Question

Suppose it is known that 8 out of the 20 teams in the sample had a season winning percentage better than 0.500.

(a) Find a 95% confidence interval for the true proportion of all teams that had a season winning percentage better than 0.500.

Answer 1

The 95% confidence interval for the true proportion of all teams that had a season winning percentage better than 0.500 is (0.1853, 0.6147).

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence interval
`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:

8 out of the 20 teams in the sample had a season winning percentage better than 0.500. This means that
`n = 20, \pi = (8)/(20) = 0.4`.

95% confidence interval

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)} = 0.4 - 1.96\sqrt{(0.6*0.4)/(20)} = 0.1853`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.4 + 1.96\sqrt{(0.6*0.4)/(20)} = 0.6147`

The 95% confidence interval for the true proportion of all teams that had a season winning percentage better than 0.500 is (0.1853, 0.6147).