In the popular TV cartoon series, "The Simpsons," the Simpson family has five members: Homer, Marge, Bart, Lisa, and Maggie. Let's involve the Simpsons in Simpson's Paradox.
Consider Homer and Bart's consumption of potato chips and donuts. Suppose that Marge is in the room 30% of the time that Bart is eating chips and 50% of the time that Bart is eating donuts. Similarly, suppose that Marge is in the room 30% of the time that Homer is eating chips and 50% of the time that Homer is eating donuts; thus she catches them eating each kind of snack equally often. Nonetheless, Marge is in the room 38% of the time that Bart is eating either donuts or chips, and Marge is in the room 42% of the time that Homer is eating either donuts or chips. Suppose that Homer and Bart never eat chips and donuts in the same snack--eating chips and eating donuts are mutually exclusive.
What fraction of the time that Bart eats either chips or donuts does he eat donuts?
What fraction of the time that Homer eats either chips or donuts does he eat donuts?
Notation:
M - Marge is in the room
C - Bart eats chips
D - Bart eats donuts.
Each probability is conditional on Bart eating either chips or donuts.
P(M) = P(M\C)P(C) + P(M\D)P(D).
rearranging the equation to solve for P(D), you get:
P(D) = [P(M)-P(M\C)P(C)]/P(M\D)
So
We don't know what P(C) is, but we do know that it is the complement of P(D), so instead of P(C) write 1-P(D)
P(D) = [.46 - .40(1-P(D))]/.60
First step: (Distribute -.40)
P(D) = [.46 - .40 + .40P(D)] / .60
Next step, combine like terms (.46-.40)
P(D) = [.06 + .40P(D)] / .60
Now write .06 and .40P(D) each as its own separate fraction over .60
P(D) = .06/.60 + .40P(D)/ .60
Step #4 - Simplify the fractions
P(D) = 1/10 + 2/3P(D)
Step #5 - sub 2/3P(D) from both sides)
1/3 P(D) = 1/10
Final step - mult both sides by 3)
P(D) = 3/10.
That's Bart. Doing the equation a similar way for homer will get you:
P(D) = [.54 - .40(1-P(D))]/.60
P(D) = [.54 - .40 + .40P(D)] / .60
P(D) = [.14 + .40P(D)] / .60
P(D) = .14/.60 + .40P(D)/ .60
P(D) = 7/30 + 2/3P(D)
1/3 P(D) = 7/30
P(D) = 21/30, or in simplest form, 7/10.
To solve this problem, we can use the concept of conditional probability.
Let's denote the event of Bart eating chips as "B", the event of Bart eating donuts as "D", and the event of Marge being in the room as "M".
We are given the following probabilities:
P(M|B) = 0.3 (Marge is in the room 30% of the time Bart is eating chips)
P(M|D) = 0.5 (Marge is in the room 50% of the time Bart is eating donuts)
P(M|B or D) = 0.38 (Marge is in the room 38% of the time Bart is eating either chips or donuts)
We can use Bayes' theorem to find the probability of Bart eating donuts when Marge is in the room:
P(D|M) = (P(M|D) * P(D)) / P(M)
P(D|M) = (0.5 * P(D)) / P(M)
Now, let's consider Homer:
P(M|B) = 0.3 (Marge is in the room 30% of the time Homer is eating chips)
P(M|D) = 0.5 (Marge is in the room 50% of the time Homer is eating donuts)
P(M|B or D) = 0.42 (Marge is in the room 42% of the time Homer is eating either chips or donuts)
Similar to Bart, we can use Bayes' theorem to find the probability of Homer eating donuts when Marge is in the room:
P(D|M) = (P(M|D) * P(D)) / P(M)
P(D|M) = (0.5 * P(D)) / P(M)
Now we need to consider how Homer and Bart's eating habits are related:
We are given that Bart and Homer never eat chips and donuts in the same snack, meaning they are mutually exclusive.
So the probability that Bart eats donuts is equal to 1 minus the probability that Bart eats chips:
P(D|B) = 1 - P(B)
Similarly, the probability that Homer eats donuts is equal to 1 minus the probability that Homer eats chips:
P(D|H) = 1 - P(H)
Now, let's find the fraction of time Bart eats either chips or donuts:
P(B or D) = P(B) + P(D) - P(B and D)
However, since Bart and Homer do not eat chips and donuts together, P(B and D) = 0.
Therefore, P(B or D) = P(B) + P(D)
Similarly, we can find the fraction of time Homer eats either chips or donuts:
P(H or D) = P(H) + P(D)
Now, let's summarize the information we have:
P(M|B) = 0.3
P(M|D) = 0.5
P(M|B or D) = 0.38
P(M|H) = 0.3
P(M|H or D) = 0.42
We also know that Marge catches them eating each kind of snack equally often, so:
P(M|B) = P(M|H)
With this information, we can proceed to calculate the required probabilities.
To solve this problem, we need to calculate the conditional probabilities for Bart and Homer eating donuts given that they are eating either chips or donuts.
Let's start by finding the fraction of the time that Bart eats either chips or donuts. We know that Marge is in the room 38% of the time that Bart is eating either donuts or chips, so we can use Marge's presence as a conditioning factor.
We can use the formula for conditional probability:
P(Donuts | (Chips or Donuts)) = P(Donuts AND (Chips or Donuts)) / P(Chips or Donuts)
Let's substitute the given values into the formula:
P(Donuts | (Chips or Donuts)) = (Marge is in the room 50% of the time Bart eats donuts) / (Marge is in the room 38% of the time Bart eats either donuts or chips)
P(Donuts | (Chips or Donuts)) = 0.5 / 0.38 ≈ 1.3158
Therefore, Bart eats donuts approximately 1.3158 times as often as he eats chips or donuts.
Now let's move on to finding the fraction of the time that Homer eats either chips or donuts. We will follow the same process as before.
P(Donuts | (Chips or Donuts)) = P(Donuts AND (Chips or Donuts)) / P(Chips or Donuts)
Substituting the given values:
P(Donuts | (Chips or Donuts)) = (Marge is in the room 50% of the time Homer eats donuts) / (Marge is in the room 42% of the time Homer eats either donuts or chips)
P(Donuts | (Chips or Donuts)) = 0.5 / 0.42 ≈ 1.1905
Therefore, Homer eats donuts approximately 1.1905 times as often as he eats chips or donuts.
To summarize:
- Bart eats donuts approximately 1.3158 times as often as he eats chips or donuts.
- Homer eats donuts approximately 1.1905 times as often as he eats chips or donuts.
To standardize data for a z test, you first subtract the _____ from each term in a data set.
a)Mean b)median c) Mode