Question

The two-way table summarizes data from an experiment comparing the effectiveness of three different diets (A, B, and C) on weight loss. Researchers randomly assigned $300$volunteer subjects to the three diets. The response variable was whether each subject lost weight over a $1-$year period.

a. Suppose we randomly select one of the subjects from the experiment. Show that the events “Diet B” and “Lost weight” are independent.

b. Copy and complete the table so that there is no association between type of diet and whether a subject lost weight.

c. Copy and complete the table so that there is an association between type of diet and whether a subject lost weight.

Short answer

Part a. The events “diet B” and “Lose weight” are independent.

Part b. Two-way table

Diet

		A	B	C	Total
Lose weight	Yes	$54$	$60$	$66$	$180$
	No	$36$	$40$	$44$	$120$
	Total	$90$	$100$	$110$	$300$

Part c. Two – way table:

Diet

		A	B	C	Total
Lose weight	Yes	$60$	$60$	$60$	$180$
	No	$30$	$40$	$50$	$120$
	Total	$90$	$100$	$110$	$300$

Step-by-step solution

Part a. Step 1. Given information

Two – way table comparing the effectiveness of three different diets (A, B, and C) on weight loss:

Part a. Step 2. Explanation

The two events are independent, if the probability of occurrence of one event does not affect the probability of occurrence of other event.

For the events “Diet B” and “Lose weight” to be independent,

The product of row total and column total, divided by the table total should be equal to the count in the table.

$$\begin{gathered} \frac{Totalcount"Yes"\times Totalcount"B"}{Tabletotalcount} \\ =\frac{180\times 100}{300} \\ =\frac{180}{3} \\ =60 \end{gathered}$$

In the table, we can see that the count in row “Yes” and column “B” is already $60$.

Thus,

The events “Diet B” and “Lose weight” are independent.

Part b. Step 1. Explanation

Two events are independent, when the probability of occurrence of one event does not affect the probability of occurrence of other event.

Then

The counts will be the product of the row total and the column total, divided by the table total provided in the bottom left corner of the table.

Calculate the counts in the two – way table:

Diet

		A	B	C	Total
Lose weight	Yes	$\frac{90\times 180}{300}$	$60$	$\frac{110\times 180}{300}$	$180$
	No	$\frac{90\times 120}{300}$	$40$	$\frac{110\times 120}{300}$	$120$
	Total	$90$	$100$	$110$	$300$

Thus,

The two – way table becomes:

Diet

		A	B	C	Total
Lose weight	Yes	$54$	role="math" localid="1664867620776" $60$	$66$	$180$
	No	$36$	role="math" localid="1664867626081" $40$	$44$	$120$
	Total	$90$	$100$	$110$	$300$

Part c. Step 1. Explanation

For two – way table, where an association exists between type of diet and whether a subject lost weight.

In this part, the count for column “A” and row “Yes” should be different from Part (b).

Suppose, if we choose the count 60 for column “A” and row “Yes” instead of 54 (in Part (b)).

Then

Put 60 in the column “A” and the row “Yes”.

And

Put the remaining counts according to the total counts of the rows and columns.

Thus,

The two – way table becomes:

Diet

		A	B	C	Total
Lose weight	Yes	$60$	$60$	$60$	$180$
	No	$30$	$40$	$50$	$120$
	Total	$90$	$100$	$110$	$300$