Question

For this problem, please round your critical value to 3 decimal places: Major television networks want to know what percentage of NBA Finals viewers also plan to watch the show that airs after it. What is the minimum sample size they need to survey to be within 3 percent of the true proportion at 99% confidence?

Answer 1

The minimum sample size they need is of 1842 viewers.

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{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

The margin of error is:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

99% confidence level

So
`\alpha = 0.01`, z is the value of Z that has a pvalue of
`1 - (0.01)/(2) = 0.995`, so
`Z = 2.575`.

Minimum sample size

n when M = 0.03.

We do not know the true proportion, so we use
`\pi = 0.5`, which is the case for which we are going to need the largest sample size.

`M = z\sqrt{(\pi(1-\pi))/(n)}`

`0.03 = 2.575\sqrt{(0.5*0.5)/(n)}`

`0.03√(n) = 2.575*0.5`

`√(n) = (2.575*0.5)/(0.03)`

`(√(n))^(2) = ((2.575*0.5)/(0.03))^(2)`

`n = 1841.8`

Rounding up

The minimum sample size they need is of 1842 viewers.