Mixing
Determine the critical region for this test at significance level $alpha = 0.05$.
$begingroup$
Let the random variable $X$ be the waiting time until the next student has to go to the toilet. Assume that $X$ has an $Exp(lambda)$ distribution with unknown $lambda$. We test $H_0 :lambda= 0.2$ against $H_1 :lambda < 0.2$, where we use $X$ as test statistic. Determine the critical region for this test at significance level $alpha = 0.05$.
In my opinion I should calculate $P(X<C)|H_0)=1-e^{-lambda x}=1-e^{-0.2x}$. So $1-e^{-0.2x}=0.05$, so $x=0.25$ and the critical region is $(-infty,0.25]$, is it the right method to solve this question?
statistics
asked Jan 23 at 20:38
Mark JaconMark Jacon
1127
$endgroup$
|
show 7 more comments
$begingroup$
Let the random variable $X$ be the waiting time until the next student has to go to the toilet. Assume that $X$ has an $Exp(lambda)$ distribution with unknown $lambda$. We test $H_0 :lambda= 0.2$ against $H_1 :lambda < 0.2$, where we use $X$ as test statistic. Determine the critical region for this test at significance level $alpha = 0.05$.
In my opinion I should calculate $P(X<C)|H_0)=1-e^{-lambda x}=1-e^{-0.2x}$. So $1-e^{-0.2x}=0.05$, so $x=0.25$ and the critical region is $(-infty,0.25]$, is it the right method to solve this question?
statistics
asked Jan 23 at 20:38
Mark JaconMark Jacon
1127
$endgroup$
1
$begingroup$
Confusing notation here. You say $Xsim Exp(X)$ with unknown $A$ then immediately start talking about $lambda$. Surely you mean $Xsim Exp(lambda)$ where $mathbb{E}(X)=1/lambda$ and $lambda$ is the unknown parameter you want to test on.
$endgroup$
– LoveTooNap29
Jan 23 at 20:55
1
$begingroup$
What do you mean by "unknown A"? What is A? And what is the sample size?
$endgroup$
– callculus
Jan 23 at 21:08
1
$begingroup$
@callculus I think that it's time that I learn how to write on latex, sorry guys, it should be a $lambda$.
$endgroup$
– Mark Jacon
Jan 23 at 21:11
1
$begingroup$
And the sample size is what, equal to 1?
$endgroup$
– callculus
Jan 23 at 21:13
2
$begingroup$
Attempt to clairfy: You seem to have only a single observation $X$ to test this hypothesis. Clearly, you want to reject when $X$ is small. In particular you want $P(X < c|lambda = .2) = 0.05.$ In R statistical software you can find $c = 0.2565$ using codeqexp(.05,.2), which returns 0.2564665, and confirm using codepexp(.2565, .2), which returns 0.05000637. In R,pexpis an exponential CDF andqexpis the quantile function (inverse CDF). For this simple problem you don't need software, but it is good to have the framework in mind.
$endgroup$
– BruceET
Jan 23 at 22:03
|
show 7 more comments
$begingroup$
Let the random variable $X$ be the waiting time until the next student has to go to the toilet. Assume that $X$ has an $Exp(lambda)$ distribution with unknown $lambda$. We test $H_0 :lambda= 0.2$ against $H_1 :lambda < 0.2$, where we use $X$ as test statistic. Determine the critical region for this test at significance level $alpha = 0.05$.
In my opinion I should calculate $P(X<C)|H_0)=1-e^{-lambda x}=1-e^{-0.2x}$. So $1-e^{-0.2x}=0.05$, so $x=0.25$ and the critical region is $(-infty,0.25]$, is it the right method to solve this question?
statistics
asked Jan 23 at 20:38
Mark JaconMark Jacon
1127
$endgroup$
Let the random variable $X$ be the waiting time until the next student has to go to the toilet. Assume that $X$ has an $Exp(lambda)$ distribution with unknown $lambda$. We test $H_0 :lambda= 0.2$ against $H_1 :lambda < 0.2$, where we use $X$ as test statistic. Determine the critical region for this test at significance level $alpha = 0.05$.
