Respuesta :
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
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;
https://brainly.com/question/24131141
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