SOLUTION: It was found that in a sample of 90 teenage boys, 70% of them have received speeding tickets. What is the 90% confidence interval of the true proportion of teenage boys who have r

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Question 855224: It was found that in a sample of 90 teenage boys, 70% of them have received speeding tickets. What is the 90% confidence interval of the true proportion of teenage boys who have received speeding tickets?
Found 2 solutions by ewatrrr, Theo:
Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website!

Hi,
n = 90, p = .70 & q = .30, m = np = .70�90 = 63
90% z =1.645
92% z = 1.751
95% z = 1.96
98% z = 2.326
99% z = 2.576
90% confidence interval
ME = 1.645
CI: 63-ME < 63 < 63 + ME

Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
i believe the answer would be as follows:
the sample proportion is .7
the sample size is 90
the formula for the standard error of a proportion is:
se = sqrt(p*q/n)
p is the sample proportion.
q is 1 - the sample proportion.
in your problem:
p .7
q = .3
n = 90
the standard error of the sample proportion would be equal to:
sqrt((.7*.3)/90) which is equal to .04830
the z-score for a 90% confidence level is equal to +/- 1.6449
this means that the confidence interval is equal to the mean proportion plus or minus 1.6449 times the standard error.
this becomes:
.7 + 1.6449 * .04830 = .7794 as the high limit.
.7 - 1.6449 * .04830 = .6206 as the low limit.
the true proportion of teenage boys in the population is estimated to be .62 to .78 after rounding to 2 decimal digits.

here's a reference, except they use a 95% confidence limit rather than a 90% confidence limit.
the procedure, however, is the same.
http://www.stat.wmich.edu/s160/book/node47.html