Respuesta :
Answer:
We have to flip the coin at least 52 times.
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]
For this problem, we have that:
The coin is fair, so [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].
How many times would we have to flip the coin in order to obtain a 99% confidence interval of width of at most .18 for the probability of flipping a head?
At least n times.
n is found when [tex]M = 0.18[/tex]. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.18 = 2.575\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.18\sqrt{n} = 2.575*0.5[/tex]
[tex]\sqrt{n} = \frac{2.575*0.5}{0.18}[/tex]
[tex](\sqrt{n})^{2} = (\frac{2.575*0.5}{0.18})^{2}[/tex]
[tex]n = 51.16[/tex]
Rounding up
We have to flip the coin at least 52 times.
