Question

In: Statistics and Probability

Caraline is interested in estimating the proportion of students at a certain college who have at least two written final exams.

Caraline is interested in estimating the proportion of students at a certain college who have at least two written final exams. She takes a random sample and finds that 60 of the 75 students she surveyed did indeed have at least 2 written finals. Compute a 99% confindence interval for her and interpret it.

Expert Solution

Point estimate = sample proportion = = x / n = 60 / 75 = 0.8

1 - = 1 - 0.8 = 0.2

Z/2 = Z0.005 = 2.576

Margin of error = E = Z / 2 * (( * (1 - )) $\sqrt$ / n)

= 2.576 ( $\sqrt$ ((0.8 * 0.2) / 75)

= 0.119

A 99% confidence interval for population proportion p is ,

± E

0.8 ± 0.119

(0.681 , 0.919)

We are 99% confident that the true proportion of students at a certain college who have at least two written final exams. between 0.681 and 0.919.

orchestra answered 2 years ago

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