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The more general definition of a p-value is

the p-value is the probability of getting a result that is at least as extreme as the observed result, provided that $H_0$ is correct.

The definition is not clear about what 'extreme' means. One example of a p-value is a p-value that defines the degree of extremeness as values that are more in favour of $H_1$. This gives the definition in your question

the p-value is the probability of getting a result that is at least as much in favour of $H_1$ as the observed result, provided that $H_0$ is correct

1. This is not the definition of a p-value but a definition of a p-value.

2. It is a bit difficult to see what they mean by

   at least as much in favour of $H_1$

   One could view this definition in terms of the likelihood ratio test which is (for simplicity we use simple hypotheses):

   $$P \left ( \frac{\mathcal{L}(H_1|X)}{\mathcal{L}(H_0|X)} \geq \frac{\mathcal{L}(H_1|x_{observed})}{\mathcal{L}(H_0|x_{observed})} \right)$$

   The $p$-value (in a likelihood ratio test) is the probability of getting a result for which the likelihood ratio of the hypotheses $H_1$ and $H_0$ is at least as much as the observed result, provided that $H_0$ is correct.

^{I call it not clear what they mean with 'at least as much in favour' because I had initially a different thought about it than the likelihood ratio}

^{- I would prefer to use phrasing in terms of that likelihood. The term 'at least as much in favour of $H_1$' confused me initially and made me think of the wrong $P \left ( {\mathcal{L}(H_1|X)}>{\mathcal{L}(H_1|x_{observed})} \right)$}

^{Example, say we have a sample $X \sim N(\mu,1)$ to test the hypotheses that are $H_0:\mu = 0$ and $H_1: \mu =2$. Let the observation be $x = 3$, then the values that are at least as much in favour for $H_1$ are in between $1$ and $3$ and the probability for that under $H_0$ is $\Phi(3)-\Phi(1) \approx 0.157$. But with the likelihood ratio test we would not consider the values between $1$ and $3$ that are more in favour of $H_1$ and instead we would consider the values $>3$ for which the outcome is relatively more in favour of $H_1$ in comparison to $H_0$.}

^{- The term 'in favour' also initially confused me because it implies that the observed result must be in favour of $H_1$ but that does not need to be the case. It can be that the values are in favour of $H_0$ (a case that often coincides with a large p-value)}

^{Example, say we have a sample $X \sim N(\mu,1)$ to test the hypotheses that are $H_0:\mu = 0$ and $H_1: \mu =2$. Let the observation be $x = 0.5$, then this is a value that is not in favour of $H_1$ (at least not compared to $H_0$).}

The more general definition of a p-value is

the p-value is the probability of getting a result that is at least as extreme as the observed result, provided that $H_0$ is correct.

The definition is not clear about what 'extreme' means. One example of a p-value is a p-value that defines the degree of extremeness as values that are more in favour of $H_1$. This gives the definition in your question

the p-value is the probability of getting a result that is at least as much in favour of $H_1$ as the observed result, provided that $H_0$ is correct

1. This is not the definition of a p-value but a definition of a p-value.

2. It is a bit difficult to see what they mean by

   at least as much in favour of $H_1$

   One could view this definition in terms of the likelihood ratio test which is (for simplicity we use simple hypotheses):

   $$P \left ( \frac{\mathcal{L}(H_1|X)}{\mathcal{L}(H_0|X)} \geq \frac{\mathcal{L}(H_1|x_{observed})}{\mathcal{L}(H_0|x_{observed})} \right)$$

   The $p$-value (in a likelihood ratio test) is the probability of getting a result for which the likelihood ratio of the hypotheses $H_1$ and $H_0$ is at least as much as the observed result, provided that $H_0$ is correct.

^{I call it not clear what they mean with 'at least as much in favour' because I had initially a different thought about it than the likelihood ratio}

^{- I would prefer to use phrasing in terms of that likelihood. The term 'at least as much in favour of $H_1$' confused me initially and made me think of the wrong $P \left ( {\mathcal{L}(H_1|X)}>{\mathcal{L}(H_1|x_{observed})} \right)$}

^{Example, say we have a sample $X \sim N(\mu,1)$ to test the hypotheses that are $H_0:\mu = 0$ and $H_1: \mu =2$. Let the observation be $x = 3$, then the values that are at least as much in favour for $H_1$ are in between $1$ and $3$ and the probability for that under $H_0$ is $\Phi(3)-\Phi(1) \approx 0.157$. But with the likelihood ratio test we would not consider the values between $1$ and $3$ that are more in favour of $H_1$ and instead we would consider the values $>3$ for which the outcome is relatively more in favour of $H_1$ in comparison to $H_0$.}

