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Sagot :
Answer:
Normality assumptions are met.
The 90% confidence interval for the proportion of adult drivers that run at least one red light in the last month is (0.4892, 0.5356). The interpretation is that we are 90% sure that the true proportion of all adult drivers than ran at least one red light in the last month is between these bounds.
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 1251 adult drivers, 641 of them have run at least one red light in the last month.
This means that [tex]n = 1251, \pi = \frac{641}{1251} = 0.5124[/tex]
Normality assumptions:
We need that: [tex]n\pi[/tex] and [tex]n(1-\pi)[/tex] are 10 or greater. So
[tex]n\pi = 1251*0.5124 = 641[/tex]
[tex]n(1-\pi) = 1251*0.4876 = 610[/tex]
So the normality assumptions are met.
90% confidence level
So [tex]\alpha = 0.1[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5124 - 1.645\sqrt{\frac{0.5124*0.4876}{1251}} = 0.4892[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5124 + 1.645\sqrt{\frac{0.5124*0.4876}{1251}} = 0.5356[/tex]
The 90% confidence interval for the proportion of adult drivers that run at least one red light in the last month is (0.4892, 0.5356). The interpretation is that we are 90% sure that the true proportion of all adult drivers than ran at least one red light in the last month is between these bounds.
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