Mixing
Type 1 error condition in one tailed statistical hypothesis test
Clash Royale CLAN TAG#URR8PPP
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Consider the following classical statistical test setup:
One assumes a coin to be unfair in the sense that heads, say, occurs more frequently than tails. Thus we set $H_0: pleqfrac12$ as null hypothesis and $H_1:p>frac12$ as alternative where $p$ is the probability for heads.
Also let X count the occurence of heads when tossing the coin $n$ times. Given $n$ and a significance level $alpha$ we get the one-tail condition
beginequation
(1)quad P(Xgeq k)leqalpha
endequation
where $P$ has a $(n,p)$-binomial distribution with $pleqfrac12$ (thus yielding the probability for rejecting $H_0$ when it's actually true).
To solve $(1)$ for $k$ it would now be common (school book) practice to set $p=frac12$ and solve $(1)$ by inversion. But this isn't correct, as we just know $pleqfrac12$.
So wouldn't it be better to rather use a distribution for "$k$ wins out of $n$ with a probability of success $leqfrac12$" and which would that appropriate distribution be?
I want to be more precise: In a more general context the maximum $alpha$ error could be defined as
beginequation
alpha_max:=max_thetainTheta_0P_theta(T(X_1,dotsc,T_n)in K)
endequation
where $T$ is some kind of test statistic, in our case counting the number of heads in a sample $X_1,dotsc,X_n$; $Theta$ is the parameter space in question (our paramter is $psimtheta$), $Theta_0$ the subspace corresponding to the null hypothesis, i.e.
beginequation
H_0: thetainTheta_0,quad H_1:thetainThetasetminusTheta_0;
endequation
and finally $K$ is the region of rejection of $H_0$, i.e.
beginequation
H_0text is rejected iff T(X_1,dotsc,T_n)in K.
endequation
So in particular we have $Theta=[0,1], Theta_0=[0,frac12]$, yielding
beginequation
alpha_max=max_pleqfrac12sum_i=k^n B_n,p(X=i),
endequation
which should now be $leq$ a given significance level.
[Definitions from http://www.wiwi.uni-muenster.de/05/download/studium/advancedstatistics/ss09/kapitel_6.pdf - couldn't find equivalent in English]
probability statistics binomial-distribution
asked Aug 2 at 15:07
Don Fuchs
234
add a comment |Â
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0
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Consider the following classical statistical test setup:
One assumes a coin to be unfair in the sense that heads, say, occurs more frequently than tails. Thus we set $H_0: pleqfrac12$ as null hypothesis and $H_1:p>frac12$ as alternative where $p$ is the probability for heads.
Also let X count the occurence of heads when tossing the coin $n$ times. Given $n$ and a significance level $alpha$ we get the one-tail condition
beginequation
(1)quad P(Xgeq k)leqalpha
endequation
where $P$ has a $(n,p)$-binomial distribution with $pleqfrac12$ (thus yielding the probability for rejecting $H_0$ when it's actually true).
To solve $(1)$ for $k$ it would now be common (school book) practice to set $p=frac12$ and solve $(1)$ by inversion. But this isn't correct, as we just know $pleqfrac12$.
So wouldn't it be better to rather use a distribution for "$k$ wins out of $n$ with a probability of success $leqfrac12$" and which would that appropriate distribution be?
I want to be more precise: In a more general context the maximum $alpha$ error could be defined as
beginequation
alpha_max:=max_thetainTheta_0P_theta(T(X_1,dotsc,T_n)in K)
endequation
where $T$ is some kind of test statistic, in our case counting the number of heads in a sample $X_1,dotsc,X_n$; $Theta$ is the parameter space in question (our paramter is $psimtheta$), $Theta_0$ the subspace corresponding to the null hypothesis, i.e.
beginequation
H_0: thetainTheta_0,quad H_1:thetainThetasetminusTheta_0;
endequation
and finally $K$ is the region of rejection of $H_0$, i.e.
beginequation
H_0text is rejected iff T(X_1,dotsc,T_n)in K.
endequation
So in particular we have $Theta=[0,1], Theta_0=[0,frac12]$, yielding
beginequation
alpha_max=max_pleqfrac12sum_i=k^n B_n,p(X=i),
endequation
which should now be $leq$ a given significance level.