In my opinion I should calculate $P(X<C)|H_0)=1-e^{-lambda x}=1-e^{-0.2x}$. So $1-e^{-0.2x}=0.05$, so $x=0.25$ and the critical region is $(-infty,0.25]$, is it the right method to solve this question?
statistics
statistics
asked Jan 23 at 20:38
Mark JaconMark Jacon
1127
asked Jan 23 at 20:38
Mark JaconMark Jacon
1127
edited Jan 23 at 21:49
Mark Jacon
asked Jan 23 at 20:38
Mark JaconMark Jacon
1127
asked Jan 23 at 20:38
Mark JaconMark Jacon
1127
asked Jan 23 at 20:38
Mark JaconMark Jacon
1127
1127
1
$begingroup$
Confusing notation here. You say $Xsim Exp(X)$ with unknown $A$ then immediately start talking about $lambda$. Surely you mean $Xsim Exp(lambda)$ where $mathbb{E}(X)=1/lambda$ and $lambda$ is the unknown parameter you want to test on.
$endgroup$
– LoveTooNap29
Jan 23 at 20:55
1
$begingroup$
What do you mean by "unknown A"? What is A? And what is the sample size?
$endgroup$
– callculus
Jan 23 at 21:08
1
$begingroup$
@callculus I think that it's time that I learn how to write on latex, sorry guys, it should be a $lambda$.
$endgroup$
– Mark Jacon
Jan 23 at 21:11
1
$begingroup$
And the sample size is what, equal to 1?
$endgroup$
– callculus
Jan 23 at 21:13
2
$begingroup$
Attempt to clairfy: You seem to have only a single observation $X$ to test this hypothesis. Clearly, you want to reject when $X$ is small. In particular you want $P(X < c|lambda = .2) = 0.05.$ In R statistical software you can find $c = 0.2565$ using codeqexp(.05,.2), which returns 0.2564665, and confirm using codepexp(.2565, .2), which returns 0.05000637. In R,pexpis an exponential CDF andqexpis the quantile function (inverse CDF). For this simple problem you don't need software, but it is good to have the framework in mind.
$endgroup$
– BruceET
Jan 23 at 22:03
|
show 7 more comments
1
$begingroup$
Confusing notation here. You say $Xsim Exp(X)$ with unknown $A$ then immediately start talking about $lambda$. Surely you mean $Xsim Exp(lambda)$ where $mathbb{E}(X)=1/lambda$ and $lambda$ is the unknown parameter you want to test on.
$endgroup$
– LoveTooNap29
Jan 23 at 20:55
1
$begingroup$
What do you mean by "unknown A"? What is A? And what is the sample size?
$endgroup$
– callculus
Jan 23 at 21:08
1
$begingroup$
@callculus I think that it's time that I learn how to write on latex, sorry guys, it should be a $lambda$.
$endgroup$
– Mark Jacon
Jan 23 at 21:11
1
$begingroup$
And the sample size is what, equal to 1?
$endgroup$
– callculus
Jan 23 at 21:13
2
$begingroup$
Attempt to clairfy: You seem to have only a single observation $X$ to test this hypothesis. Clearly, you want to reject when $X$ is small. In particular you want $P(X < c|lambda = .2) = 0.05.$ In R statistical software you can find $c = 0.2565$ using codeqexp(.05,.2), which returns 0.2564665, and confirm using codepexp(.2565, .2), which returns 0.05000637. In R,pexpis an exponential CDF andqexpis the quantile function (inverse CDF). For this simple problem you don't need software, but it is good to have the framework in mind.
$endgroup$
– BruceET
Jan 23 at 22:03
1
1
$begingroup$
Confusing notation here. You say $Xsim Exp(X)$ with unknown $A$ then immediately start talking about $lambda$. Surely you mean $Xsim Exp(lambda)$ where $mathbb{E}(X)=1/lambda$ and $lambda$ is the unknown parameter you want to test on.
$endgroup$
– LoveTooNap29
Jan 23 at 20:55
$begingroup$
Confusing notation here. You say $Xsim Exp(X)$ with unknown $A$ then immediately start talking about $lambda$. Surely you mean $Xsim Exp(lambda)$ where $mathbb{E}(X)=1/lambda$ and $lambda$ is the unknown parameter you want to test on.
$endgroup$
– LoveTooNap29
Jan 23 at 20:55
1
1
$begingroup$
What do you mean by "unknown A"? What is A? And what is the sample size?
$endgroup$
– callculus
Jan 23 at 21:08
$begingroup$
What do you mean by "unknown A"? What is A? And what is the sample size?
$endgroup$
– callculus
Jan 23 at 21:08
1
1
$begingroup$
@callculus I think that it's time that I learn how to write on latex, sorry guys, it should be a $lambda$.
