Easy hypothesis testing in discrete case (uniform distribution)
$begingroup$
We have an estimator $hat{X}$ of $N$ which takes values in ${1,2,cdots,N}$ with the following mass function:
$$P_N(hat{X}=k) = left(frac{k}{N}right)^n-left(frac{k-1}{N}right)^{n}.$$
We are given the hypothesis $$H_0:N=20, H_1:N=22$$
and we know that $Nin{20,22}.$
I am not sure how to find the rejection region and the boundary of the region knowing that the significance level of the test is $alpha = 0.05.$
I think that we reject when $N=22.$ But I am not sure how to account for $alpha$ in this computation. Any hints will be much appreciated.
statistics
asked Jan 13 at 20:51
model_checkermodel_checker
4,14621931
$endgroup$
add a comment |
$begingroup$
We have an estimator $hat{X}$ of $N$ which takes values in ${1,2,cdots,N}$ with the following mass function:
$$P_N(hat{X}=k) = left(frac{k}{N}right)^n-left(frac{k-1}{N}right)^{n}.$$
We are given the hypothesis $$H_0:N=20, H_1:N=22$$
and we know that $Nin{20,22}.$
I am not sure how to find the rejection region and the boundary of the region knowing that the significance level of the test is $alpha = 0.05.$
I think that we reject when $N=22.$ But I am not sure how to account for $alpha$ in this computation. Any hints will be much appreciated.
statistics
asked Jan 13 at 20:51
model_checkermodel_checker
4,14621931
$endgroup$
add a comment |
$begingroup$
We have an estimator $hat{X}$ of $N$ which takes values in ${1,2,cdots,N}$ with the following mass function:
$$P_N(hat{X}=k) = left(frac{k}{N}right)^n-left(frac{k-1}{N}right)^{n}.$$
We are given the hypothesis $$H_0:N=20, H_1:N=22$$
and we know that $Nin{20,22}.$
I am not sure how to find the rejection region and the boundary of the region knowing that the significance level of the test is $alpha = 0.05.$
I think that we reject when $N=22.$ But I am not sure how to account for $alpha$ in this computation. Any hints will be much appreciated.
statistics
asked Jan 13 at 20:51
model_checkermodel_checker
4,14621931
$endgroup$
We have an estimator $hat{X}$ of $N$ which takes values in ${1,2,cdots,N}$ with the following mass function:
$$P_N(hat{X}=k) = left(frac{k}{N}right)^n-left(frac{k-1}{N}right)^{n}.$$
We are given the hypothesis $$H_0:N=20, H_1:N=22$$
and we know that $Nin{20,22}.$
I am not sure how to find the rejection region and the boundary of the region knowing that the significance level of the test is $alpha = 0.05.$
I think that we reject when $N=22.$ But I am not sure how to account for $alpha$ in this computation. Any hints will be much appreciated.
statistics
statistics
asked Jan 13 at 20:51
model_checkermodel_checker
4,14621931
asked Jan 13 at 20:51
model_checkermodel_checker
4,14621931
edited Jan 13 at 22:49
model_checker
asked Jan 13 at 20:51
model_checkermodel_checker
4,14621931
asked Jan 13 at 20:51
model_checkermodel_checker
4,14621931
asked Jan 13 at 20:51
model_checkermodel_checker
4,14621931
4,14621931
add a comment |
add a comment |
1 Answer
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$begingroup$
I suspect you mean to say you would reject $H_0$ when $hat{X}=22$. You would also reject $H_0$ when $hat{X}=21$ since that too is inconsistent with $N=20$.
You take $alpha$ into account in deciding whether to reject $H_0$ when $hat{X}=20$ or particular smaller values. For that you need some calculations
- If $H_0$ is true and $N=20$, then $P_{20}(hat{X} ge k) = left(frac{20}{20}right)^n-left(frac{k-1}{20}right)^{n}$.
- You want this to be as big as possible but less than or equal to $alpha=0.05$.
- Solving $left(frac{20}{20}right)^n-left(frac{k-1}{20}right)^{n} le 0.05$ requires $left(frac{k-1}{20}right)^{n} ge 0.95$ and so $kge 1+20sqrt[n]{0.95}$
- This then gives a rejection region when $hat{X} ge 1+20sqrt[n]{0.95}$
Note that the answer depends on $n$, which you have not specified.
In practice, since $hat{X}$ is an integer and by the two hypotheses cannot exceed $22$, the rejection region for $alpha=0.05$ would be when $hat{X} in {20,21,22}$ when $n=1$ and $hat{X} in {21,22}$ when $n gt 1$
answered Jan 14 at 8:33
HenryHenry
101k482168
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1 Answer
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$begingroup$
I suspect you mean to say you would reject $H_0$ when $hat{X}=22$. You would also reject $H_0$ when $hat{X}=21$ since that too is inconsistent with $N=20$.
You take $alpha$ into account in deciding whether to reject $H_0$ when $hat{X}=20$ or particular smaller values. For that you need some calculations
- If $H_0$ is true and $N=20$, then $P_{20}(hat{X} ge k) = left(frac{20}{20}right)^n-left(frac{k-1}{20}right)^{n}$.
