Confidence intervals

Question

We are interested in estimating the proportion of students from a local trade school who find a job within one year of graduation. We survey and found that 52 of 63 randomly sampled graduates found jobs. The graduating class was 714 students. Calculate a 95% confidence level for the proportion of students who found jobs within one year of graduation.

Answer 1

Here we are trying to estimate the population mean of students who received job offers from a sample of students. As we are working with proportions (# of students / sample) the best way to model the individual behavior is by a Bernoulli random variable, or a coin that every person in our sample flips with the probability of heads equalling p. Reminder that the key statistics of a binomial are it's mean and variance defined below:

mean = p

variance = (1-p)(p)

When summing up individual coin flips, we get a Binomial where for a sample of n students each flip a coin

mean = np

variance = (1-p)(p)/n

We're in the home stretch now. With a mean and variance, we can calculate the 95% confidence interval of our distribution using a z-score with standard deviation equalling the square root of the variance.

p = p' ± Z√[(1-p')(p')/n]

p' (sample probability) = 52/63 = 0.8253

Z (95% interval) = 1.96

p = 0.8254 ± 1.96√[(.1746)(.8254)/63]

p = 0.8254 ± 1.96√.002287

p = 0.8254 ± 1.96(0.0478)

p = (0.7317, 0.9191)

Multiplying that by our total population of 714 students, we expect the true number of students who found jobs to be between 522 and 656 with 95% confidence.