Question

We saw in Chapter 13, Exercise 36 that First USA tested the effectiveness of a double miles campaign by recently sending out offers to a random sample of 50,000 card- holders. Of those, 1184 registered for the promotion. Even though this is nearly a 2.4% rate, a staff member suspects that the success rate for the full campaign will be no different than the standard 2% rate that they are used to seeing in similar campaigns. What do you predict?

Answer 1

The success rate for the full campaign will be different than the standard rate of 2%.

Explanation:

To predict the results conduct a hypothesis test for single proportion.

• Assumptions:

Assume that the significance level of the test is
`\alpha =5\%\ or\ 0.05`.

• The hypothesis is defined as follows:

`H_(0):` The the success rate for the full campaign will be no different than the standard rate, i.e. p = 0.02

`H_(a):` The the success rate for the full campaign will be different than the standard rate, i.e. p ≠ 0.02

• According to the Central limit theorem as the sample size is large, i.e, n = 50,000 > 30, the sampling distribution of sample proportion is normally distributed with mean
  `p` and standard deviation
  `\sqrt{(p(1-p))/(n) }`. Then the test statistic is defined as:

`z=\frac{\hat p-p}{\sqrt{(p(1-p))/(n) }}`

Compute the value of the test statistic as follows:

`z=\frac{\hat p-p}{\sqrt{(p(1-p))/(n) }}\\=\frac{0.024-0.02}{\sqrt{(0.02(1-0.02))/(50000) }}\\=6.3887\\\approx 6.39`

• Decision Rule:

The hypothesis test is two tailed. Then the for 5% level of significance the rejection region is defined as:
`[-1.96\leq Z\leq 1.96]`, i.e if the test statistic value lies out of this region then the null hypothesis will be rejected.

The calculated value of the test statistic is z = 6.39.

That is,
`z=6.39>1.96`.

Thus, we may reject the null hypothesis.

• Conclusion:

As the null hypothesis is rejected at 5% level of significance this implies that the success rate for the full campaign will be different than the standard rate of 2%.