Candy maker makes child, and adult bags of jelly beans with different color mixes. The company claims, the child size is 30% red, and the adult size is 15% red. Assume their claim is true. Suppose we take a random sample of 50 from the child's mix and a separate random sample of 100 from the adult mix. What is the probability that the proportion of red in the child sample is less than or equal to the proportion of red in the adult sample?
Please help, we have never been over how to find the probability of a comparison of proportions before, and apparently it is on tomorrow's midterm!
Best Answer
The elegant answer by @Graham Kemp would be correct if the question were whether the number in the of reds in the sample of 50 is less than the number in the sample of 100. But the question was about proportions.
A simulation of a million repetitions of this experiment confirms that the probability of the former is about 0.539 (last place in some doubt). For proportions the number is much different: about 0.023. This makes sense because there are (said to be) proportionately more reds in child's population.
My guess is you are expected to use normal approximations for these proportions.
For the child mix the number is $X \sim Bin(50,.3),$ so $E(X) = 50(.3) = 15$ and $V(X) = 15(.7) = 10.5$ For proportions $E(\hat p_c) = (X/n) = .3$ and $V(\hat p_c) = (.3)(.7)/50 = 0.0042.$
For proportions in the adult mix $E(\hat p_a) = .15$ and $V(\hat p_a) = (.15)(.85)/100 = 0.001275.$
You are interested in the difference in proportions, which can be found by subtracting the means and (by independence) adding the variances. $E(\hat p_c - \hat p_a) = .3-.15 = .15$ and $V(\hat p_c - \hat p_a) = .0042 + .001275 = 0.005475.$
The two binomial proportions are roughly normally distributed, and so their difference is also. The question is whether the difference in sample proportions is less than 0.
With software this can be computed directly as 0.0213. On an exam you would have to standardize (subtracting the mean from 0 and dividing the difference by the standard deviation) and look up the result in normal tables. Some accuracy is lost in the necessary rounding. I got 0.0212.
The sample sizes are a little small for normal approximations to be accurate, and in any case they can't be trusted beyond the second decimal place. But the agreement with simulation results (which suffer only from simulation approximation) is encouraging. In any case, about the only computational option for an in-class exam without computers is to use normal tables.
The bottom line is that it is extremely unlikely the proportion of reds drawn from the child mix would be smaller than the proportion drawn from the adult mix.
The R computer code for the simulation is as follows: