Answer:
I just realized wasn't answering the question in the answer box.
Not sure if you guys still need this but:
a) This is an experiment, because there are 2 groups randomly assigned to a treatment.
b) i. A larger percent of people who got ad 1 clicked it and made the purchase, 22/100, than the amount who clicked ad 2 and made the purchase, 16/100.
ii. A larger percent of people who got ad 2 clicked the ad overall 20+16= 36 so 36/100 than the percent of those who got as 1 and clicked it 22+8=30 so 30/100.
c) There is more than one way to do this, but I usually do it this way:
p(made a purchase | ad 2)= 16/100= 0.16
p(made a purchase)= 38/200= 0.19
p(made a purchase | ad 1)= 22/100= 0.11
The the two events are NOT independent, because 0.19 is NOT= 0.22 is NOT= 0.16. Therefore, the ad that the customer receives DOES affect the probability that they make the purchase.
d) When working with a sample proportion, the np>10 n(1-p)>10 rule is an important condition for normal approximation. In order to use a z-test to calculate the p-value the sample needs to be normal.
e) p1= 22/100= 0.22
p2= 16/100= 0.16
p1-p2= 0.22-0.16= 0.06
In this case the p-value means there is a 27.95% probability of getting a difference of sample proportions .06 or greater OR -.06 or lower.
f) The company should fail to reject the null hypothesis-- in this case there is NOT convincing evidence to support that there is a difference between the amount of people who click and as and purchase something and the amount of people who click ad 2 and purchase something.