[Definitions from http://www.wiwi.uni-muenster.de/05/download/studium/advancedstatistics/ss09/kapitel_6.pdf - couldn't find equivalent in English]
probability statistics binomial-distribution
asked Aug 2 at 15:07
Don Fuchs
234
Jetzt musst du nur noch eine konkrete Frage zu der Ergänzung stellen. Hast du eingentlich schon mal beim Binomialtest in Wiki vorbei geschaut?
– callculus
Aug 4 at 16:00
Wieso du hier ein maximales $alpha$ suchst ist unklar. Beim Hypothesentest ist dieser von vorneherein schon fest gelegt, wie du auch bei meiner Ungleichung in der Antwort siehst.
– callculus
Aug 4 at 16:14
Ich verstehe es jetzt so: Das Signifikanzniveau (meist mit $alpha$ bezeichnet) gibt eine obere Schranke für die W., einen Fehler 1. Art zu begehen, sagen $alpha_1$. Da man aber unter Annahme von $H_0$ $p$ nicht genau kennt (man hat nur eine Ungleichung), schätzt man $alpha_1$ durch $alpha_max$ ab.
– Don Fuchs
Aug 4 at 17:46
Und aus Monotoniegründen ist nun die Summe $sum B_n,p(X=i)$ maximal für $p=p_0=frac12$ (bei festem $k$) (die W. für mind. $k$ mal Kopf steigt mit der Erfolgwahrscheinlichkeit).
– Don Fuchs
Aug 4 at 17:54
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider the following classical statistical test setup:
One assumes a coin to be unfair in the sense that heads, say, occurs more frequently than tails. Thus we set $H_0: pleqfrac12$ as null hypothesis and $H_1:p>frac12$ as alternative where $p$ is the probability for heads.
Also let X count the occurence of heads when tossing the coin $n$ times. Given $n$ and a significance level $alpha$ we get the one-tail condition
beginequation
(1)quad P(Xgeq k)leqalpha
endequation
where $P$ has a $(n,p)$-binomial distribution with $pleqfrac12$ (thus yielding the probability for rejecting $H_0$ when it's actually true).
To solve $(1)$ for $k$ it would now be common (school book) practice to set $p=frac12$ and solve $(1)$ by inversion. But this isn't correct, as we just know $pleqfrac12$.
So wouldn't it be better to rather use a distribution for "$k$ wins out of $n$ with a probability of success $leqfrac12$" and which would that appropriate distribution be?
I want to be more precise: In a more general context the maximum $alpha$ error could be defined as
beginequation
alpha_max:=max_thetainTheta_0P_theta(T(X_1,dotsc,T_n)in K)
endequation
where $T$ is some kind of test statistic, in our case counting the number of heads in a sample $X_1,dotsc,X_n$; $Theta$ is the parameter space in question (our paramter is $psimtheta$), $Theta_0$ the subspace corresponding to the null hypothesis, i.e.
beginequation
H_0: thetainTheta_0,quad H_1:thetainThetasetminusTheta_0;
endequation
and finally $K$ is the region of rejection of $H_0$, i.e.
beginequation
H_0text is rejected iff T(X_1,dotsc,T_n)in K.
endequation
So in particular we have $Theta=[0,1], Theta_0=[0,frac12]$, yielding
beginequation
alpha_max=max_pleqfrac12sum_i=k^n B_n,p(X=i),
endequation
which should now be $leq$ a given significance level.
[Definitions from http://www.wiwi.uni-muenster.de/05/download/studium/advancedstatistics/ss09/kapitel_6.pdf - couldn't find equivalent in English]
probability statistics binomial-distribution
asked Aug 2 at 15:07
Don Fuchs
234
Consider the following classical statistical test setup:
One assumes a coin to be unfair in the sense that heads, say, occurs more frequently than tails. Thus we set $H_0: pleqfrac12$ as null hypothesis and $H_1:p>frac12$ as alternative where $p$ is the probability for heads.
Also let X count the occurence of heads when tossing the coin $n$ times. Given $n$ and a significance level $alpha$ we get the one-tail condition
beginequation
(1)quad P(Xgeq k)leqalpha
endequation
where $P$ has a $(n,p)$-binomial distribution with $pleqfrac12$ (thus yielding the probability for rejecting $H_0$ when it's actually true).
To solve $(1)$ for $k$ it would now be common (school book) practice to set $p=frac12$ and solve $(1)$ by inversion. But this isn't correct, as we just know $pleqfrac12$.
So wouldn't it be better to rather use a distribution for "$k$ wins out of $n$ with a probability of success $leqfrac12$" and which would that appropriate distribution be?
I want to be more precise: In a more general context the maximum $alpha$ error could be defined as
beginequation
alpha_max:=max_thetainTheta_0P_theta(T(X_1,dotsc,T_n)in K)
endequation
where $T$ is some kind of test statistic, in our case counting the number of heads in a sample $X_1,dotsc,X_n$; $Theta$ is the parameter space in question (our paramter is $psimtheta$), $Theta_0$ the subspace corresponding to the null hypothesis, i.e.
beginequation
H_0: thetainTheta_0,quad H_1:thetainThetasetminusTheta_0;
endequation
and finally $K$ is the region of rejection of $H_0$, i.e.
beginequation
H_0text is rejected iff T(X_1,dotsc,T_n)in K.
endequation
So in particular we have $Theta=[0,1], Theta_0=[0,frac12]$, yielding
beginequation
alpha_max=max_pleqfrac12sum_i=k^n B_n,p(X=i),
endequation
which should now be $leq$ a given significance level.
[Definitions from http://www.wiwi.uni-muenster.de/05/download/studium/advancedstatistics/ss09/kapitel_6.pdf - couldn't find equivalent in English]
probability statistics binomial-distribution
asked Aug 2 at 15:07
Don Fuchs
234
edited Aug 4 at 15:09
asked Aug 2 at 15:07
Don Fuchs
234
asked Aug 2 at 15:07
Don Fuchs
234
asked Aug 2 at 15:07
Don Fuchs
234
234
Jetzt musst du nur noch eine konkrete Frage zu der Ergänzung stellen. Hast du eingentlich schon mal beim Binomialtest in Wiki vorbei geschaut?
– callculus
Aug 4 at 16:00
Wieso du hier ein maximales $alpha$ suchst ist unklar. Beim Hypothesentest ist dieser von vorneherein schon fest gelegt, wie du auch bei meiner Ungleichung in der Antwort siehst.
– callculus
Aug 4 at 16:14
Ich verstehe es jetzt so: Das Signifikanzniveau (meist mit $alpha$ bezeichnet) gibt eine obere Schranke für die W., einen Fehler 1. Art zu begehen, sagen $alpha_1$. Da man aber unter Annahme von $H_0$ $p$ nicht genau kennt (man hat nur eine Ungleichung), schätzt man $alpha_1$ durch $alpha_max$ ab.
– Don Fuchs
Aug 4 at 17:46
Und aus Monotoniegründen ist nun die Summe $sum B_n,p(X=i)$ maximal für $p=p_0=frac12$ (bei festem $k$) (die W. für mind. $k$ mal Kopf steigt mit der Erfolgwahrscheinlichkeit).
– Don Fuchs
Aug 4 at 17:54
add a comment |Â
Jetzt musst du nur noch eine konkrete Frage zu der Ergänzung stellen. Hast du eingentlich schon mal beim Binomialtest in Wiki vorbei geschaut?
– callculus
Aug 4 at 16:00
Wieso du hier ein maximales $alpha$ suchst ist unklar. Beim Hypothesentest ist dieser von vorneherein schon fest gelegt, wie du auch bei meiner Ungleichung in der Antwort siehst.
– callculus
Aug 4 at 16:14
Ich verstehe es jetzt so: Das Signifikanzniveau (meist mit $alpha$ bezeichnet) gibt eine obere Schranke für die W., einen Fehler 1. Art zu begehen, sagen $alpha_1$. Da man aber unter Annahme von $H_0$ $p$ nicht genau kennt (man hat nur eine Ungleichung), schätzt man $alpha_1$ durch $alpha_max$ ab.
– Don Fuchs
Aug 4 at 17:46
Und aus Monotoniegründen ist nun die Summe $sum B_n,p(X=i)$ maximal für $p=p_0=frac12$ (bei festem $k$) (die W. für mind. $k$ mal Kopf steigt mit der Erfolgwahrscheinlichkeit).
