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I have been presented with the following question:

An attorney claims that more than 25% of all lawyers advertise. A sample of 200 lawyers in a certain city showed that 63 had used some form of advertisement. At $\alpha$ = 0.05 is there enough evidence to support the attorney's claim? Use the P-value method.

I don't know what method I should use to solve it, because it seems as though a standard deviation is needed in order to generate a test-statistic. I believe a z-score test will be used but I can't be positive. Can anyone shed light on this situation?


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i think a p-test should be used. any suggestions? – franklin May 7 '12 at 21:39
So, you can think of this as binomial sampling with $N = 200$ and some unknown $p$. We calculate $\hat p = 63 / 200$. It happens that the variance (equivalently the standard deviation) of the binomial distribution only depends on $N$ and $p$ through a formula that I'm sure you have in your notes and hence can be estimated using $\hat p$ and used to construct a Z test. – guy May 7 '12 at 21:44
The function $\sqrt{p(1-p)}$ is roughly constant for a broad range of values of $p$. – André Nicolas May 7 '12 at 21:56
up vote 3 down vote accepted

This is a simple binomial hypothesis test.

You should have the null hypothesis that the underlying probability ($\pi$) is exactly 25%. (Sometimes this might be stated as less than or equal to 25%, but during the test we'll assume that this null hypothesis is true by using the 25% value.)

The alternative hypothesis is that the underlying probability is greater than 25% (matching the attorney's claim)

Now you want to work out the probability of the (right-hand) tail, based on the observed result. That is: P(X$\geq$63). This is called the p-value, as you require. Here, you work out the p-value according to a Binomial distribution matching the null hypothesis [X~B(200,0.25)].

The point of the p-value is to decide whether this result could reasonably be generated by the null hypothesis.

If this resulting p-value is smaller or equal to (unlikelier) than $\alpha$=5%, then the result is "significant" - the result does not appear to be compatible with the null hypothesis, so therefore we'd accept the alternative hypothesis. (So, there is evidence the attorney's claim is correct)

If the resulting p-value is bigger than 5%, the result is "not significant" - the result appers to be compatible with the null hypothesis, so we accept the null hypothesis. (So, there is no evidence the attorney's claim is correct)

Edited To respond to your idea: a z-score would only apply to a test if we use a Normal distribution. Since this test is using a Binomial distribution, we don't really have any concept of a z-score here.

This idea is quite subtle so feel free to ask questions if you're not sure.

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why is it not, as @guy suggests, valid to use $\hat{p}$ to construct a z-test – franklin May 7 '12 at 23:56
It seems like @guy is suggesting an alternative method using a confidence interval, which would be valid as well. I should note that z can appear if we use a Normal distribution to approximate this Binomial distribution. – Ronald May 8 '12 at 0:00
@Ronald My idea was to use the normal approximation to the binomial, which is typically what is taught in introductory courses for testing the sampling proportion. So we would form a test statistic of $Z = \frac{\sqrt n (\hat p - p_0)}{\sqrt{\hat p (1 - \hat p)}}$ with $p_0 = .25$ and $n = 200$; you can also get a valid test statistic by replacing $\hat p$ with $p_0$ in the denominator. Both methods are accurate in the big; the latter method works reasonable in the small as well. – guy May 8 '12 at 0:43
@guy That's cool, makes sense, and I agree with what you're saying. But, I don't personally like this approximation - in the modern day, it's not (typically) necessary and it creates yet another unintuitive test statistic that students find unmemorable. – Ronald May 9 '12 at 0:01
@Ronald I disagree that it's not memorable; it's the obvious thing to do once you know the CLT, which is mandatory. I would also argue that it is desirable to teach these sorts approximations since they are exceedingly common in practice. – guy May 9 '12 at 4:36

Since $63/200 \gt 25\%$, any sensible approach should say the evidence strengthens the claim, though the evidence does not guarantee it.

I suspect you are intended to do something like the following, though I dislike this approach myself.

Set the (one-sided) null hypothesis as "25% or fewer lawyers advertise": then the probability that as many as 63 or more out of 200 are found in the sample is 0.023 or less and so, since this is less than 0.05, you might reject the null hypothesis at this confidence level in favour of the claim.

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Your first sentence is correct, but evidence can never guarantee something probabilistic, no matter whether it strengthens or weakens the claim. Hypothesis testing allows us to come to some decision in practice, although it's useful to note that this is not a guarantee either. – Ronald May 7 '12 at 23:44

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