# In what situations should I use and not use a pooled estimator for $\hat{p}$

In a question, it says that a true-false exam is used to discriminate between well-prepared students and poorly prepared students. There are $\frac{205}{250}$ well-prepared students and $\frac{137}{250}$ poorly prepared students who answered a certain item in the exam correctly.

The goal is to do a hypothesis test to see whether the given item in the test that is answered correctly can be expected to be at least $15\%$ higher among well-prepared students than among poorly prepared students.

So, $p_1=\frac{205}{250}$ and $p_2=\frac{137}{250}$.

$H_0: p_1-p_2=0.15\\H_1:p_1-p_2 > 0.15$

I know I could just use the formula: $$Z=\frac { p_{ 1 }-p_{ 2 }-\delta }{ \sqrt { \frac { p_{ 1 }(1-p_{ 1 }) }{ n_{ 1 } } +\frac { p_{ 2 }(1-p_{ 2 }) }{ n_{ 2 } } } }$$

But there is something here that I am very confuse about and that is: should I use a pooled estimator for a $\hat { p }$?

The two values of $p_1$ and $p_2$ are merely sample proportions and are not the true population proportion. In a usual two-proportion hypothesis, I would just use a pooled estimator for a $\hat{p}$ to get the ${ \sigma }_{ \hat { p } }$. So, what I would do is:

$\hat { p } =\frac { X_{ 1 }+X_{ 2 } }{ n_{ 1 }+n_{ 2 } } =\frac { 205+137 }{ 250+250 } =\frac { 171 }{ 250 } \\ { \sigma }_{ \hat { p } }=\sqrt { \frac { \hat { p } (1-\hat { p } ) }{ n_{ 1 } } +\frac { \hat { p } (1-\hat { p } ) }{ n_{ 2 } } } =\sqrt { \frac { \frac { 171 }{ 250 } (1-\frac { 171 }{ 250 } ) }{ 250 } +\frac { \frac { 171 }{ 250 } (1-\frac { 171 }{ 250 } ) }{ 250 } } =0.04158\\ Z=\frac { \frac { 205 }{ 250 } -\frac { 137 }{ 250 } -0.15 }{ { \sigma }_{ \hat { p } } } =2.9341$

However, the answer given to this question did not utilise the pooled estimator. Instead, it just uses back the sample proportion for the ${ \sigma }_{ \hat { p } }$. So, the answer given is written this way: ${ \sigma }_{ \hat { p_1 } -\hat{p_2} }=\sqrt { \frac { p_{ 1 }(1-p_{ 1 }) }{ n_{ 1 } } +\frac { p_{ 2 }(1-p_{ 2 }) }{ n_{ 2 } } } =\sqrt { \frac { \frac { 205 }{ 250 } (1-\frac { 205 }{ 250 } ) }{ 250 } +\frac { \frac { 137 }{ 250 } (1-\frac { 137 }{ 250 } ) }{ 250 } } =0.03976\\ Z=\frac { \frac { 205 }{ 250 } -\frac { 137 }{ 250 } -0.15 }{ { \sigma }_{ \hat { p_1 } -\hat{p_2} } } =3.0684$

Should I use a pooled estimator for $\hat{p}$ in this case? From my understanding, I use the pooled estimator for $\hat{p}$ when I don't have the true population proportion values. In this question, the given values are only the sampled proportions. But often, I wouldn't know the true population proportion values in the first place and therefore, I would always end up using the pooled estimator if I rely on my understanding.

So, in what situation should I then be using a pooled estimator and what other situation should I then not be using a pooled estimator?

• N.B. you might get better help with this question at stats.SE – Ronald Apr 20 '12 at 13:06
• @Ronald: Thank you for your flag - however since there's already an accepted answer and statistics is still essentially on-topic here, it seems that the recommendation is to leave it here. – Zev Chonoles Apr 21 '12 at 16:33
• do the answers not get migrated with the question, then? My concern is that people at stats.SE may be able to provide better/more correct insight. – Ronald Apr 21 '12 at 17:12

## 1 Answer

My intuition is: in this question you are testing the difference between proportions. Your null hypothesis assumption is that the proportions are not the same. So, I would say this means the standard deviation of the proportion cannot be estimated by pooling in this way (since, there are two proportions that are assumed to be different!)

Pooling the standard deviation is possible in the case where the proportions are equal in the Null Hypothesis. Pooling is then permitted (and offers more accuracy), because of the assumption that the Null Hypothesis is true and (therefore) that the proportions are equal - so your assumption is that there is only one true proportion that applies to both samples. In that case, $p_{1} - p_2 = 0$, we treat the two samples effectively as though they come from the same population.

If we are assuming the proportions differ by 0.15, $p_1 - p_2 = 0.15$, there may be an alternative way of pooling the two samples, but this would perhaps be over-complicated. The answer given seems to be correct.

Quote: "Likewise, if we have null hypothesis of the form p1 = p2 + k , our assumption is that the proportions are different, so there is no to estimate by pooling..." (sic)

• ohh...In the above case, $H_0: p_1-p_2=0.15$, therefore there is a difference between the proportions in the null hypothesis. If in another case say, we still want to test the difference between the proportions in the null hypothesis but now, with just $H_0: p_1-p_2=0$. Then in this case, I am still testing for the difference, but since $p_1-p_2=0$ => $p_1=p_2$, should I use a pooled variance for $H_0: p_1-p_2=0$? – xenon Apr 20 '12 at 13:22
• I just refreshed the page and realised you had an update, which already have answered my question in my previous comment. Thank you much! – xenon Apr 20 '12 at 13:38

## protected by Zev ChonolesApr 22 '16 at 4:03

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