Question

The Internet is affecting us all in many different ways, so there are many reasons for estimating the proportion of adults who use it. Assume that a manager for E-­‐Bay wants to determine the current percentage of U.S. adults who now use the Internet. How many adults must be surveyed in order to be 95% confident that the sample percentage is in error by no more than three percentage points? a. b. In 2006, 73% of adults used the Internet. No known possible value of the proportion.

Answer 1

Answer:

a) At least 842 adults must be surveyed.

b) At least 1068 adults must be surveyed.

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 $Z = 1.96$.

a) 73% of adults used the Internet.

At least n adults must be surveyed.

n is found when $M = 0.03, \pi = 0.73$

So

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

$0.03 = 1.96\sqrt{\frac{0.73*0.27}{n}}$

$0.03\sqrt{n} = 1.96\sqrt{0.73*0.27}$

$\sqrt{n} = \frac{1.96\sqrt{0.73*0.27}}{0.03}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.73*0.27}}{0.03})^{2}$

$n = 841.3$

Rounding up

At least 842 adults must be surveyed.

b. No known possible value of the proportion.

Same as above, the difference as the since we do not know the value of the proportion, we use $\pi = 0.5$, which is when the largest sample size is going to be needed.

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

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

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

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

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

$n = 1067.11$

Rounding up

At least 1068 adults must be surveyed.