Question

Suppose that prior to conducting a coin-flipping experiment, we suspect that the coin is fair. How many times would we have to flip the coin in order to obtain a 96.5% confidence interval of width of at most .12 for the probability of flipping a head

Answer 1

We have to flip the coin 78 times.

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)}`

We suspect that the coin is fair.

This means that
`\pi = 0.5`

96.5% confidence level

So
`\alpha = 0.035`, z is the value of Z that has a pvalue of
`1 - (0.035)/(2) = 0.9825`, so
`Z = 2.108`.

How many times would we have to flip the coin in order to obtain a 96.5% confidence interval of width of at most .12 for the probability of flipping a head?

n times, and n is found when M = 0.12. So

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

`0.12 = 2.108\sqrt{(0.5*0.5)/(n)}`

`0.12√(n) = 2.108*0.5`

`√(n) = (2.108*0.5)/(0.12)`

`(√(n))^2 = ((2.108*0.5)/(0.12))^2`

`n = 77.1`

Rounding up

We have to flip the coin 78 times.