found by the following formula:

If overall (experiment-wise) we a probability of only .05, (we call that αEW = .05) that we conclude in there are differences in at least one of the tests when there is not, then α for each individual test must be reduced as follows:

Thus if we are making multiple comparisons (for all possibilities), the value of α used in LSD should be α = αEW/C, i.e. LSDEW is found by:


STEP 1:

By Excel: First Copy and Paste the Averages ('s) and the Groups, in that order to another set of cells on the spreadsheet.

STEP 2: Order the averages by doing a Z to A sort on cells G13:H17.

STEP 3: Calculate the number of comparisons C by k(k-1)/2.

STEP 4: Calculate α to be used in each test by α = αEW/C.

STEP 5: Calculate the LSD(EW)’s by formula above.

In this case we only need to calculate one since the ni’s are equal.

Thus any two sample means that differ by more than 17.2355 indicates that the corresponding population means differ.

Battery E:

150 -140 = 10 --- cannot conclude a difference between E and D

150 - 121.6667 = 28.3333 -- can conclude E differs from A (and C and B).

Battery D:

140 - 121.6667 = 18.3333 -- can conclude D differs from A (and C and B).

Battery A:

121.6667 - 100 = 21.667 -- can conclude A differs from C (and B).

Battery C:

100 - 90 = 10 --- cannot conclude a difference between C and B