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Eduardwww [97]
4 years ago
14

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. Develop a 95% confidence interval to estimate the true proportion of students who watch more than 10 hours of television each week. The confidence interval is
A. .533 to .717 B. .564 to .686 C. .552 to .698 D. .551 to .739
(B) .564 to .686 for this. I need help on the question below please........

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?

A. 761
B. 1001
C. 1488
D. 1728
Mathematics
1 answer:
const2013 [10]4 years ago
7 0

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 \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:

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 n = 240, \pi = \frac{150}{240} = 0.625

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}} = 0.625 - 1.96\sqrt{\frac{0.625*0.375}{240}} = 0.564

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.625 + 1.96\sqrt{\frac{0.625*0.375}{240}} = 0.686

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 z = 2.575

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

So

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}

(0.03\sqrt{240+n})^{2} = (2.575*\sqrt{0.625*0.375})^{2}

0.0009(n + 240) = 1.5541

0.0009n + 0.216 = 1.5541

n = 1488

The correct answer is:

C. 1488

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