Finding the confidence level

Question

The following question is on a study guide for a test:

A candidate who is running for Student Body President wants to know if more Juniors vote than Seniors. The hope is to interpret this information to know who to campaign towards.

The student body candidate wants to know the answer to his question. You access last years records and find 3750 Junior voted and only 3900 Seniors voted. Using excel statistics, you discovered determine that 38.8% of the Juniors and 37.5% of the Seniors voted in last years election.

Based on last years election, what is the highest confidence level for which you can claim that more Juniors vote than Seniors? You are solving for the highest confidence level in which you can reject the Null hypothesis.

Answer: 87%

As you can see, I have the answer. I just don't know the process for finding that answer. Can someone walk me through the step by step please?

Answer 1

This is a bit of a strange question, but here goes.

You are essentially asked to compare proportions in two samples: juniors and seniors. In both samples, you are looking at the proportion of voters $P$.

Let $P_j=0.388$ and $P_s=0.375$ be the respective proportions of juniors and seniors.

You want to test (reject) the null hypothesis that $H_0: P_j - P_s \leq 0$ (you are only interested in Juniors voting more, not differently, so this will be a one-tailed test). The alternative is $H_1: P_j - P_s > 0$.

What you first need to realize that these $P_x$ are Bernoulli random variables. Hence, their variance is $Var_x = P_x(1-P_x) \forall x \in \{j,s\}$

Then the standard error of the difference in proportions is $SE(D) = \sqrt{\frac{Var_j}{n_j} + \frac{Var_s}{n_s}}$.

Now here's where it gets weird (in my opinion, I might be wrong). By knowing the number of people who voted and their proportions in the population, you also know the whole population size. I.e., you've sampled the whole population. So $n_j$ and $n_s$ should be 9665 and 10400, respectively. But if you then do the math with these numbers, you get too high a confidence level.

So let's go with the supplied 3750 and 3900. Using those numbers, you get that $SE(D) = \sqrt{0.00012} = 0.01111$. So the resulting z-score is $\frac{(0.388-0.375)}{0.01111} = 1.17019$.

Plugging this number into the cumulative normal, you get $P(D > 0) = 0.87904$, which is almost what you should've gotten (the 87% in your result could've arisen through rounding somewhere midway through).

Anyway, I believe this was the intended calculation process, but I'm not sure I see the reason behind using the voter numbers as population sizes. Or am I completely wrong somewhere?