Respuesta :
Answer:
B. .564 to .686
C. 1488
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:
Professor York randomly surveyed 240 students at Oxnard University, and found that 150 of the students surveyed watch more than 10 hours of television weekly. This means that [tex]n = 240, \pi = \frac{150}{240} = 0.625[/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}} = 0.625 - 1.96\sqrt{\frac{0.625*0.375}{240}} = 0.564[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.625 + 1.96\sqrt{\frac{0.625*0.375}{240}} = 0.686[/tex]
The correct answer is:
B. .564 to .686
How many additional students would Professor York have to sample to estimate the proportion of all Oxnard University students who watch more than 10 hours of television each week within+/- 3% with 99% confidence?
With 99% confidence level, we have [tex]z = 2.575[/tex]
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
So
[tex]0.03 = 2.575*\sqrt{\frac{0.625*0.375}{240+n}}
[tex]0.03\sqrt{240+n} = 2.575*\sqrt{0.625*0.375}[/tex]
[tex](0.03\sqrt{240+n})^{2} = (2.575*\sqrt{0.625*0.375})^{2}[/tex]
[tex]0.0009(n + 240) = 1.5541[/tex]
[tex]0.0009n + 0.216 = 1.5541[/tex]
[tex]n = 1488[/tex]
The correct answer is:
C. 1488
