Question

An online company is hoping to sell a new product by placing a pop-up advertisement on their website. There are two different designs for the advertisement, and the company would like to determine which one is more effective, as measured by clicks and purchases. For the first 200 visitors to the updated website, half were randomly assigned to receive advertisement 1 and half to receive advertisement 2. The two-way table summarizes the results:

Answer 1

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.