Question

An experimenter flips a coin 100 times and gets 59 heads. Find the 98% confidence interval for the probability of flipping a head with this coin.

Answer 1

Answer:

The 98% confidence interval for the probability of flipping a head with this coin is (0.4756, 0.7044).

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 z-score that has a p-value of $1 - \frac{\alpha}{2}$.

An experimenter flips a coin 100 times and gets 59 heads.

This means that $n = 100, \pi = \frac{59}{100} = 0.59$

98% confidence level

So $\alpha = 0.02$, z is the value of Z that has a p-value of $1 - \frac{0.02}{2} = 0.99$, so $Z = 2.327$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.59 - 2.327\sqrt{\frac{0.59*0.41}{100}} = 0.4756$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.59 + 2.327\sqrt{\frac{0.59*0.41}{100}} = 0.7044$

The 98% confidence interval for the probability of flipping a head with this coin is (0.4756, 0.7044).