Question

A programmer plans to develop a new software system. In planning for the operating system that he will​ use, he needs to estimate the percentage of computers that use a new operating system. How many computers must be surveyed in order to be 90​% confident that his estimate is in error by no more than one percentage point?
a. Assume that nothing is known about the percentage of computers with new operating systems. n = ?
b. Assume that a recent survey suggests that 99% of computers use a new operating system. n = ?

Answer 1

Answer:

a) $n = 9604$

b) n = 381

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 $Z = 1.645$.

a. Assume that nothing is known about the percentage of computers with new operating systems. n = ?

When we do not know the proportion, we use $\pi = 0.5$, which is when we are going to need the largest sample size.

The sample size is n when M = 0.01.

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

$0.01 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.01\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.01}$

$(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.01})^{2}$

$n = 9604$

b. Assume that a recent survey suggests that 99% of computers use a new operating system. n = ?

Now we have that $\pi = 0.99$. So

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

$0.01 = 1.96\sqrt{\frac{0.99*0.01}{n}}$

$0.01\sqrt{n} = 1.96*\sqrt{0.99*0.01}$

$\sqrt{n} = \frac{1.96*\sqrt{0.99*0.01}}{0.01}$

$(\sqrt{n})^{2} = (\frac{1.96*\sqrt{0.99*0.01}}{0.01})^{2}$

$n = 380.3$

Rouding up

n = 381