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 90​% confident that the sample percentage is within five percentage points of the true population percentage for all sales transactions.

Answer 1

Answer:

We need at least 271 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:

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

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

Minimum sample size within five percentage points.

n when M = 0.05.

We dont know the proportion, so we use $\pi = 0.5$, which would be the proportion requiring the largest sample size

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

$0.05 = 1.645\sqrt{\frac{0.5*0.5}{n}}$

$\0.05\sqrt{n} = 1.645*0.5$

$\sqrt{n} = 16.45$

$(\sqrt{n})^{2} = (16.45)^{2}$

$n = 270.6$

Rouding up

We need at least 271 transactions