Determine the probability of committing a Type II Error.
$begingroup$
Suppose X is uniformly distributed on the interval $[0;mu]$, with $mu$ unknown. The null hypothesis is that $mu = 2.5$ and the alternative hypothesis is that $mu geq 2.5$.
We test the hypothesis by sampling $X_1$ and $X_2$ from $X$ and taking the maximum of the two as our test statistic $T$.
We decide to reject $H_0$ in favor of $H_1$ when $T geq 2$.
Suppose that the real value of $mu$ is equal to $3$.
Determine the Probability of committing a Type II Error.
I have some problem to compute the Probability of committing a Type II Error in this exercise, how should I start it, do I have to convert to N(0,1) distribution?
statistics
asked Jan 9 at 12:12
FTACFTAC
2649
$endgroup$
add a comment |
$begingroup$
Suppose X is uniformly distributed on the interval $[0;mu]$, with $mu$ unknown. The null hypothesis is that $mu = 2.5$ and the alternative hypothesis is that $mu geq 2.5$.
We test the hypothesis by sampling $X_1$ and $X_2$ from $X$ and taking the maximum of the two as our test statistic $T$.
We decide to reject $H_0$ in favor of $H_1$ when $T geq 2$.
Suppose that the real value of $mu$ is equal to $3$.
Determine the Probability of committing a Type II Error.
I have some problem to compute the Probability of committing a Type II Error in this exercise, how should I start it, do I have to convert to N(0,1) distribution?
statistics
asked Jan 9 at 12:12
FTACFTAC
2649
$endgroup$
add a comment |
$begingroup$
Suppose X is uniformly distributed on the interval $[0;mu]$, with $mu$ unknown. The null hypothesis is that $mu = 2.5$ and the alternative hypothesis is that $mu geq 2.5$.
We test the hypothesis by sampling $X_1$ and $X_2$ from $X$ and taking the maximum of the two as our test statistic $T$.
We decide to reject $H_0$ in favor of $H_1$ when $T geq 2$.
Suppose that the real value of $mu$ is equal to $3$.
Determine the Probability of committing a Type II Error.
I have some problem to compute the Probability of committing a Type II Error in this exercise, how should I start it, do I have to convert to N(0,1) distribution?
statistics
asked Jan 9 at 12:12
FTACFTAC
2649
$endgroup$
Suppose X is uniformly distributed on the interval $[0;mu]$, with $mu$ unknown. The null hypothesis is that $mu = 2.5$ and the alternative hypothesis is that $mu geq 2.5$.
We test the hypothesis by sampling $X_1$ and $X_2$ from $X$ and taking the maximum of the two as our test statistic $T$.
We decide to reject $H_0$ in favor of $H_1$ when $T geq 2$.
Suppose that the real value of $mu$ is equal to $3$.
Determine the Probability of committing a Type II Error.
I have some problem to compute the Probability of committing a Type II Error in this exercise, how should I start it, do I have to convert to N(0,1) distribution?
statistics
statistics
asked Jan 9 at 12:12
FTACFTAC
2649
asked Jan 9 at 12:12
FTACFTAC
2649
asked Jan 9 at 12:12
FTACFTAC
2649
asked Jan 9 at 12:12
FTACFTAC
2649
asked Jan 9 at 12:12
FTACFTAC
2649
2649
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Hint:
Let $beta$ denote the probability of a type II error under the assumption that $mu = 3$. This means $T < 2$ although values up to $mu = 3$ can be assumed. Then
$beta = P_{mu = 3}(T <2) = P_{mu = 3}([0,2]times[0,2])$
where
$P_{mu = 3}$ is the uniform distribution on the square $[0,3]times[0,3]$.
Some more info:
Note that you are dealing with squares, as the sample consists of two (independent) random variables. So, you need to consider the squares $2^2$ and $3^2$. Then, you get the correct results.
Maybe you may draw the region $T<2$ on the square with side length $3$ to get a visual grip of what you are calculating.
answered Jan 9 at 12:42
trancelocationtrancelocation
14.2k1829
$endgroup$
$begingroup$
Thanks for the Hint, so I have to compute P(T<2|$mu$=3) right? My problem is that I don't know how to include the real value of $mu$ that is equal to 3 and the $mu=2.5$. I think that the result should be $frac{2-0}{3-0}=0.66$ but the correct answer is 0.44, where I'm wrong? maybe I should consider also the value 2.5
$endgroup$
– FTAC
Jan 9 at 13:15
1
$begingroup$
You do not need the value $2.5$ here as we are dealing with a type II error: accepting $H_0$ although $H_1: mu = 3 >2.5$ is true. Since for each $mu > 2.5$ the probability of the type II error changes, the specific $mu = 3$ is indicated.
$endgroup$
– trancelocation
Jan 9 at 13:20
1
$begingroup$
@FabioTaccaliti : I added some more info for clarification. Hope this helps.
$endgroup$
– trancelocation
Jan 9 at 13:25
$begingroup$
Thank you a lot! Now is clear!
