Question

Beth wants to determine a 80 percent confidence interval for the true proportion of high school students in the area who attend their home basketball games. how large of a sample must she have to get a margin of error less than 0.04? [if no estimate is known for p, let p^ = 0.5]

Answer 1

Answer:

We need to sample at least 256 high school students.

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 for the interval is:

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

80% confidence level

So $\alpha = 0.2$, z is the value of Z that has a pvalue of $1 - \frac{0.2}{2} = 0.9$, so $Z = 1.28$.

How large of a sample must she have to get a margin of error less than 0.04?

We need a sample of at least n high school students, in which n is found when M = 0.04.

We use $\pi = 0.5$, since no estimate is know. So

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

$0.04 = 1.28\sqrt{\frac{0.5*0.5}{n}}$

$0.04\sqrt{n} = 1.28*0.5$

$\sqrt{n} = \frac{1.28*0.5}{0.04}$

$(\sqrt{n})^{2} = (\frac{1.28*0.5}{0.04})^{2}$

$n = 256$

We need to sample at least 256 high school students.

Answer 2

TO solve this question on "how large of a sample must Beth have to get a margin of error less than 0.04", we will use the margin of error formula:

$E=Z\sqrt{\frac{\hat p(1-\hat p)}{n}}$

Here, for a 80% confidence level, $Z=1.28$

$\hat p =0.5$ (given)

Thus, margin of error, E=0.04

rearranging we get:

$n=(\frac{Z}{E})^2\hat p(1-\hat p)$

Plugging in gives us:

$n=(\frac{1.28}{0.04})^2\times 0.5(1-0.5)=1024\times0.25=256$

Thus, Beth's sample for the true proportion of high school students in the area who attend their home basketball games must have 256 students.