- You want this to be as big as possible but less than or equal to $alpha=0.05$.
- Solving $left(frac{20}{20}right)^n-left(frac{k-1}{20}right)^{n} le 0.05$ requires $left(frac{k-1}{20}right)^{n} ge 0.95$ and so $kge 1+20sqrt[n]{0.95}$
- This then gives a rejection region when $hat{X} ge 1+20sqrt[n]{0.95}$
Note that the answer depends on $n$, which you have not specified.
In practice, since $hat{X}$ is an integer and by the two hypotheses cannot exceed $22$, the rejection region for $alpha=0.05$ would be when $hat{X} in {20,21,22}$ when $n=1$ and $hat{X} in {21,22}$ when $n gt 1$
answered Jan 14 at 8:33
HenryHenry
101k482168
$endgroup$
add a comment |
$begingroup$
I suspect you mean to say you would reject $H_0$ when $hat{X}=22$. You would also reject $H_0$ when $hat{X}=21$ since that too is inconsistent with $N=20$.
You take $alpha$ into account in deciding whether to reject $H_0$ when $hat{X}=20$ or particular smaller values. For that you need some calculations
- If $H_0$ is true and $N=20$, then $P_{20}(hat{X} ge k) = left(frac{20}{20}right)^n-left(frac{k-1}{20}right)^{n}$.
- You want this to be as big as possible but less than or equal to $alpha=0.05$.
- Solving $left(frac{20}{20}right)^n-left(frac{k-1}{20}right)^{n} le 0.05$ requires $left(frac{k-1}{20}right)^{n} ge 0.95$ and so $kge 1+20sqrt[n]{0.95}$
- This then gives a rejection region when $hat{X} ge 1+20sqrt[n]{0.95}$
Note that the answer depends on $n$, which you have not specified.
In practice, since $hat{X}$ is an integer and by the two hypotheses cannot exceed $22$, the rejection region for $alpha=0.05$ would be when $hat{X} in {20,21,22}$ when $n=1$ and $hat{X} in {21,22}$ when $n gt 1$
answered Jan 14 at 8:33
HenryHenry
101k482168
$endgroup$
add a comment |
$begingroup$
I suspect you mean to say you would reject $H_0$ when $hat{X}=22$. You would also reject $H_0$ when $hat{X}=21$ since that too is inconsistent with $N=20$.
You take $alpha$ into account in deciding whether to reject $H_0$ when $hat{X}=20$ or particular smaller values. For that you need some calculations
- If $H_0$ is true and $N=20$, then $P_{20}(hat{X} ge k) = left(frac{20}{20}right)^n-left(frac{k-1}{20}right)^{n}$.
- You want this to be as big as possible but less than or equal to $alpha=0.05$.
- Solving $left(frac{20}{20}right)^n-left(frac{k-1}{20}right)^{n} le 0.05$ requires $left(frac{k-1}{20}right)^{n} ge 0.95$ and so $kge 1+20sqrt[n]{0.95}$
- This then gives a rejection region when $hat{X} ge 1+20sqrt[n]{0.95}$
Note that the answer depends on $n$, which you have not specified.
In practice, since $hat{X}$ is an integer and by the two hypotheses cannot exceed $22$, the rejection region for $alpha=0.05$ would be when $hat{X} in {20,21,22}$ when $n=1$ and $hat{X} in {21,22}$ when $n gt 1$
answered Jan 14 at 8:33
HenryHenry
101k482168
$endgroup$
I suspect you mean to say you would reject $H_0$ when $hat{X}=22$. You would also reject $H_0$ when $hat{X}=21$ since that too is inconsistent with $N=20$.
You take $alpha$ into account in deciding whether to reject $H_0$ when $hat{X}=20$ or particular smaller values. For that you need some calculations
- If $H_0$ is true and $N=20$, then $P_{20}(hat{X} ge k) = left(frac{20}{20}right)^n-left(frac{k-1}{20}right)^{n}$.
- You want this to be as big as possible but less than or equal to $alpha=0.05$.
- Solving $left(frac{20}{20}right)^n-left(frac{k-1}{20}right)^{n} le 0.05$ requires $left(frac{k-1}{20}right)^{n} ge 0.95$ and so $kge 1+20sqrt[n]{0.95}$
- This then gives a rejection region when $hat{X} ge 1+20sqrt[n]{0.95}$
Note that the answer depends on $n$, which you have not specified.
In practice, since $hat{X}$ is an integer and by the two hypotheses cannot exceed $22$, the rejection region for $alpha=0.05$ would be when $hat{X} in {20,21,22}$ when $n=1$ and $hat{X} in {21,22}$ when $n gt 1$
answered Jan 14 at 8:33
HenryHenry
101k482168
answered Jan 14 at 8:33
HenryHenry
101k482168
answered Jan 14 at 8:33
HenryHenry
101k482168
answered Jan 14 at 8:33
HenryHenry
101k482168
101k482168
add a comment |
add a comment |
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