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Sagot :
Answer:
The 99% confidence interval for the percent of all the students at that college who have at least $1000 in credit card debt is (20.11%, 32.17%).
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].
In a simple random sample of 352 students at a college, 92 reported that they have at least $1000 of credit card debt.
This means that [tex]n = 352, \pi = \frac{92}{352} = 0.2614[/tex]
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a p-value 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}} = 0.2614 - 2.575\sqrt{\frac{0.2614*0.7386}{352}} = 0.2011[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2614 + 2.575\sqrt{\frac{0.2614*0.7386}{352}} = 0.3217[/tex]
As percent:
0.2011*100% = 20.11%
0.3217*100% = 32.17%.
The 99% confidence interval for the percent of all the students at that college who have at least $1000 in credit card debt is (20.11%, 32.17%).
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