Question

In a random sample of n = 500 families owning television sets in the city of Hamilton, Canada, it is found that x = 340 subscribe to HBO. How large a sample is required if we want to be 95% confident that our estimate of p is within 0.02 of the true value?

Answer 1

Answer:

We need a sample of at least 2090 families if we want to be 95% confident that our estimate of p is within 0.02 of the true value.

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 interval $1-\alpha$, 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 of the interval is:

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

In this problem, we have that:

$n = 500, x = 340, p = \frac{340}{500} = 0.68$

95% confidence interval

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 large a sample is required if we want to be 95% confident that our estimate of p is within 0.02 of the true value?

The margin of error decreases when the sample size increases. So we need a sample of at least n people when M = 0.02 for the estimate of p being within 0.02 of the value.

So

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

$0.02 = 1.96\sqrt{\frac{0.68*0.32}{n}}$

$0.02\sqrt{n} = 0.9143$

$\sqrt{n} = \frac{0.9143}{0.02}$

$\sqrt{n} = 45.71$

$\sqrt{n}^{2} = (45.71)^{2}$

$n = 2089.83$

We need a sample of at least 2090 families if we want to be 95% confident that our estimate of p is within 0.02 of the true value.