$endgroup$
– Mark Jacon
Jan 23 at 21:11
$begingroup$
@callculus I think that it's time that I learn how to write on latex, sorry guys, it should be a $lambda$.
$endgroup$
– Mark Jacon
Jan 23 at 21:11
1
1
$begingroup$
And the sample size is what, equal to 1?
$endgroup$
– callculus
Jan 23 at 21:13
$begingroup$
And the sample size is what, equal to 1?
$endgroup$
– callculus
Jan 23 at 21:13
2
2
$begingroup$
Attempt to clairfy: You seem to have only a single observation $X$ to test this hypothesis. Clearly, you want to reject when $X$ is small. In particular you want $P(X < c|lambda = .2) = 0.05.$ In R statistical software you can find $c = 0.2565$ using code
qexp(.05,.2), which returns 0.2564665, and confirm using code pexp(.2565, .2), which returns 0.05000637. In R, pexp is an exponential CDF and qexp is the quantile function (inverse CDF). For this simple problem you don't need software, but it is good to have the framework in mind.$endgroup$
– BruceET
Jan 23 at 22:03
$begingroup$
Attempt to clairfy: You seem to have only a single observation $X$ to test this hypothesis. Clearly, you want to reject when $X$ is small. In particular you want $P(X < c|lambda = .2) = 0.05.$ In R statistical software you can find $c = 0.2565$ using code
qexp(.05,.2), which returns 0.2564665, and confirm using code pexp(.2565, .2), which returns 0.05000637. In R, pexp is an exponential CDF and qexp is the quantile function (inverse CDF). For this simple problem you don't need software, but it is good to have the framework in mind.$endgroup$
– BruceET
Jan 23 at 22:03
|
show 7 more comments
1 Answer
1
active
oldest
votes
$begingroup$
In fact, any region $Rsubset mathbb R$ such that
$$P_{lambda=0.2}(Xin R)=0.05$$
is a correct answer, in theory. But some of these answers would be absurd in practice.
Since the alternative hypothesis is that $lambda<0.2$, and since this is equivalent to say that $E(X)>frac1{0.2}=5$, we see that it is only reasonable to reject $H_0$ when $X$ takes values sensibly greater than $5$. That is, the critical region should be
$$(C,infty)$$
for $Cinmathbb R$ such that
$$P(Xin (C,infty))=P(X>C)=1-(1-e^{-0.2C})=e^{-0.2C}=0.05.$$
So $C=-5ln(0.05)approx 14.98$ and the critical region is $$(14.98,+infty).$$
answered Jan 23 at 23:33
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
$begingroup$
why you assume the $E(X)$ and not the value of lambda? The critical region should not be the area where we reject $H_0$ in favour of $H_1$, so where $lambda<0.2$?
$endgroup$
– Mark Jacon
Jan 24 at 8:17
1
$begingroup$
By definition, the critical region is any region $R$ such that $$P(Xin R)=alpha$$ if $H_0$ is true. So both mine or your answer can be technically right. But for the test to make sense in practice (or, more rigorously, to have good properties, like so-called unbiasedness), you have to build $R$ using those values which are more likely to occur if $H_1$ is true. And the case is that the smaller $lambda$ is, the greater the values of $X$ expected. This is the case for the exponential distribution, and that's why it is reasonable a critical region that contains big, not small, values.
$endgroup$
– Alejandro Nasif Salum
Jan 27 at 0:24
1
$begingroup$
I mean, let's assume that $H_0$ is false, because actually $lambda=0.0001$. The expected value of $X$ would be $10,000$, so you want to choose a critical region that includes the greatest values. If you chose a critical region such as $(0,0.25)$ or so, it is almost sure that you will fail to reject $H_0$ (because those values are most likely not going to occur, given that $E(X)=10,000$), even when it is really obvious that $lambda$ is much smaller than what is assumed in $H_0$. That is, $H_0$ is evidently false, but the test would say that you have to keep this absurd hypothesis.
$endgroup$
– Alejandro Nasif Salum
Jan 27 at 0:29
add a comment |
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$begingroup$
In fact, any region $Rsubset mathbb R$ such that
$$P_{lambda=0.2}(Xin R)=0.05$$
is a correct answer, in theory. But some of these answers would be absurd in practice.
