SOLUTION: how do I find a 95% confidence interval for the true proportion of all medical students who plan to work in a rural community. given the population proportiob p of 380 randomlty se

Question 1060701: how do I find a 95% confidence interval for the true proportion of all medical students who plan to work in a rural community. given the population proportiob p of 380 randomlty selected medical students, 21 said that they planned to work in a rural community.
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
P = 21/380 = 0.0553 approximately 0.06, n = 380
:
standard error of the sample proportion = square root(0.06 * (1-0.06) / 380) = 0.0122
:
95% confidence interval(C.I.)
alpha(a) = 1 - (95/100) = 0.05
critical probability(p*) = 1 - (0.05/2) = 0.975
critical value(cv) = 1.96 (use z table to find cv with p*)
margin of error(ME) = 1.96 * 0.0122 = 0.0239
95% C.I. = p + or - 0.0239
:

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