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Sagot :
Answer:
The 99% confidence interval for the proportion of drug-related deaths that were caused by legally prescribed drugs is (0.841, 0.943).
As a percentage, the confidence interval is (84.1%, 94.3%).
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].
223 out of 250 deaths were caused by legally prescribed drugs.
This means that:
[tex]n = 250, \pi = \frac{223}{250} = 0.892[/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}} = 0.892 - 2.575\sqrt{\frac{0.892*0.108}{250}} = 0.841[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.892 + 2.575\sqrt{\frac{0.892*0.108}{250}} = 0.943[/tex]
The 99% confidence interval for the proportion of drug-related deaths that were caused by legally prescribed drugs is (0.841, 0.943).
As a percent:
Multiply the proportions by 100%.
0.841*100% = 84.1%
0.943*100%= 94.3%
As a percentage, the confidence interval is (84.1%, 94.3%).
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