Since the alternative hypothesis is that $lambda<0.2$, and since this is equivalent to say that $E(X)>frac1{0.2}=5$, we see that it is only reasonable to reject $H_0$ when $X$ takes values sensibly greater than $5$. That is, the critical region should be
$$(C,infty)$$
for $Cinmathbb R$ such that
$$P(Xin (C,infty))=P(X>C)=1-(1-e^{-0.2C})=e^{-0.2C}=0.05.$$
So $C=-5ln(0.05)approx 14.98$ and the critical region is $$(14.98,+infty).$$
answered Jan 23 at 23:33
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
$begingroup$
why you assume the $E(X)$ and not the value of lambda? The critical region should not be the area where we reject $H_0$ in favour of $H_1$, so where $lambda<0.2$?
$endgroup$
– Mark Jacon
Jan 24 at 8:17
1
$begingroup$
By definition, the critical region is any region $R$ such that $$P(Xin R)=alpha$$ if $H_0$ is true. So both mine or your answer can be technically right. But for the test to make sense in practice (or, more rigorously, to have good properties, like so-called unbiasedness), you have to build $R$ using those values which are more likely to occur if $H_1$ is true. And the case is that the smaller $lambda$ is, the greater the values of $X$ expected. This is the case for the exponential distribution, and that's why it is reasonable a critical region that contains big, not small, values.
$endgroup$
– Alejandro Nasif Salum
Jan 27 at 0:24
1
$begingroup$
I mean, let's assume that $H_0$ is false, because actually $lambda=0.0001$. The expected value of $X$ would be $10,000$, so you want to choose a critical region that includes the greatest values. If you chose a critical region such as $(0,0.25)$ or so, it is almost sure that you will fail to reject $H_0$ (because those values are most likely not going to occur, given that $E(X)=10,000$), even when it is really obvious that $lambda$ is much smaller than what is assumed in $H_0$. That is, $H_0$ is evidently false, but the test would say that you have to keep this absurd hypothesis.
$endgroup$
– Alejandro Nasif Salum
Jan 27 at 0:29
add a comment |
$begingroup$
In fact, any region $Rsubset mathbb R$ such that
$$P_{lambda=0.2}(Xin R)=0.05$$
is a correct answer, in theory. But some of these answers would be absurd in practice.
Since the alternative hypothesis is that $lambda<0.2$, and since this is equivalent to say that $E(X)>frac1{0.2}=5$, we see that it is only reasonable to reject $H_0$ when $X$ takes values sensibly greater than $5$. That is, the critical region should be
$$(C,infty)$$
for $Cinmathbb R$ such that
$$P(Xin (C,infty))=P(X>C)=1-(1-e^{-0.2C})=e^{-0.2C}=0.05.$$
So $C=-5ln(0.05)approx 14.98$ and the critical region is $$(14.98,+infty).$$
answered Jan 23 at 23:33
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
$begingroup$
why you assume the $E(X)$ and not the value of lambda? The critical region should not be the area where we reject $H_0$ in favour of $H_1$, so where $lambda<0.2$?
$endgroup$
– Mark Jacon
Jan 24 at 8:17
1
$begingroup$
By definition, the critical region is any region $R$ such that $$P(Xin R)=alpha$$ if $H_0$ is true. So both mine or your answer can be technically right. But for the test to make sense in practice (or, more rigorously, to have good properties, like so-called unbiasedness), you have to build $R$ using those values which are more likely to occur if $H_1$ is true. And the case is that the smaller $lambda$ is, the greater the values of $X$ expected. This is the case for the exponential distribution, and that's why it is reasonable a critical region that contains big, not small, values.
$endgroup$
– Alejandro Nasif Salum
Jan 27 at 0:24
1
$begingroup$
I mean, let's assume that $H_0$ is false, because actually $lambda=0.0001$. The expected value of $X$ would be $10,000$, so you want to choose a critical region that includes the greatest values. If you chose a critical region such as $(0,0.25)$ or so, it is almost sure that you will fail to reject $H_0$ (because those values are most likely not going to occur, given that $E(X)=10,000$), even when it is really obvious that $lambda$ is much smaller than what is assumed in $H_0$. That is, $H_0$ is evidently false, but the test would say that you have to keep this absurd hypothesis.
$endgroup$
– Alejandro Nasif Salum
Jan 27 at 0:29
add a comment |
$begingroup$
In fact, any region $Rsubset mathbb R$ such that
$$P_{lambda=0.2}(Xin R)=0.05$$
is a correct answer, in theory. But some of these answers would be absurd in practice.
