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Sagot :
Answer:
The 98% confidence interval for the probability of flipping a head with this coin is (0.4756, 0.7044).
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 z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
An experimenter flips a coin 100 times and gets 59 heads.
This means that [tex]n = 100, \pi = \frac{59}{100} = 0.59[/tex]
98% confidence level
So [tex]\alpha = 0.02[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.02}{2} = 0.99[/tex], so [tex]Z = 2.327[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.59 - 2.327\sqrt{\frac{0.59*0.41}{100}} = 0.4756[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.59 + 2.327\sqrt{\frac{0.59*0.41}{100}} = 0.7044[/tex]
The 98% confidence interval for the probability of flipping a head with this coin is (0.4756, 0.7044).
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