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

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

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)`.

For this problem, we have that:

`n = 25, \pi = (12)/(25) = 0.48`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

The lower limit of this interval is:

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

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.48 + 1.96\sqrt{(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