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.

  • 1
    $\begingroup$ Because this distribution is not Normal--it is Binomial--the answers you are getting could be misleading in some applications. The crucial idea is that the Binomial distribution is discrete. That matters, because there are only a small number of proportions you are likely to see. Thus, $24.54\%$ of the time the sample proportion will be $35/140=0.25$ or less and $30.93\%$ of the time the proportion will be $36/140=0.257$ or less. That's an appreciable gap in proportions--and neither is exactly $30\%$. $\endgroup$ – whuber Jul 24 '16 at 16:37

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.

  • $\begingroup$ Yes, you made a mistake there. I am guessing the table you saw was the one here on the bottom of the page. This table only gives the area from 0 to a positive number. So you cannot use that to find quantile values. $\endgroup$ – Greenparker Jul 24 '16 at 16:21
  • $\begingroup$ Oh I see, I looked for the z-score value for 0.3, when I should've been looking for 0.5-0.3 = 0.2, for the proportion under the limit on the cumulative z-score table. Thanks! $\endgroup$ – P A N Jul 24 '16 at 16:21

I got a result of 0.26 when I did the calculation you suggest (which appears to be correct). It may have been a calculation/rounding error on your part?


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