Question

A community college wants to estimate the proportion of
students who are active in student government. A random sample of
224 students showed 56 were active in student government. Create a
confidence interval at 90% confidence for the proportion of all
students in that college who were active in student
government.

Answer #1

Let be the sample proportion.

The sample proportion is = 56/224 = 0.25

90% Confidence interval is given by

Two sided critical z value at 90% alpha level = Z_{c} =
Z_{0.05} = 1.64

(From the standard normal statistical table)

The confidence interval is given by

± E = 0.25 ± 1.64 * sqrt( 0.25*(1 – 0.25)/224) = (0.203,0.297)

The confidence interval at 90% confidence for the proportion of all students in that college who were active in student government is (0.203, 0.297).

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