Question

The television landscape has changed as networks compete with streaming services such as Netflix and Amazon. Suppose that CBS plans to conduct interviews with television viewers in an attempt to estimate the proportion of viewers in the 18–49 age group who watch “most” of their television on network television as opposed to streaming services. CBS wishes to have 95% confidence and a margin of error in its estimate of ±0.03. What sample size is required?

Answer 1

Answer:

A sample size of 1068 is required.

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 $Z = 1.96$.

CBS wishes to have 95% confidence and a margin of error in its estimate of ±0.03. What sample size is required?

We need a sample size of n, and is found when M = 0.03.

We dont have an estimate for the true proportion, so we use $\pi = 0.5$, which is when the largest sample size will be needed.

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

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

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

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

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

$n = 1067.1$

Rounding up,

A sample size of 1068 is required.