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Sagot :
Answer:
The 99% confidence interval is between 62.36%(lower bound) and 89.64%(upper bound).
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].
A sample of 65 students from the freshmen class is used and a mean score of 76% correct is obtained.
This means that [tex]n = 65, \pi = 0.76[/tex]
99% confidence level
So [tex]\alpha = 0.005[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.005}{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}} = 0.76 - 2.575\sqrt{\frac{0.76*0.24}{65}} = 0.6236[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.76 + 2.575\sqrt{\frac{0.76*0.24}{65}} = 0.8964[/tex]
0.6236*100 = 62.36%
0.8964*100 = 89.64%
The 99% confidence interval is between 62.36%(lower bound) and 89.64%(upper bound).
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