Respuesta :
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 [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
Z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
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 [tex]n = 20, \pi = \frac{8}{20} = 0.4[/tex].
95% confidence interval
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 - 1.96\sqrt{\frac{0.6*0.4}{20}} = 0.1853[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 + 1.96\sqrt{\frac{0.6*0.4}{20}} = 0.6147[/tex]
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).
