Question

A die is rolled 25 times and 12 evens are observed. Calculate and interpret a 95% confidence interval to estimate the true proportion of evens rolled on a die.

Answer 1

Answer:

The 95% confidence interval to estimate the true proportion of evens rolled on a die is (0.2842, 0.6758). This means that we are 95% sure that for the entire population of dies, the true proportion of evens rolled on a die is between 0.2842 and 0.6758

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 $1 - \frac{\alpha}{2}$.

For this problem, we have that:

$n = 25, \pi = \frac{12}{25} = 0.48$

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 $Z = 1.96$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.48 - 1.96\sqrt{\frac{0.48*0.52}{25}} = 0.2842$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.48 + 1.96\sqrt{\frac{0.48*0.52}{25}} = 0.6758$

The 95% confidence interval to estimate the true proportion of evens rolled on a die is (0.2842, 0.6758). This means that we are 95% sure that for the entire population of dies, the true proportion of evens rolled on a die is between 0.2842 and 0.6758