Question 629:

Asked on February 17, 2009

Tags: ANOVA

1

Answer:

No answer provided yet.

For this test we want to compare more than 2 means so we would use a 1-Way ANOVA. The Null hypothesis is that there is no difference between any means. The alternative hypothesis is that at least one mean is different than the other two.

 

The computations for the ANOVA are simple in concept, but a bit tedious to do by hand and should usually be left to a spreadsheet or other software program. The concept is that we want to compare the variation within each store with the variation BETWEEN each store. Ideally, we want low variation within each store and high variation between the stores, this would led us to conclude there is a difference. This ratio of with and between group variation makes up the test statistics F.

 

Below is the Minitab Output

One-way ANOVA: Store A, Store B, Store C

 

Source  DF      SS     MS     F      P

Factor   2   682.8  341.4  5.52  0.015

Error   16   989.0   61.8

Total   18  1671.8

 

S = 7.862   R-Sq = 40.84%   R-Sq(adj) = 33.45%

 

 

                           Individual 95% CIs For Mean Based on

                           Pooled StDev

Level    N    Mean  StDev  ----+---------+---------+---------+-----

Store A  6  22.500  5.788        (---------*---------)

Store B  8  32.750  9.633                        (--------*-------)

Store C  5  19.000  6.557  (----------*----------)

                           ----+---------+---------+---------+-----

                            14.0      21.0      28.0      35.0

 

Pooled StDev = 7.862

 

 

We see that our F statistics is 5.52 and has a p-value of .015 associated with it. The p-value is found by using a table of F-values with 2 degrees of freedom in the numerators (for the between group variation) and 16 degrees of freedom for the within group variation. Notice that the within group variation is called "Error," that's because, as I alluded to earlier, we want lower variability within each group, a higher amount of variability is indicative of more error in measurement.  So we see the p-value of .015 is less than the alpha of value of .05. This means we would reject the null hypothesis and conclude there is a difference between at least one of the means. A look at the confidence intervals suggests this difference is between store B and Store C and most likely between Store A and Store B. We could also conduct a post-hoc test between all mean pairs to confirm this suspicion.

 

In conclusion:

The critical F for this test is 3.6337, which is found using the excel function =FINV(.05,2,16).

The test statistics is 5.54, which is above the critical F, so we see that we reject the null.

The p-value is .015

We reject the null hypothesis