Question

You want to see if three different cafes yield different costs for a lunch meal for a week. You randomly select five measurements from trials on an automated driving machine for each cafe. At the 0.05 significance level, is there a difference among cafes in mean lunch meal cost? Test this claim showing all calculation steps clearly in the Ms World file (25P) and draw an ANOVA Table for the solution

Cafe 1 35 23 34 34 39

Cafe 2 20 28 26 22 18

Cafe 3 25 25 34 38 42

Answer 1

Null Hypothesis: There is no difference between cafes in mean lunch cost.

Alternative Hypothesis: There is a difference between at least one pair of cafes.

Level of significance:

Let us calulate the Totals and squares for getting teh sum of squares in the ANOVA table.

						Total
Cafe 1	35	23	34	34	39	165
Cafe 2	20	28	26	22	18	114
Cafe 3	25	25	34	38	42	164
						443

Let us denote be denoted by the observation at the   cafe.

There are k=3 cafes and n=3*5=15 observations in total.

1. Correction factor:

2. Total sum of squares:

We need the individual square and the sum. Let us use the following table of squares and then add up teh values:

Cafe 1	1225	529	1156	1156	1521
Cafe 2	400	784	676	484	324
Cafe 3	625	625	1156	1444	1764
						13869

3. Sum of squares due to cafe:

4. Error Sum of squares:

5. Degrees of freedom:

Total=15-1=14

Cafes:3-1=2

Error df=14-2=12

We shall now form the ANOVA table:

Source of Variation	SS	df	MS	F	P-value	F crit
Between Cafes	340.1333	2	170.0667	4.579892	0.033267	3.885294
Error	445.6	12	37.13333
Total	785.7333	14

Decision: From the ANOVA table above, we see that the p-value is 0.033<0.05 and hence we reject the Null hypothesis. Hence, we conclude that there is enough evidence to claim that there is a significant difference between at least two cafes.