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Sagot :
Answer:
0.0101 < p < 0.1499
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].
Out of 100 adult residents sampled, 8 had kids. Based on this, construct a 99%.
This means that [tex]n = 100, \pi = \frac{8}{100} = 0.08[/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}} = 0.08 - 2.575\sqrt{\frac{0.08*0.92}{100}} = 0.0101[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.08 + 2.575\sqrt{\frac{0.08*0.92}{100}} = 0.1499[/tex]
Express your answer in tri-inequality form.
0.0101 < p < 0.1499
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