To standardize data for a z test, you first subtract the _____ from each term in a data set.
a)Mean b)median c) Mode
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?
2 answers
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.
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.
1 answer