Question

Many states are carefully considering steps that would help them collect sales taxes on items purchased through the Internet. How many randomly selected sales transactions must be surveyed to determine the percentage that transpired over the​ Internet? Assume that we want to be 95​% confident that the sample percentage is within three percentage points of the true population percentage for all sales transactions.

Answer 1

Answer:

We need to select at least 1068 sales transactions.

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:

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

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 $Z = 1.96$.

How many randomly selected sales transactions must be surveyed to determine the percentage that transpired over the​ Internet?

We need to survey at least n sales transactions.

Within three percentage points, so M = 0.03.

We do not know the population proportion, so we estimate it at $\pi = 0.5$, which is when we are going to need the largest sample size.

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

$0.03 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.03\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.03}$

$(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.03})^{2}$

$n = 1067.1$

Rounding up

1068

We need to select at least 1068 sales transactions.