^{- The term 'in favour' also initially confused me because it implies that the observed result must be in favour of $H_1$ but that does not need to be the case. It can be that the values are in favour of $H_0$.}

^{Example, say we have a sample $X \sim N(\mu,1)$ to test the hypotheses that are $H_0:\mu = 0$ and $H_1: \mu =10$. Let the observation be $x = 3$, then this is a value that is not in favour of $H_1$ (at least not compared to $H_0$).}

1. It is not the definition of a p-value but a definition of a p-value.

2. It is a bit difficult to see what they mean by

   at least as much in favour of $H_1$

   One could view this definition in terms of the likelihood ratio test which is (for simplicity we use simple hypotheses):

   $$P \left ( \frac{\mathcal{L}(H_1|X)}{\mathcal{L}(H_0|X)} \geq \frac{\mathcal{L}(H_1|x_{observed})}{\mathcal{L}(H_0|x_{observed})} \right)$$

   The $p$-value (in a likelihood ratio test) is the probability of getting a result for which the likelihood ratio of the hypotheses $H_1$ and $H_0$ is at least as much as the observed result, provided that $H_0$ is correct.

More about that in favour term:

• I would prefer to use phrasing in terms of that likelihood. The term 'at least as much in favour of $H_1$' confused me initially and made me think of the wrong $P \left ( {\mathcal{L}(H_1|X)}>{\mathcal{L}(H_1|x_{observed})} \right)$

  Example, say we have a sample $X \sim N(\mu,1)$ to test the hypotheses that are $H_0:\mu = 0$ and $H_1: \mu =2$. Let the observation be $x = 3$, then the values that are at least as much in favour for $H_1$ are in between $1$ and $3$ and the probability for that under $H_0$ is $\Phi(3)-\Phi(1) \approx 0.157$. But with the likelihood ratio test we would not consider the values between $1$ and $3$ that are more in favour of $H_1$ and instead we would consider the values $>3$ for which the outcome is relatively more in favour of $H_1$ in comparison to $H_0$.

• The term 'in favour' also initially confused me because it implies that the observed result must be in favour of $H_1$ but that does not need to be the case. It can be that the values are in favour of $H_0$ (a case that often coincides with a large p-value)

  Example, say we have a sample $X \sim N(\mu,1)$ to test the hypotheses that are $H_0:\mu = 0$ and $H_1: \mu =2$. Let the observation be $x = 0.5$, then this is a value that is not in favour of $H_1$ (at least not compared to $H_0$).

The more general definition of a p-value is

the p-value is the probability of getting a result that is at least as extreme as the observed result, provided that $H_0$ is correct.

The definition is not clear about what 'extreme' means. One example of a p-value is a p-value that defines the degree of extremeness as values that are more in favour of $H_1$. This gives the definition in your question

the p-value is the probability of getting a result that is at least as much in favour of $H_1$ as the observed result, provided that $H_0$ is correct

1. This is not the definition of a p-value but a definition of a p-value.

2. It is a bit difficult to see what they mean by

   at least as much in favour of $H_1$

   One could view this definition in terms of the likelihood ratio test which is (for simplicity we use simple hypotheses):

   $$P \left ( \frac{\mathcal{L}(H_1|X)}{\mathcal{L}(H_0|X)} \geq \frac{\mathcal{L}(H_1|x_{observed})}{\mathcal{L}(H_0|x_{observed})} \right)$$

   The $p$-value (in a likelihood ratio test) is the probability of getting a result for which the likelihood ratio of the hypotheses $H_1$ and $H_0$ is at least as much as the observed result, provided that $H_0$ is correct.

^{I call it not clear what they mean with 'at least as much in favour' because I had initially a different thought about it than the likelihood ratio}

^{- I would prefer to use phrasing in terms of that likelihood. The term 'at least as much in favour of $H_1$' confused me initially and made me think of the wrong $P \left ( {\mathcal{L}(H_1|X)}>{\mathcal{L}(H_1|x_{observed})} \right)$}

^{Example, say we have a sample $X \sim N(\mu,1)$ to test the hypotheses that are $H_0:\mu = 0$ and $H_1: \mu =2$. Let the observation be $x = 3$, then the values that are at least as much in favour for $H_1$ are in between $1$ and $3$ and the probability for that under $H_0$ is $\Phi(3)-\Phi(1) \approx 0.157$. But with the likelihood ratio test we would not consider the values between $1$ and $3$ that are more in favour of $H_1$ and instead we would consider the values $>3$ for which the outcome is relatively more in favour of $H_1$ in comparison to $H_0$.}