– Don Fuchs
Aug 4 at 17:54
Jetzt musst du nur noch eine konkrete Frage zu der Ergänzung stellen. Hast du eingentlich schon mal beim Binomialtest in Wiki vorbei geschaut?
– callculus
Aug 4 at 16:00
Jetzt musst du nur noch eine konkrete Frage zu der Ergänzung stellen. Hast du eingentlich schon mal beim Binomialtest in Wiki vorbei geschaut?
– callculus
Aug 4 at 16:00
Wieso du hier ein maximales $alpha$ suchst ist unklar. Beim Hypothesentest ist dieser von vorneherein schon fest gelegt, wie du auch bei meiner Ungleichung in der Antwort siehst.
– callculus
Aug 4 at 16:14
Wieso du hier ein maximales $alpha$ suchst ist unklar. Beim Hypothesentest ist dieser von vorneherein schon fest gelegt, wie du auch bei meiner Ungleichung in der Antwort siehst.
– callculus
Aug 4 at 16:14
Ich verstehe es jetzt so: Das Signifikanzniveau (meist mit $alpha$ bezeichnet) gibt eine obere Schranke für die W., einen Fehler 1. Art zu begehen, sagen $alpha_1$. Da man aber unter Annahme von $H_0$ $p$ nicht genau kennt (man hat nur eine Ungleichung), schätzt man $alpha_1$ durch $alpha_max$ ab.
– Don Fuchs
Aug 4 at 17:46
Ich verstehe es jetzt so: Das Signifikanzniveau (meist mit $alpha$ bezeichnet) gibt eine obere Schranke für die W., einen Fehler 1. Art zu begehen, sagen $alpha_1$. Da man aber unter Annahme von $H_0$ $p$ nicht genau kennt (man hat nur eine Ungleichung), schätzt man $alpha_1$ durch $alpha_max$ ab.
– Don Fuchs
Aug 4 at 17:46
Und aus Monotoniegründen ist nun die Summe $sum B_n,p(X=i)$ maximal für $p=p_0=frac12$ (bei festem $k$) (die W. für mind. $k$ mal Kopf steigt mit der Erfolgwahrscheinlichkeit).
– Don Fuchs
Aug 4 at 17:54
Und aus Monotoniegründen ist nun die Summe $sum B_n,p(X=i)$ maximal für $p=p_0=frac12$ (bei festem $k$) (die W. für mind. $k$ mal Kopf steigt mit der Erfolgwahrscheinlichkeit).
– Don Fuchs
Aug 4 at 17:54
add a comment |Â
1 Answer
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Both Null hypothesis are possible. The crucial point is the definition of the alternative hypothesis, $H_1$. This definition is unique as you can see at the table below. $$beginarray hline &H_0 &H_1 \ hline texttttwo-tailed & p=p_0 &pneq p_0 \ hline textttright-tailed & p=p_0 textor pleq p_0 &p>p_0 \ hline textttleft-tailed & p=p_0 textor pgeq p_0 &p<p_0 \ hline endarray$$
For the right-tailed case you evaluate the the smallest value of $c$, where
$$sum_i=c^n B(i| p_0,n)leq alpha$$
Then the critical range is $c, c+1, ldots, n $.
answered Aug 2 at 16:21
callculus
16.2k31427
Alright, but still the question remains why we use $p_0$ in your sum in the cases where $H_0:pleq p_0$ (or $geq$): In these cases we do not know the underlying probability distibution to calculate $P(H_0text is trueland H_0text is rejected)$.
– Don Fuchs
Aug 2 at 18:03
In your case $p_0=frac12$
– callculus
Aug 2 at 18:06
Of course, but what if $p=frac14<frac12=p_0$. Assuming $H_0$, for all we know this could be the case.
