Respuesta :
Answer:
a) The 95% CI for the proportion of people that watched Super Bowl is (0.4531, 0.6183).
b) 48.1% = 0.481 is part of the confidence interval, which means that this prognosis does not contradict the findings.
Step-by-step explanation:
a) Construct the 95% CI for the proportion of people that watched Super Bowl.
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [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].
75 out of 140 watched the Super Bowl:
This means that [tex]n = 140, \pi = \frac{75}{140} = 0.5357[/tex]
95% confidence level
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}} = 0.5357 - 1.96\sqrt{\frac{0.5357*0.4643}{140}} = 0.4531[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5357 + 1.96\sqrt{\frac{0.5357*0.4643}{140}} = 0.6183[/tex]
The 95% CI for the proportion of people that watched Super Bowl is (0.4531, 0.6183).
b) The prognosis stated that the proportion of people watching the game is 48.1%. Does this prognosis contradict your findings
48.1% = 0.481 is part of the confidence interval, which means that this prognosis does not contradict the findings.
