Mathematics: A credit card company is about to send out a mailing to test the market for a new credit card. From that sample, they want to estimate the true proportion of people who will sign up for the card nation-wide. A pilot study suggests that about 0.5% of the people receiving the offer will accept it. To be within a tenth of a percentage point (0.001) of the true rate with 95% confidence, how big does the test mailing have to be?

I think the answere is 100 because i took a guessStep-by-step explanation:...

A credit card company is about to send out a - id6170659, kenishajohnson116

Ответ:

chrisdelreal86

24.01.2022

Mathematics

We need a mailing list of at least 191112 people.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

For this problem, we have that:

$\pi = 0.005$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so Z = 1.96 .

To be within a tenth of a percentage point (0.001) of the true rate with 95% confidence, how big does the test mailing have to be?

They need at least n people

n is found when M = 0.001 . So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.001 = 1.96\sqrt{\frac{0.005*0.995}{n}}$

$0.001\sqrt{n} = 1.96\sqrt{0.005*0.995}$

$\sqrt{n} = \frac{1.96\sqrt{0.005*0.995}}{0.001}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.005*0.995}}{0.001})^{2}$

n = 19111.96

We need a mailing list of at least 191112 people.

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