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81 students at a college were asked whether they had completed their required English 101 course, and 53 students said "yes". Construct the 99% confidence interval for the proportion of students at the college who have completed their required English 101 course. Enter your answers as decimals (not percents) accurate to three decimal places. The Confidence Interval is ( , )

Sagot :

Answer:

[tex](0.518,0.790)[/tex]

Step-by-step explanation:

Use the formula [tex]CI=\hat{p}\pm z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n} }[/tex] where [tex]\hat{p}[/tex] is the sample proportion, [tex]n[/tex] is the sample size, and [tex]z^*[/tex] is the corresponding z-value for a given confidence level.

We know that [tex]\hat{p}=\frac{53}{81}[/tex], [tex]n=81[/tex], and [tex]z^*=2.5758[/tex] for a 99% confidence level.

Therefore, the 99% confidence interval is [tex](0.518,0.790)[/tex]

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