## Question 625:

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 Cell Type output 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: A, B, C **

Source DF SS MS F P

Factor 2 80.11 40.06 9.44 0.002

Error 15 63.67 4.24

Total 17 143.78

S = 2.060 R-Sq = 55.72% R-Sq(adj) = 49.81%

Individual 95% CIs For Mean Based on

Pooled StDev

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

A 6 123.83 2.04 (------*-------)

B 6 123.00 2.00 (------*------)

C 6 127.83 2.14 (------*-------)

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

122.5 125.0 127.5 130.0

Pooled StDev = 2.06

We see that our F statistics is 9.44 and has a p-value of .002 associated with it. The p-value is found by using a table of F-values with 2 degrees of freedom in the numerator (for the between group variation) and 15 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 .002 is less than the typical alpha of value of .05 as well as less than an alpha of .01. Such a low p-value indicates that the difference between means we observe has less than a 1% chance of being due to chance alone. We would reject the null hypothesis and conclude there is a difference between at least one of the Cell-Type means. A look at the confidence intervals suggests this difference is between Type C and A and between Type C and 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 3.682, which is found using the excel function =FINV(.05,2,15), assuming an alpha of .05

The test statistics is 9.44, which is above the critical F value, so we see that we reject the null.

The p-value is .002

We reject the null hypothesis