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Sagot :
Answer:
The 99% confidence interval for p in this case is (0.3317, 0.5883).
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].
Randomly selects 100 students from the school and asks the President to name each one. The President is able to correctly name 46 of the students.
This means that:
[tex]n = 100, \pi = \frac{46}{100} = 0.46[/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].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.46 - 2.575\sqrt{\frac{0.46*0.54}{100}} = 0.3317[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.46 + 2.575\sqrt{\frac{0.46*0.54}{100}} = 0.5883[/tex]
The 99% confidence interval for p in this case is (0.3317, 0.5883).
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