Question

1. A major credit card company is planning a new offer for their current cardholders. The company will give double airline miles on purchases for the next 6 months if the cardholder goes online and registers for this offer. To test the effectiveness of this campaign, the company recently sent out offers to a random sample of 50,000 cardholders. Of those, 1184 registered. A staff member suspects that the success rate for the full campaign will be more than the standard 2% rate that they are used to seeing in similar campaigns. What do you predict?

1. What are the hypotheses? (2 pts)

2. Is the condition for Normal Approximation met? Be sure to explain / show your calculations. (3 pts)

Do you think the rate would increase if they use this fundraising campaign? (5 pts)
Fully explain your reasoning, including all calculations & sketches – along with a sentence form explanation of your findings. (If you need extra space, ask your instructor for a blank sheet)

Answer 1

(a)

Prediction-

We observe registration rate .

Thus we can predict that the success rate for the full campaign will be more than the standard 2% rate that were used to observe in similar campaigns.

(b)

Normal approximation-

We define registration of a cardholder as success (for the company). Registering of a cardholder is independent of registration of other cardholders. Thus the given situation can be treated as a Binomial process. Suppose, random variable X denotes number of cardholders registered.

We observe that,

• Number of success
• Number of failures

Thus both conditions of Normal approximation of Binomial distribution are met.

(c)

Calculations (using hypothesis test) for effectiveness of the fundraising campaign-

We have to perform one sample proportion test.

We have to test for null hypothesis

against the alternative hypothesis

Our test statistic is given by

Here,

Number of observations

[Using R-code '1-pnorm(5.877664)']

We reject our null hypothesis if , level of significance

We generally test for level of significance 0.10, 0.05, 0.01 or something like these.

So, we reject our null hypothesis.

Hence, based on the given data we can conclude that there is significant evidence that success rate for the full campaign will be more than the standard 2% rate that were used to see in similar campaigns.