We saw that for paired differences we had the hypothesis that

        H0:  m1 = m2

        H1:   m1 m2   

What if there are many types and we want to see if it makes a difference which we look at?  We could test the appropriate hypothesis for each pair, but if we test for enough pairs, then we are bound to find two different ones even if they are not any (Eventually we will get unlucky).  For example if there were 12 means that we wanted to test to see if they were all the same, then we would have to test C(12,2)  =  66 pairs.  If we used a level of significance of 0 .05, then if the populations actually had the same means, then we could expect that on average we would reject the null hypothesis (66)(0.05)  =  3.3 times.  We need a test where it is rare to make the mistake of saying that a pair of means differ when they really are the same. 

  Instead we will do an ANOVA (ANalysis Of VAriance) test for multiple means.  We call each population a treatment.  The test statistic turns out to be the "F" statistic that we saw when we looked at when comparing variances.  The calculation to arrive at the test statistic is quite complicated, hence we will assume that either the reader will be looking at a textbook for this, or rely on a computer.

In order to use ANOVA, one must make the following (not always reasonable) assumptions.

        1.  Each of the populations follows approximately normal distribution.

        2.  Each sample is randomly selected and independent of every other sample.

        3.  The standard deviations of all the populations are approximately equal to each other.

Since we are looking at an F statistic, we need the numerator and denominator degrees of freedom.  Let N be the total sample size, that is, the sum of each of the sample sizes.  Let k be the number of treatments (number of samples).  Then

        d.f.N  =  k - 1  (numerator degrees of freedom)

        d.f.D  =  N - k   (denominator degrees of freedom)

ANOVA is always a right tailed test, hence the table will give the true P-value (we never need to to multiply by 2).



Does it make a difference which type of car we buy in terms of cost of maintenance?  We test 10 American, 20 Japanese, 30 Korean, and 44 German cars.  Suppose that the F-statistic was computed to be 3.2.  what can be concluded at the 0.05 level of significance?



We have

        H0:   m1 = m2 =  m3 = m4

        H1:   mi mj for some i not j (At least two are different)

We use two types of degrees of freedom: 


        N = the total of all the n


        k = the number of samples.  

In our case 

        N = 10 + 20 + 30 + 44 = 104 


        k = 4 

then we have

        k - 1 = numerator degrees of freedom and

        N - k = denominator degrees of freedom

In our example we have

        3 numerator degrees of freedom and

        100 denominator degrees of freedom.

Since F = 3.2, we use the table to find that

        0.025 < P < 0.050

In particular, the P-value is smaller than the level of significance, hence we can reject the null hypothesis and accept the alternative hypothesis. Hence, we can conclude that it does make a difference which car one buys in terms of average cost of maintenance.  


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