Since the alternative hypothesis is that $lambda<0.2$, and since this is equivalent to say that $E(X)>frac1{0.2}=5$, we see that it is only reasonable to reject $H_0$ when $X$ takes values sensibly greater than $5$. That is, the critical region should be
$$(C,infty)$$
for $Cinmathbb R$ such that
$$P(Xin (C,infty))=P(X>C)=1-(1-e^{-0.2C})=e^{-0.2C}=0.05.$$
So $C=-5ln(0.05)approx 14.98$ and the critical region is $$(14.98,+infty).$$
answered Jan 23 at 23:33
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
In fact, any region $Rsubset mathbb R$ such that
$$P_{lambda=0.2}(Xin R)=0.05$$
is a correct answer, in theory. But some of these answers would be absurd in practice.
Since the alternative hypothesis is that $lambda<0.2$, and since this is equivalent to say that $E(X)>frac1{0.2}=5$, we see that it is only reasonable to reject $H_0$ when $X$ takes values sensibly greater than $5$. That is, the critical region should be
$$(C,infty)$$
for $Cinmathbb R$ such that
$$P(Xin (C,infty))=P(X>C)=1-(1-e^{-0.2C})=e^{-0.2C}=0.05.$$
So $C=-5ln(0.05)approx 14.98$ and the critical region is $$(14.98,+infty).$$
answered Jan 23 at 23:33
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Jan 23 at 23:33
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Jan 23 at 23:33
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Jan 23 at 23:33
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
4,765118
$begingroup$
why you assume the $E(X)$ and not the value of lambda? The critical region should not be the area where we reject $H_0$ in favour of $H_1$, so where $lambda<0.2$?
$endgroup$
– Mark Jacon
Jan 24 at 8:17
1
$begingroup$
By definition, the critical region is any region $R$ such that $$P(Xin R)=alpha$$ if $H_0$ is true. So both mine or your answer can be technically right. But for the test to make sense in practice (or, more rigorously, to have good properties, like so-called unbiasedness), you have to build $R$ using those values which are more likely to occur if $H_1$ is true. And the case is that the smaller $lambda$ is, the greater the values of $X$ expected. This is the case for the exponential distribution, and that's why it is reasonable a critical region that contains big, not small, values.
$endgroup$
– Alejandro Nasif Salum
Jan 27 at 0:24
1
$begingroup$
I mean, let's assume that $H_0$ is false, because actually $lambda=0.0001$. The expected value of $X$ would be $10,000$, so you want to choose a critical region that includes the greatest values. If you chose a critical region such as $(0,0.25)$ or so, it is almost sure that you will fail to reject $H_0$ (because those values are most likely not going to occur, given that $E(X)=10,000$), even when it is really obvious that $lambda$ is much smaller than what is assumed in $H_0$. That is, $H_0$ is evidently false, but the test would say that you have to keep this absurd hypothesis.
$endgroup$
– Alejandro Nasif Salum
Jan 27 at 0:29
add a comment |
$begingroup$
why you assume the $E(X)$ and not the value of lambda? The critical region should not be the area where we reject $H_0$ in favour of $H_1$, so where $lambda<0.2$?
$endgroup$
– Mark Jacon
Jan 24 at 8:17
1
$begingroup$
By definition, the critical region is any region $R$ such that $$P(Xin R)=alpha$$ if $H_0$ is true. So both mine or your answer can be technically right. But for the test to make sense in practice (or, more rigorously, to have good properties, like so-called unbiasedness), you have to build $R$ using those values which are more likely to occur if $H_1$ is true. And the case is that the smaller $lambda$ is, the greater the values of $X$ expected. This is the case for the exponential distribution, and that's why it is reasonable a critical region that contains big, not small, values.
$endgroup$
– Alejandro Nasif Salum
Jan 27 at 0:24
1
$begingroup$
I mean, let's assume that $H_0$ is false, because actually $lambda=0.0001$. The expected value of $X$ would be $10,000$, so you want to choose a critical region that includes the greatest values. If you chose a critical region such as $(0,0.25)$ or so, it is almost sure that you will fail to reject $H_0$ (because those values are most likely not going to occur, given that $E(X)=10,000$), even when it is really obvious that $lambda$ is much smaller than what is assumed in $H_0$. That is, $H_0$ is evidently false, but the test would say that you have to keep this absurd hypothesis.
