Respuesta :
Answer:
We have to flip the coin 78 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]
We suspect that the coin is fair.
This means that [tex]\pi = 0.5[/tex]
96.5% confidence level
So [tex]\alpha = 0.035[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.035}{2} = 0.9825[/tex], so [tex]Z = 2.108[/tex].
How many times would we have to flip the coin in order to obtain a 96.5% confidence interval of width of at most .12 for the probability of flipping a head?
n times, and n is found when M = 0.12. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.12 = 2.108\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.12\sqrt{n} = 2.108*0.5[/tex]
[tex]\sqrt{n} = \frac{2.108*0.5}{0.12}[/tex]
[tex](\sqrt{n})^2 = (\frac{2.108*0.5}{0.12})^2[/tex]
[tex]n = 77.1[/tex]
Rounding up
We have to flip the coin 78 times.
