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This appeared on a friends Stats 101 exam:

You have a random sample of 2000 respondents who's mean age is 47.3. A friend of yours, (who is knowledgeable about such things) claims that republican voters are, on average, older and demonstrate a mean age of 50. Using your data, demonstrate if he is correct, or incorrect to a 90% confidence level

The natural reaction for most of us here is that there isn't enough data in the question to even guess at the accuracy of the statement. When asked, the professor said "It isn't relevant to the problem".

Are we being trolled or is there a way, mathematically speaking, to state that there's insufficient information for a conclusion... or /gasp/ is there actually an answer ?

edit
I just now saw that there is a stackexchange specifically for Stats :( If someone has the ability, I don't mind this being moved to a more appropriate forum.

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Are the respondents Republican? –  a little don May 3 '11 at 17:57

2 Answers 2

up vote 1 down vote accepted

So you want the confidence interval to be of the following form:

$$50 \pm 1.65 \cdot \frac{s}{\sqrt{2000}}$$ where $\frac{s}{\sqrt{2000}}$ is the standard error of the mean. We need to find $s$. Given the data, we can't find $s$ exactly. But we can probably use a minimum variance unbiased estimator. One such estimator for $s$ is the following: $$K_n = \sqrt{\frac{n-1}{2}} \frac{\Gamma(\frac{n-1}{2})}{\Gamma(\frac{n}{2})}$$ where $\Gamma(n)$ is the gamma function. Also, we don't know the number of republicans in the sample. So this interval just gives a tighter estimate (i.e. less conservative).

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Thanks PEV, not only gave an answer but helped me better understand the problem! –  JHarnach May 3 '11 at 19:55
    
How can $K_n$ approximately estimate $s$ when it doesn't even use the data? I think the question expects one to test whether the average age is actually $50$ (since your knowledgeable friend says Republicans are aged $50$) and so $\frac{\bar X - 50}{\sqrt{2000}} \sim N(0, \sigma^2)$ approximately assuming that Republicans are the same age as everyone else. But there is no information in the question about $\sigma^2$ so we are stuck as far as I can see, unless one makes a heroic assumption about the variance of ages within a population. –  guy Mar 22 '12 at 19:37
    
Major typo in my last comment, should be $\sqrt{2000} (\bar X - 50)$. $\sigma^2 := \mbox{Var } X_1$ obviously. –  guy Mar 22 '12 at 19:54

I'm sure the average age of Republican voters, together with a confidence level, can be found somewhere on the internet. Find these figures and see whether your friend is right. For all we have been told, the 2000 respondents may have been drawn from the subscribers to a fishermen's magazine.

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