Question

The following table provides the data of the pumpkins that were used at two halloween parties.

Party A	Party B
Mean weight (kg)	4.5	Mean weight (kg)	6
Sum of all deviations	12	Sum of all deviations	18
Total number of pumpkins used	50	Total number of pumpkins used	75

Are the pumpkins at one party significantly heavier than the pumpkins at the other party?

Answer 1

Party A:
Mean = 4.5
$\text{MAD}=\frac{\text{Sum of all deviations}}{\text{Total number of pumpkins used}}=\frac{12}{50}=0.24$

Party B:
Mean = 6
$\text{MAD}=\frac{\text{Sum of all deviations}}{\text{Total number of pumpkins used}}=\frac{18}{75}=0.24$
Thus, the MAD for both parties A and B are the same.

We can describe the variation in weight by expressing the difference in means as a multiple of MAD:
$=\frac{\text{Difference in mean weight}}{\text{MAD}}=\frac{6-4.5}{0.24}=\frac{1.5}{0.24}=6.25>2$

The quotient is more than 2, so the pumpkins at one party are significantly heavier than the pumpkins at the other party.