Question

A study shows that 66% of all dog owners greet their dog before they greet their spouse or children when they get home from work. Assume that the claim is true and each arrival home is independent.

a.) In a series of arriving home from work, what is the probability that the first time a person doesn’t greet their dog first is on the 5th arrival home.

b.) What is the probability that a person greets their dog first on 4 or fewer of the next 10 arrivals home?

c.) Suppose that a person goes on a vacation for two weeks. When they return they greet their dog first only two out of ten days. Is this evidence that their rate is now less than 66%?

Answer 1

a.) The probability that a person greets their dog first on the first four arrivals home is (66/100)^4 = 0.260056. The probability that they don't greet their dog first on the fifth arrival home is (100-66)/100 = 0.34. The probability that they greet their dog first on the first four arrivals home and don't greet their dog first on the fifth arrival home is 0.260056 * 0.34 = 0.08823904.

b.) The probability that a person greets their dog first on 4 or fewer of the next 10 arrivals home is the same as the probability that they greet their dog first on the first 4 arrivals home and don't greet their dog first on any of the next 6 arrivals home. This probability is 0.08823904 * (1-0.66)^6 = 0.0031139488.

c.) No, this is not necessarily evidence that their rate is now less than 66%. The fact that they greeted their dog first on only two out of ten days could be due to chance. To determine whether there is significant evidence that their rate has changed, you would need to perform a statistical test.