Problem description
If a population proportion is 0.28, and if the sample size is 140, 30% of the time the sample proportion will be less than what value if you are taking random samples?
This is a sample proportion problem for a binomial sample distribution. Sample distributions tends to become normally distributed when enough samples are taken.
$ p = 0.28 $
$ n = 140 $
$ P \ (\hat{p} < x) = 30 \% $
My reasoning has been to find the $z$-score for the left part of the normal curve, up to the proportion value $ P = 30 \% $. According to a $ z $-score table, the $ z $-value for $ 0.2995 $ is $z = -0.84$.
Using the z-score formula for sample proportions:
$$ z = \frac{\hat{p}-p}{\sqrt{\frac{p*q}{n}}} = -0.84 = \frac{\hat{p}-0.28}{\sqrt{\frac{0.28*(1-0.28)}{140}}}$$
We solve the equation for $ \hat{p} $, which should give $ \hat{p} = 0.24 $.
However, my solution sheet says the correct answer should be $ 0.26 $.
Have I solved the problem incorrectly in identifying $ \hat{p} = 0.24 $?
I realize that we are actually looking for the value of $ x $, however I'm not sure if my logic is correct to assume that $ x = \hat{p} $ for these limit values.
Best Answer
It looks like you made an error looking at the $z$-score table. The $z$-value for .30 is around -.525. Using that you will get the right answer.