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Helga [31]
4 years ago
9

We saw in Chapter 13, Exercise 36 that First USA tested the effectiveness of a double miles campaign by recently sending out off

ers 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?
Mathematics
1 answer:
Alchen [17]4 years ago
4 0

Answer:

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

Step-by-step explanation:

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

  • <u>Assumptions</u>:

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. <em>p</em> = 0.02

H_{a}: The the success rate for the full campaign will be different than the standard rate, i.e. <em>p</em> ≠ 0.02

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

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

Compute the value of the test statistic as follows:

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

  • <u>Decision Rule</u>:

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 <em>z</em> = 6.39.

That is, z=6.39>1.96.

Thus, we may reject the null hypothesis.

  • <u>Conclusion</u>:

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%.

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