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Sagot :
Answer:
The 95% confidence interval for the population proportion of people in New Hampshire with cholesterol levels over 200 is (0.3216, 0.4784).
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].
You are told that a random sample of 150 people from Manchester New Hampshire have been given cholesterol tests, and 60 of these people had levels over the "safe" count of 200.
This means that [tex]n = 150, \pi = \frac{60}{150} = 0.4[/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].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 - 1.96\sqrt{\frac{0.4*0.6}{150}} = 0.3216[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 + 1.96\sqrt{\frac{0.4*0.6}{150}} = 0.4784[/tex]
The 95% confidence interval for the population proportion of people in New Hampshire with cholesterol levels over 200 is (0.3216, 0.4784).
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