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Sagot :
Answer:
The margin of error for the 95% confidence interval used to estimate the population proportion is of 0.0209.
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].
The margin of error is of:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In a clinical test with 2161 subjects, 1214 showed improvement from the treatment.
This means that [tex]n = 2161, \pi = \frac{1214}{2161} = 0.5618[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
Margin of error:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]M = 1.96\sqrt{\frac{0.5618*0.4382}{2161}}[/tex]
[tex]M = 0.0209[/tex]
The margin of error for the 95% confidence interval used to estimate the population proportion is of 0.0209.
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