Question

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

Answer 1

Answer:

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).

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 interval $1-\alpha$, 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:

8 out of the 20 teams in the sample had a season winning percentage better than 0.500. This means that $n = 20, \pi = \frac{8}{20} = 0.4$.

95% confidence interval

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 - 1.96\sqrt{\frac{0.6*0.4}{20}} = 0.1853$

The upper limit of this interval is:

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

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).