Question

in a random sample of 29 graduate students, 10 of them were unemployed. contruct a 95% confidence interval for the true proportion of graduate students who failed to secure a job.

Answer 1

p = 10/29 = 0.345

Z for 95% confidence interval = Z_0.025 = 1.96

Confidence interval = (p + Z_0.025 * sqrt(p * (1 - p) / n))

                                = (0.345 + 1.96 * sqrt(0.345 * (1 - 0.345) / 29))

                                = (0.345 + 0.173)

                                = (0.172 , 0.518)

                                = (0.172 < p < 0.518)