^{- The term 'in favour' also initially confused me because it implies that the observed result must be in favour of $H_1$ but that does not need to be the case. It can be that the values are in favour of $H_0$ (a case that often coincides with a large p-value)}

^{Example, say we have a sample $X \sim N(\mu,1)$ to test the hypotheses that are $H_0:\mu = 0$ and $H_1: \mu =2$. Let the observation be $x = 0.5$, then this is a value that is not in favour of $H_1$ (at least not compared to $H_0$).}

It is a bit difficult to see what they mean by

at least as much in favour of $H_1$

One could view this definition in terms of the likelihood ratio test which is (for simplicity we use simple hypotheses):

$$P \left ( \frac{\mathcal{L}(H_1|X)}{\mathcal{L}(H_0|X)} \geq \frac{\mathcal{L}(H_1|x_{observed})}{\mathcal{L}(H_0|x_{observed})} \right)$$

$p$-value is the probability of getting a result for which the likelihood ratio of the hypotheses $H_1$ and $H_0$ is at least as much as the observed result, provided that $H_0$ is correct.

• I would prefer to use phrasing in terms of that likelihood. The term 'at least as much in favour of $H_1$' confused me initially and made me think of the wrong $P \left ( {\mathcal{L}(H_1|X)}>{\mathcal{L}(H_1|x_{observed})} \right)$

  Example, say we have a sample $X \sim N(\mu,1)$ to test the hypotheses that are $H_0:\mu = 0$ and $H_1: \mu =2$. Let the observation be $x = 3$, then the values that are at least as much in favour for $H_1$ are in between $1$ and $3$ and the probability for that under $H_0$ is $\Phi(3)-\Phi(1) \approx 0.157$. But with the likelihood ratio test we would not consider the values between $1$ and $3$ that are more in favour of $H_1$ and instead we would consider the values $>3$ for which the outcome is relatively more in favour of $H_1$ in comparison to $H_0$.

• The term 'in favour' also initially confused me because it implies that the observed result must be in favour of $H_1$ but that does not need to be the case.

•  

  Some problem with this definition is that it only works for the likelihood ratio test with simple hypotheses. For instance, it already does not work when one tests $H_0: \mu = 0$ versus $H_1: \mu \neq 0$

In conclusion: this definition of the p-value is not correct*. But, a slightly altered version in terms of likelihood ratio (such that it relates to relatively more in favour) might be correct provided that this is about the case of simple hypotheses.

^{*Technically it is a correct p-value (not the p-value), but it is a weird way to compute a p-value as the example in the first point shows (the example with $p=0.157$).}

1. It is not the definition of a p-value but a definition of a p-value.

2. It is a bit difficult to see what they mean by

   at least as much in favour of $H_1$

   One could view this definition in terms of the likelihood ratio test which is (for simplicity we use simple hypotheses):

   $$P \left ( \frac{\mathcal{L}(H_1|X)}{\mathcal{L}(H_0|X)} \geq \frac{\mathcal{L}(H_1|x_{observed})}{\mathcal{L}(H_0|x_{observed})} \right)$$

   The $p$-value (in a likelihood ratio test) is the probability of getting a result for which the likelihood ratio of the hypotheses $H_1$ and $H_0$ is at least as much as the observed result, provided that $H_0$ is correct.

More about that in favour term:

• I would prefer to use phrasing in terms of that likelihood. The term 'at least as much in favour of $H_1$' confused me initially and made me think of the wrong $P \left ( {\mathcal{L}(H_1|X)}>{\mathcal{L}(H_1|x_{observed})} \right)$

  Example, say we have a sample $X \sim N(\mu,1)$ to test the hypotheses that are $H_0:\mu = 0$ and $H_1: \mu =2$. Let the observation be $x = 3$, then the values that are at least as much in favour for $H_1$ are in between $1$ and $3$ and the probability for that under $H_0$ is $\Phi(3)-\Phi(1) \approx 0.157$. But with the likelihood ratio test we would not consider the values between $1$ and $3$ that are more in favour of $H_1$ and instead we would consider the values $>3$ for which the outcome is relatively more in favour of $H_1$ in comparison to $H_0$.

• The term 'in favour' also initially confused me because it implies that the observed result must be in favour of $H_1$ but that does not need to be the case. It can be that the values are in favour of $H_0$ (a case that often coincides with a large p-value)

  Example, say we have a sample $X \sim N(\mu,1)$ to test the hypotheses that are $H_0:\mu = 0$ and $H_1: \mu =2$. Let the observation be $x = 0.5$, then this is a value that is not in favour of $H_1$ (at least not compared to $H_0$).