Respuesta :
Answer:
D The report's claim is false because the margin of error is 3.87%
Step-by-step explanation:
The correct option is:
A. The report's claim is false because the margin of error is 2.94%.
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To solve this question, we have to find the 95% confidence interval for the percentage of citizens using public transportation with a monthly subscription.
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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].
658 out of 1000 said they paid using a monthly subscription as opposed to paying for each ride individually.
This means that:
[tex]n = 1000, \pi = \frac{658}{1000} = 0.658[/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}} = 0.658 - 1.96\sqrt{\frac{0.658*0.342}{1000}} = 0.6285[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.658 + 1.96\sqrt{\frac{0.658*0.342}{1000}} = 0.6874[/tex]
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As percentages, we have that:
- 0.6285*100% = 62.85%
- 0.6874*100% = 68.74%
Thus, the interpretation is that we can can be 95% confident that the percentage of citizens using public transportation with a monthly subscription is between 62.85% and 68.74%, and not 63.33% and 68.27%.
The correct margin of error is 68.74 - 65.8 = 2.94%, and thus, the correct option is:
A.
The report's claim is false because the margin of error is 2.94%.
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