– Don Fuchs
Aug 2 at 18:13
@DonFuchs We don´t know the real value of $p$-before and after the test. The only statement we can make that is the following. If the estimated value of $pcdot n$ is in the interval $c, c+1, ldots, n $ we do not accept the Null hypothesis with a statistictial significance of $alpha$. Or we do not reject the alternative hypothesis $pgeq frac12$ with a statistictial significance of $alpha$
– callculus
Aug 2 at 18:26
Sorry, but that doesn't convince. The point is: What we do in your sum above is to calculate $P(text$H_0$ is truelandtext$H_0$ rejected)$ (which is the type 1 or $alpha$ error) without really taking the condition $text$H_0$ is true$ into account (by additionaly assuming $p$ not to be less than $p_0$).
– Don Fuchs
Aug 4 at 14:08
 |Â
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1 Answer
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1 Answer
1
active
oldest
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active
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active
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Both Null hypothesis are possible. The crucial point is the definition of the alternative hypothesis, $H_1$. This definition is unique as you can see at the table below. $$beginarray hline &H_0 &H_1 \ hline texttttwo-tailed & p=p_0 &pneq p_0 \ hline textttright-tailed & p=p_0 textor pleq p_0 &p>p_0 \ hline textttleft-tailed & p=p_0 textor pgeq p_0 &p<p_0 \ hline endarray$$
For the right-tailed case you evaluate the the smallest value of $c$, where
$$sum_i=c^n B(i| p_0,n)leq alpha$$
Then the critical range is $c, c+1, ldots, n $.
answered Aug 2 at 16:21
callculus
16.2k31427
Alright, but still the question remains why we use $p_0$ in your sum in the cases where $H_0:pleq p_0$ (or $geq$): In these cases we do not know the underlying probability distibution to calculate $P(H_0text is trueland H_0text is rejected)$.
– Don Fuchs
Aug 2 at 18:03
In your case $p_0=frac12$
– callculus
Aug 2 at 18:06
Of course, but what if $p=frac14<frac12=p_0$. Assuming $H_0$, for all we know this could be the case.
– Don Fuchs
Aug 2 at 18:13
@DonFuchs We don´t know the real value of $p$-before and after the test. The only statement we can make that is the following. If the estimated value of $pcdot n$ is in the interval $c, c+1, ldots, n $ we do not accept the Null hypothesis with a statistictial significance of $alpha$. Or we do not reject the alternative hypothesis $pgeq frac12$ with a statistictial significance of $alpha$
– callculus
Aug 2 at 18:26
Sorry, but that doesn't convince. The point is: What we do in your sum above is to calculate $P(text$H_0$ is truelandtext$H_0$ rejected)$ (which is the type 1 or $alpha$ error) without really taking the condition $text$H_0$ is true$ into account (by additionaly assuming $p$ not to be less than $p_0$).
– Don Fuchs
Aug 4 at 14:08
 |Â
show 3 more comments
up vote
0
down vote
Both Null hypothesis are possible. The crucial point is the definition of the alternative hypothesis, $H_1$. This definition is unique as you can see at the table below. $$beginarray hline &H_0 &H_1 \ hline texttttwo-tailed & p=p_0 &pneq p_0 \ hline textttright-tailed & p=p_0 textor pleq p_0 &p>p_0 \ hline textttleft-tailed & p=p_0 textor pgeq p_0 &p<p_0 \ hline endarray$$
For the right-tailed case you evaluate the the smallest value of $c$, where
$$sum_i=c^n B(i| p_0,n)leq alpha$$
Then the critical range is $c, c+1, ldots, n $.
answered Aug 2 at 16:21
callculus
16.2k31427
Alright, but still the question remains why we use $p_0$ in your sum in the cases where $H_0:pleq p_0$ (or $geq$): In these cases we do not know the underlying probability distibution to calculate $P(H_0text is trueland H_0text is rejected)$.
– Don Fuchs
Aug 2 at 18:03
In your case $p_0=frac12$
– callculus
Aug 2 at 18:06
Of course, but what if $p=frac14<frac12=p_0$. Assuming $H_0$, for all we know this could be the case.
– Don Fuchs
Aug 2 at 18:13
@DonFuchs We don´t know the real value of $p$-before and after the test. The only statement we can make that is the following. If the estimated value of $pcdot n$ is in the interval $c, c+1, ldots, n $ we do not accept the Null hypothesis with a statistictial significance of $alpha$. Or we do not reject the alternative hypothesis $pgeq frac12$ with a statistictial significance of $alpha$
– callculus
Aug 2 at 18:26
Sorry, but that doesn't convince. The point is: What we do in your sum above is to calculate $P(text$H_0$ is truelandtext$H_0$ rejected)$ (which is the type 1 or $alpha$ error) without really taking the condition $text$H_0$ is true$ into account (by additionaly assuming $p$ not to be less than $p_0$).
