7.10: Collaborative Activity

For this collaborative activity you will need a standard deck of cards, a ten-sided die, a twenty-sided die, and someone in your group who is good at shuffling cards. We will simulate an experiment where a group of patients with a minor medical condition are recruited for a medical study. Each patient will either receive a standard treatment or a new treatment where which treatment each individual receives will be randomly determined by drawing cards from the shuffled deck of cards. According to the treatment determined by the card draw, a dice roll will then determine whether the treatment was successful. The data gathered from the experiment will be used to calculate probabilities, odds, and odds ratios.

To begin, designate an individual in your group as the person who will shuffle the deck of cards and will successively draw cards from the top of the deck. Another individual will oversee rolling the dice, and a third individual will write down the simulated data on the provided data sheet shown in Table 7.7.

Table 7.7 Data collection table for the collaborative activity.

Replication	Card	Dice Roll	Treatment	Outcome
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25

The activity begins by shuffling the deck of cards. Shuffle them at least five times so that the order of the cards is mixed up well. Drawing from the deck will then be approximately equivalent to random draws from the deck of cards.

Next, repeat the following process for twenty-five replications:

1. Selected a card from the top of the deck and write the in the Card column of the data table on the line corresponding to the current replication. You can us a two-character abbreviation to write down the results using 2–10, J, Q, K, and A to represent the value of the card and C, D, H, and S to represent the card suit. For example, 2H will represent the two of hearts.

2. If the card that has been selected is a heart or a diamond, roll the twenty-sided die, and record the result in the Dice Roll column of the data table on the line corresponding to the current replication.

3. If the card that has been selected is a club or a spade, then roll the ten-sided die, and write the result in the Dice Roll column of the data table on the line corresponding to the current replication.

Table 7.8 gives an example of what the first two columns of the data table for ten hypothetical replications of this process. From the table we can observe that the first card drawn from the top of the deck was the five of spades, which has been represented by 5S on the table in the Card column for the first replication. Because this card is from the spade suit, the ten-sided die was rolled and the resulting roll was 3, which has been recorded in the Dice Roll column for the first replication. The second card drawn from the top of the deck was the two of hearts, which has been represented by 2H on the table in the Card column for the second replication. Because this card is from the heart suit, the twenty-sided die was rolled and the resulting roll was 5, which has been recorded in the Dice Roll column for the second replication. The remainder of the table was filled in for the remaining eight replications of this process.

Table 7.8 An example of ten replications of the collaborative activity. For the Card column the card value (A,2–10,J,Q,K) is given first and the suite second (C = club, D = diamond, H = heart, S = spade). The first card, marked by 5S represents the five of spades.

Replication	Card	Dice Roll	Treatment	Outcome
1	5S	3	N	C
2	2H	9	O	N
3	KC	8	N	N
4	QD	3	O	C
5	KD	16	O	N
6	AC	4	N	C
7	8D	20	O	N
8	10H	18	O	N
9	4D	19	O	N
10	9C	2	N	C

The next step is to translate the card draws and dice rolls into what is simulated in our medical study. Each replication represents an individual in the study. The suit of the card draw represents whether they are given the standard treatment or the new treatment for their medical condition. Each replication where a drawn card is either a club or spade corresponds to individuals who received the new treatment. Therefore, each time either a club or a spade is drawn, write N in the Treatment column of the data table in the line corresponding to that replication. The remaining replications, where either a diamond or a heart is drawn, corresponds to individuals who received the old treatment. Therefore, each time either a diamond or a heart is drawn, write O in the Treatment column of the data table in the line corresponding to that replication. In Table 7.8 we can observe that the first card was a spade so that N has been written in the Treatment column for the first replication. Similarly, the second card was a heart so that O has been written in the Treatment column for the second replication. The remaining replications in the table are treated in a similar manner.

Next, we need to determine whether the patient is cured or not. This will be determined by the dice roll. Regardless of whether the ten or the twenty-sided die is used we will conclude that the individual is cured of the medical condition if the dice roll is 7 or lower. For each replication where the dice roll is less than or equal to 7, write C in the Outcome column of the data table to symbolize that the individual is cured. If the die roll is 8 or larger, write N in the Outcome column of the data table to symbolize that the individual is not cured. In Table 7.8 we can observe that the first die roll was 3 so that C has been written in the Outcome column for the first replication. Similarly, the second die roll was 5 so that C has been written in the Outcome column for the second replication. For the third replication the die roll was 8 so that an N has been written in the Outcome column for the third replication. The remaining replications in the table are treated in a similar manner.

Calculations:

1. After the data table has been completed, construct a table of the same form as Table 7.3. Use the old treatment group as the control group and the new treatment group as the treatment group. Hence, using the notation in Table 7.3, the value of \(a\), will be equal to the observed number of times that a person with the new treatment was not cured, the value of \(b\) will be equal to the number of times that a person with the new treatment was cured, the value of \(c\) will be equal to the number of times that a person with the old treatment was not cured, and the value of \(d\) will be equal to the number of times that a person with the old treatment was cured. Table 7.9 shows this table for the example data given in Table 7.8.

Table 7.9 Data for comparing the cure of the condition for the new treatment and the old treatment groups for the data given in Table 7.8.

	Condition Cured?
	No	Yes
New Treatment	1	3
Old Treatment	5	1

2. Next, we will compute the probabilities and the odds that the old and new treatments cured the medical condition based on what was observed in the experiment. To compute the observed probability that the new treatment cured the medical condition, take the number of individuals cured by the new treatment and divide it by the number of people who received the new treatment. A similar calculation is used to compute the observed probability that the old treatment cured the medical condition. For the example data given in Tables 7.8 and 7.9, the observed probability that the new treatment cured the medical condition is \(3/4\), and for the old treatment the observed probability is \(1/6\). Using the methods outlined in this chapter, the odds that the new treatment cured the medical condition are 3 to 1, and the odds that the old treatment cured the medical condition are 1 to 5.
3. Finally, we will compute the observed odds ratio for not being cured with the new treatment compared to the old treatment using the formula given in this chapter. For the example data given in Tables 7.8 and 7.9, the observed odds ratio is

\[\text{odds ratio} = \frac{a\times d}{b\times c} = \frac{1\times 1}{3\times 5}=\frac{1}{15} \nonumber \]

Questions:

1. Because we know how the random mechanisms work that created the data, we know the values of all the probabilities associated with the different outcomes in the data. Recall that everyone received the new treatment if the corresponding card drawn from the deck was a club or a spade, and they received the old treatment otherwise. Assuming that the cards are drawn at random and that each card has an equal chance of being drawn for everyone, what are the probabilities associated with being assigned the new and the old treatments? Convert these probabilities into chances and odds.

2. Suppose that an individual has been assigned to the new treatment. Based on how the data was generated, what is the probability that they will not be cured? What is the probability that they will be cured? Convert these probabilities into chances and odds.

3. Suppose that an individual has been assigned to the old treatment. Based on how the data was generated, what is the probability that they will not be cured? What is the probability that they will be cured? Convert these probabilities into chances and odds.

4. Using the probabilities and odds given above, compute the odds ratio for not being cured with the new treatment compared to the old treatment using the formula given in this chapter.

5. Now compare the probabilities, odds, and odds ratio computed from the observed data with what was computed based on the random model in the problems above. Comment on the differences and explain why the values computed from the observed data differ from what was computed using the true random model for the data.

6. What do the odds ratios indicate about how well the new treatment does when compared to the old treatment? Does this conclusion make sense based on how the date were created?