The manager of a department store wants to determine the proportion of customers who use the store's credit card when making a purchase. What sample size should he select so that at 96% confidence the error will not be more than 10%

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".

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

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 96% of confidence, our significance level would be given by [tex]\alpha=1-0.96=0.04[/tex] and [tex]\alpha/2 =0.02[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-2.05, z_{1-\alpha/2}=2.05[/tex]

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

[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)

Since we don't have prior info about the proportion we assume that the estimation for this is [tex] \hat p = 0.5[/tex]

We have that [tex]ME =\pm 0.1[/tex] (10%) and we are interested in order to find the value of n, if we solve n from equation (a) we got:

[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)

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

[tex]n=\frac{0.5(1-0.5)}{(\frac{0.1}{2.05})^2}=105.06[/tex]

And rounded up we have that n=106