– Don Fuchs
Aug 4 at 14:08
 |Â
show 3 more comments
up vote
0
down vote
up vote
0
down vote
Both Null hypothesis are possible. The crucial point is the definition of the alternative hypothesis, $H_1$. This definition is unique as you can see at the table below. $$beginarray hline &H_0 &H_1 \ hline texttttwo-tailed & p=p_0 &pneq p_0 \ hline textttright-tailed & p=p_0 textor pleq p_0 &p>p_0 \ hline textttleft-tailed & p=p_0 textor pgeq p_0 &p<p_0 \ hline endarray$$
For the right-tailed case you evaluate the the smallest value of $c$, where
$$sum_i=c^n B(i| p_0,n)leq alpha$$
Then the critical range is $c, c+1, ldots, n $.
answered Aug 2 at 16:21
callculus
16.2k31427
Both Null hypothesis are possible. The crucial point is the definition of the alternative hypothesis, $H_1$. This definition is unique as you can see at the table below. $$beginarray hline &H_0 &H_1 \ hline texttttwo-tailed & p=p_0 &pneq p_0 \ hline textttright-tailed & p=p_0 textor pleq p_0 &p>p_0 \ hline textttleft-tailed & p=p_0 textor pgeq p_0 &p<p_0 \ hline endarray$$
For the right-tailed case you evaluate the the smallest value of $c$, where
$$sum_i=c^n B(i| p_0,n)leq alpha$$
Then the critical range is $c, c+1, ldots, n $.
answered Aug 2 at 16:21
callculus
16.2k31427
answered Aug 2 at 16:21
callculus
16.2k31427
answered Aug 2 at 16:21
callculus
16.2k31427
answered Aug 2 at 16:21
callculus
16.2k31427
16.2k31427
Alright, but still the question remains why we use $p_0$ in your sum in the cases where $H_0:pleq p_0$ (or $geq$): In these cases we do not know the underlying probability distibution to calculate $P(H_0text is trueland H_0text is rejected)$.
– Don Fuchs
Aug 2 at 18:03
In your case $p_0=frac12$
– callculus
Aug 2 at 18:06
Of course, but what if $p=frac14<frac12=p_0$. Assuming $H_0$, for all we know this could be the case.
– Don Fuchs
Aug 2 at 18:13
@DonFuchs We don´t know the real value of $p$-before and after the test. The only statement we can make that is the following. If the estimated value of $pcdot n$ is in the interval $c, c+1, ldots, n $ we do not accept the Null hypothesis with a statistictial significance of $alpha$. Or we do not reject the alternative hypothesis $pgeq frac12$ with a statistictial significance of $alpha$
– callculus
Aug 2 at 18:26
Sorry, but that doesn't convince. The point is: What we do in your sum above is to calculate $P(text$H_0$ is truelandtext$H_0$ rejected)$ (which is the type 1 or $alpha$ error) without really taking the condition $text$H_0$ is true$ into account (by additionaly assuming $p$ not to be less than $p_0$).
– Don Fuchs
Aug 4 at 14:08
 |Â
show 3 more comments
Alright, but still the question remains why we use $p_0$ in your sum in the cases where $H_0:pleq p_0$ (or $geq$): In these cases we do not know the underlying probability distibution to calculate $P(H_0text is trueland H_0text is rejected)$.
– Don Fuchs
Aug 2 at 18:03
In your case $p_0=frac12$
– callculus
Aug 2 at 18:06
Of course, but what if $p=frac14<frac12=p_0$. Assuming $H_0$, for all we know this could be the case.
– Don Fuchs
Aug 2 at 18:13
@DonFuchs We don´t know the real value of $p$-before and after the test. The only statement we can make that is the following. If the estimated value of $pcdot n$ is in the interval $c, c+1, ldots, n $ we do not accept the Null hypothesis with a statistictial significance of $alpha$. Or we do not reject the alternative hypothesis $pgeq frac12$ with a statistictial significance of $alpha$
– callculus
Aug 2 at 18:26
Sorry, but that doesn't convince. The point is: What we do in your sum above is to calculate $P(text$H_0$ is truelandtext$H_0$ rejected)$ (which is the type 1 or $alpha$ error) without really taking the condition $text$H_0$ is true$ into account (by additionaly assuming $p$ not to be less than $p_0$).
