Question

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?

Answer 1

We need a mailing list of at least 191112 people.

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{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

The margin of error is:

`M = z\sqrt{(\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 - (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{(\pi(1-\pi))/(n)}`

`0.001 = 1.96\sqrt{(0.005*0.995)/(n)}`

`0.001√(n) = 1.96√(0.005*0.995)`

`√(n) = (1.96√(0.005*0.995))/(0.001)`

`(√(n))^(2) = ((1.96√(0.005*0.995))/(0.001))^(2)`

`n = 19111.96`

We need a mailing list of at least 191112 people.

Answer 2

`n=(0.005(1-0.005))/(((0.001)/(1.96))^2)=19111.96`

And rounded up we have that n=19112

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by
`\alpha=1-0.95=0.05` and
`\alpha/2 =0.025`. And the critical value would be given by:

`z_(\alpha/2)=-1.96, z_(1-\alpha/2)=1.96`

The margin of error for the proportion interval is given by this formula:

`ME=z_(\alpha/2)\sqrt{(\hat p (1-\hat p))/(n)}` (a)

And on this case we have that
`ME =\pm 0.001` and we are interested in order to find the value of n, if we solve n from equation (a) we got:

`n=(\hat p (1-\hat p))/(((ME)/(z))^2)` (b)

And replacing into equation (b) the values from part a we got:

`n=(0.005(1-0.005))/(((0.001)/(1.96))^2)=19111.96`

And rounded up we have that n=19112