$endgroup$
– Alejandro Nasif Salum
Jan 27 at 0:29
$begingroup$
why you assume the $E(X)$ and not the value of lambda? The critical region should not be the area where we reject $H_0$ in favour of $H_1$, so where $lambda<0.2$?
$endgroup$
– Mark Jacon
Jan 24 at 8:17
$begingroup$
why you assume the $E(X)$ and not the value of lambda? The critical region should not be the area where we reject $H_0$ in favour of $H_1$, so where $lambda<0.2$?
$endgroup$
– Mark Jacon
Jan 24 at 8:17
1
1
$begingroup$
By definition, the critical region is any region $R$ such that $$P(Xin R)=alpha$$ if $H_0$ is true. So both mine or your answer can be technically right. But for the test to make sense in practice (or, more rigorously, to have good properties, like so-called unbiasedness), you have to build $R$ using those values which are more likely to occur if $H_1$ is true. And the case is that the smaller $lambda$ is, the greater the values of $X$ expected. This is the case for the exponential distribution, and that's why it is reasonable a critical region that contains big, not small, values.
$endgroup$
– Alejandro Nasif Salum
Jan 27 at 0:24
$begingroup$
By definition, the critical region is any region $R$ such that $$P(Xin R)=alpha$$ if $H_0$ is true. So both mine or your answer can be technically right. But for the test to make sense in practice (or, more rigorously, to have good properties, like so-called unbiasedness), you have to build $R$ using those values which are more likely to occur if $H_1$ is true. And the case is that the smaller $lambda$ is, the greater the values of $X$ expected. This is the case for the exponential distribution, and that's why it is reasonable a critical region that contains big, not small, values.
$endgroup$
– Alejandro Nasif Salum
Jan 27 at 0:24
1
1
$begingroup$
I mean, let's assume that $H_0$ is false, because actually $lambda=0.0001$. The expected value of $X$ would be $10,000$, so you want to choose a critical region that includes the greatest values. If you chose a critical region such as $(0,0.25)$ or so, it is almost sure that you will fail to reject $H_0$ (because those values are most likely not going to occur, given that $E(X)=10,000$), even when it is really obvious that $lambda$ is much smaller than what is assumed in $H_0$. That is, $H_0$ is evidently false, but the test would say that you have to keep this absurd hypothesis.
$endgroup$
– Alejandro Nasif Salum
Jan 27 at 0:29
$begingroup$
I mean, let's assume that $H_0$ is false, because actually $lambda=0.0001$. The expected value of $X$ would be $10,000$, so you want to choose a critical region that includes the greatest values. If you chose a critical region such as $(0,0.25)$ or so, it is almost sure that you will fail to reject $H_0$ (because those values are most likely not going to occur, given that $E(X)=10,000$), even when it is really obvious that $lambda$ is much smaller than what is assumed in $H_0$. That is, $H_0$ is evidently false, but the test would say that you have to keep this absurd hypothesis.
$endgroup$
– Alejandro Nasif Salum
Jan 27 at 0:29
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1
$begingroup$
Confusing notation here. You say $Xsim Exp(X)$ with unknown $A$ then immediately start talking about $lambda$. Surely you mean $Xsim Exp(lambda)$ where $mathbb{E}(X)=1/lambda$ and $lambda$ is the unknown parameter you want to test on.
$endgroup$
– LoveTooNap29
Jan 23 at 20:55
1
$begingroup$
What do you mean by "unknown A"? What is A? And what is the sample size?
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– callculus
Jan 23 at 21:08
1
$begingroup$
@callculus I think that it's time that I learn how to write on latex, sorry guys, it should be a $lambda$.
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– Mark Jacon
Jan 23 at 21:11
1
$begingroup$
And the sample size is what, equal to 1?
$endgroup$
– callculus
Jan 23 at 21:13
2
$begingroup$
Attempt to clairfy: You seem to have only a single observation $X$ to test this hypothesis. Clearly, you want to reject when $X$ is small. In particular you want $P(X < c|lambda = .2) = 0.05.$ In R statistical software you can find $c = 0.2565$ using code
qexp(.05,.2), which returns 0.2564665, and confirm using codepexp(.2565, .2), which returns 0.05000637. In R,pexpis an exponential CDF andqexpis the quantile function (inverse CDF). For this simple problem you don't need software, but it is good to have the framework in mind.$endgroup$
– BruceET
Jan 23 at 22:03