– Don Fuchs
Aug 4 at 14:08
Alright, but still the question remains why we use $p_0$ in your sum in the cases where $H_0:pleq p_0$ (or $geq$): In these cases we do not know the underlying probability distibution to calculate $P(H_0text is trueland H_0text is rejected)$.
– Don Fuchs
Aug 2 at 18:03
Alright, but still the question remains why we use $p_0$ in your sum in the cases where $H_0:pleq p_0$ (or $geq$): In these cases we do not know the underlying probability distibution to calculate $P(H_0text is trueland H_0text is rejected)$.
– Don Fuchs
Aug 2 at 18:03
In your case $p_0=frac12$
– callculus
Aug 2 at 18:06
In your case $p_0=frac12$
– callculus
Aug 2 at 18:06
Of course, but what if $p=frac14<frac12=p_0$. Assuming $H_0$, for all we know this could be the case.
– Don Fuchs
Aug 2 at 18:13
Of course, but what if $p=frac14<frac12=p_0$. Assuming $H_0$, for all we know this could be the case.
– Don Fuchs
Aug 2 at 18:13
@DonFuchs We don´t know the real value of $p$-before and after the test. The only statement we can make that is the following. If the estimated value of $pcdot n$ is in the interval $c, c+1, ldots, n $ we do not accept the Null hypothesis with a statistictial significance of $alpha$. Or we do not reject the alternative hypothesis $pgeq frac12$ with a statistictial significance of $alpha$
– callculus
Aug 2 at 18:26
@DonFuchs We don´t know the real value of $p$-before and after the test. The only statement we can make that is the following. If the estimated value of $pcdot n$ is in the interval $c, c+1, ldots, n $ we do not accept the Null hypothesis with a statistictial significance of $alpha$. Or we do not reject the alternative hypothesis $pgeq frac12$ with a statistictial significance of $alpha$
– callculus
Aug 2 at 18:26
Sorry, but that doesn't convince. The point is: What we do in your sum above is to calculate $P(text$H_0$ is truelandtext$H_0$ rejected)$ (which is the type 1 or $alpha$ error) without really taking the condition $text$H_0$ is true$ into account (by additionaly assuming $p$ not to be less than $p_0$).
– Don Fuchs
Aug 4 at 14:08
Sorry, but that doesn't convince. The point is: What we do in your sum above is to calculate $P(text$H_0$ is truelandtext$H_0$ rejected)$ (which is the type 1 or $alpha$ error) without really taking the condition $text$H_0$ is true$ into account (by additionaly assuming $p$ not to be less than $p_0$).
– Don Fuchs
Aug 4 at 14:08
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Jetzt musst du nur noch eine konkrete Frage zu der Ergänzung stellen. Hast du eingentlich schon mal beim Binomialtest in Wiki vorbei geschaut?
– callculus
Aug 4 at 16:00
Wieso du hier ein maximales $alpha$ suchst ist unklar. Beim Hypothesentest ist dieser von vorneherein schon fest gelegt, wie du auch bei meiner Ungleichung in der Antwort siehst.
– callculus
Aug 4 at 16:14
Ich verstehe es jetzt so: Das Signifikanzniveau (meist mit $alpha$ bezeichnet) gibt eine obere Schranke für die W., einen Fehler 1. Art zu begehen, sagen $alpha_1$. Da man aber unter Annahme von $H_0$ $p$ nicht genau kennt (man hat nur eine Ungleichung), schätzt man $alpha_1$ durch $alpha_max$ ab.
– Don Fuchs
Aug 4 at 17:46
Und aus Monotoniegründen ist nun die Summe $sum B_n,p(X=i)$ maximal für $p=p_0=frac12$ (bei festem $k$) (die W. für mind. $k$ mal Kopf steigt mit der Erfolgwahrscheinlichkeit).
– Don Fuchs
Aug 4 at 17:54