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

Answer:

We have to flip the coin 78 times.

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:

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

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 $1 - \frac{0.035}{2} = 0.9825$, 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

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

$0.12 = 2.108\sqrt{\frac{0.5*0.5}{n}}$

$0.12\sqrt{n} = 2.108*0.5$

$\sqrt{n} = \frac{2.108*0.5}{0.12}$

$(\sqrt{n})^2 = (\frac{2.108*0.5}{0.12})^2$

$n = 77.1$

Rounding up

We have to flip the coin 78 times.