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?