(1 point) Dylan wants to determine a 99 percent confidence interval for the true proportion p of high school students in the area who attend their home basketball games. Out of n randomly selected students he finds that that exactly half attend their home basketball games. About how large would n have to be to get a margin of error less than 0.01 for p

Answers

Answer 1

Answer:

n would have to be at least 16577

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 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].

The margin of error is:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

Out of n randomly selected students he finds that that exactly half attend their home basketball games.

This means that [tex]\pi = 0.5[/tex]

99% confidence level

So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].

About how large would n have to be to get a margin of error less than 0.01 for p

A sample size of n is needed. n is found when M = 0.01. So

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]0.01 = 2.575\sqrt{\frac{0.5*0.5}{n}}[/tex]

[tex]0.01\sqrt{n} = 2.575*0.5[/tex]

[tex]\sqrt{n} = \frac{2.575*0.5}{0.01}[/tex]

[tex](\sqrt{n})^2 = (\frac{2.575*0.5}{0.01})^2[/tex]

[tex]n = 16576.6[/tex]

Rounding up

n would have to be at least 16577

Answer 2

When the error margin is smaller than 0.01, the value of n will be 16576.6.

What is a confidence interval?

The mean of your estimate plus and minus the range of that estimate constitutes a confidence interval.

Given that;

Confidence interval = 99 %

Randomly selected students,n

margin of error,M0.01

The margin of error for the given information is;

[tex]\rm M = z \frac{\sigma(1-\sigma)}{n} \\\\\ 0.01 = 2.575 \frac{0.5(0.5)}{n} \\\\ 0.01 \sqrt n = 2.575 \times 0.5 \\\\ \sqrt n = \frac{2.575 \times 0.5}{0.01} \\\\ n = 16576.6[/tex]

Hence, the value of n for which the margin of error is less than 0.01 will be 16576.6.

To learn more about the confidence interval refer;

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To know more about mode click here

https://brainly.com/question/300591

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