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I am following a course on statistics and I am having a lot of trouble with this question:

(To avoid clutter I will skip the context of the question and just give the given values.)

Compare two proportions $p_1$ and $p_2$, given $n_1 = 250$, $x_1 = 219$ and $n_2 = 200$, $x_2 = 154$.

  1. Does this data show that $p_1$ is greater than $p_2$? Determine your answer with $\alpha = 0.01$

  2. Calculate the chance on a Type-II error given the true difference between $p_1$ and $p_2$ is $0.15$, so $p_1 - p_2 = 0.15$.

More info then this was not given.

The $\alpha$ for the Type-II error is not specified so I suppose I need to use $\alpha = 0.01$. The first question is not that hard but is mostly the second question where I get stuck. I have tried every formula I could find but it always seemed like there was insufficient information to be able to calculate the chance on the Type-II error.

I bump into the same kind of problem when I need to calculate the minimum required sample size (for a two proportion test). So I think there is something fundamental I am overlooking when doing these kind of exercises.

Is there someone who can help me?

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    $\begingroup$ Calculation of Type 2 errors often (but not always) involves some form of conditional probability (think about how a Type 2 error might be defined for the context). $\endgroup$
    – Red Five
    Commented Jun 15 at 23:23
  • $\begingroup$ "The $\alpha$ for the Type-II error is not specified so I suppose I need to use $\alpha = 0.01$" seems a strange thing to say. In (2) you are given $\alpha = 0.05$ (for a Type I error assuming the hypothesis $p_1=p_2$) and then asked to find the chance of a Type II error of wrongly failing to reject this hypothesis if in fact $p_1-p_2=0.15$ $\endgroup$
    – Henry
    Commented Jun 16 at 2:07
  • $\begingroup$ @Henry you are right, this was a mistake, the $\alpha = 0.05$ is wrong. I will edit the question. Thanks! $\endgroup$
    – Felix Laga
    Commented Jun 16 at 12:17

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The question, as you described it has (at least) 3 issues: 3 "small-ish" ones, and a major one.

  1. Small-ish one: it is not clear what the sample sizes should be for this power computation. Should we use the same 250 and 200 from the first part? Let's assume we should, given no other indication
  2. Small-ish one. Does the sample with the larger size have a larger proportion? Or the other way around? The first part of the question implies that p1 (with the larger sample size) has the larger proportion. And the 2nd part says $p_1-p_2=0.15$, not $=-0.15$? So let's assume that $p_1$ corresponds to the largest proportion with a sample size of 250.
  3. Small-ish one. It is not clear which test you will be using (Fisher, $\chi^2,?$). I will assume Fisher. Nor is it clear if you are running it single or double sided. While trying to show p1 is larger that p2 implies one-sided, I will use double-sided, for generality.
  4. Major one. All we know is that $p_1-p_2=.15$. That is not sufficient to answer the question. The power of the test will depend on whether $p_1=.65$, or $p_1=.5$, etc. Using G*Power, , with $p_1=.65, p_2=.5$, you get a power of .884. But with $p_1=.8, p_2=.65$, your power is .941, and with $p_1=.95, p_2=.80$, the power is 0.999, and is the same as for $p_1=.2, p_2=.05$. The power is greater if the proportions are extreme, and lower if the proportions are closer to .5.

So, no, the problem does nto have a single answer (or rather, it is ill-posed).

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