$endgroup$
– FTAC
Jan 9 at 13:48
add a comment |
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$begingroup$
Hint:
Let $beta$ denote the probability of a type II error under the assumption that $mu = 3$. This means $T < 2$ although values up to $mu = 3$ can be assumed. Then
$beta = P_{mu = 3}(T <2) = P_{mu = 3}([0,2]times[0,2])$
where
$P_{mu = 3}$ is the uniform distribution on the square $[0,3]times[0,3]$.
Some more info:
Note that you are dealing with squares, as the sample consists of two (independent) random variables. So, you need to consider the squares $2^2$ and $3^2$. Then, you get the correct results.
Maybe you may draw the region $T<2$ on the square with side length $3$ to get a visual grip of what you are calculating.
answered Jan 9 at 12:42
trancelocationtrancelocation
14.2k1829
$endgroup$
$begingroup$
Thanks for the Hint, so I have to compute P(T<2|$mu$=3) right? My problem is that I don't know how to include the real value of $mu$ that is equal to 3 and the $mu=2.5$. I think that the result should be $frac{2-0}{3-0}=0.66$ but the correct answer is 0.44, where I'm wrong? maybe I should consider also the value 2.5
$endgroup$
– FTAC
Jan 9 at 13:15
1
$begingroup$
You do not need the value $2.5$ here as we are dealing with a type II error: accepting $H_0$ although $H_1: mu = 3 >2.5$ is true. Since for each $mu > 2.5$ the probability of the type II error changes, the specific $mu = 3$ is indicated.
$endgroup$
– trancelocation
Jan 9 at 13:20
1
$begingroup$
@FabioTaccaliti : I added some more info for clarification. Hope this helps.
$endgroup$
– trancelocation
Jan 9 at 13:25
$begingroup$
Thank you a lot! Now is clear!
$endgroup$
– FTAC
Jan 9 at 13:48
add a comment |
$begingroup$
Hint:
Let $beta$ denote the probability of a type II error under the assumption that $mu = 3$. This means $T < 2$ although values up to $mu = 3$ can be assumed. Then
$beta = P_{mu = 3}(T <2) = P_{mu = 3}([0,2]times[0,2])$
where
$P_{mu = 3}$ is the uniform distribution on the square $[0,3]times[0,3]$.
Some more info:
Note that you are dealing with squares, as the sample consists of two (independent) random variables. So, you need to consider the squares $2^2$ and $3^2$. Then, you get the correct results.
Maybe you may draw the region $T<2$ on the square with side length $3$ to get a visual grip of what you are calculating.
answered Jan 9 at 12:42
trancelocationtrancelocation
14.2k1829
$endgroup$
$begingroup$
Thanks for the Hint, so I have to compute P(T<2|$mu$=3) right? My problem is that I don't know how to include the real value of $mu$ that is equal to 3 and the $mu=2.5$. I think that the result should be $frac{2-0}{3-0}=0.66$ but the correct answer is 0.44, where I'm wrong? maybe I should consider also the value 2.5
$endgroup$
– FTAC
Jan 9 at 13:15
1
$begingroup$
You do not need the value $2.5$ here as we are dealing with a type II error: accepting $H_0$ although $H_1: mu = 3 >2.5$ is true. Since for each $mu > 2.5$ the probability of the type II error changes, the specific $mu = 3$ is indicated.
$endgroup$
– trancelocation
Jan 9 at 13:20
1
$begingroup$
@FabioTaccaliti : I added some more info for clarification. Hope this helps.
$endgroup$
– trancelocation
Jan 9 at 13:25
$begingroup$
Thank you a lot! Now is clear!
$endgroup$
– FTAC
Jan 9 at 13:48
add a comment |
$begingroup$
Hint:
Let $beta$ denote the probability of a type II error under the assumption that $mu = 3$. This means $T < 2$ although values up to $mu = 3$ can be assumed. Then
$beta = P_{mu = 3}(T <2) = P_{mu = 3}([0,2]times[0,2])$
where
$P_{mu = 3}$ is the uniform distribution on the square $[0,3]times[0,3]$.
Some more info:
Note that you are dealing with squares, as the sample consists of two (independent) random variables. So, you need to consider the squares $2^2$ and $3^2$. Then, you get the correct results.
Maybe you may draw the region $T<2$ on the square with side length $3$ to get a visual grip of what you are calculating.
answered Jan 9 at 12:42
trancelocationtrancelocation
14.2k1829
$endgroup$
Hint:
Let $beta$ denote the probability of a type II error under the assumption that $mu = 3$. This means $T < 2$ although values up to $mu = 3$ can be assumed. Then
$beta = P_{mu = 3}(T <2) = P_{mu = 3}([0,2]times[0,2])$
where
$P_{mu = 3}$ is the uniform distribution on the square $[0,3]times[0,3]$.
Some more info:
Note that you are dealing with squares, as the sample consists of two (independent) random variables. So, you need to consider the squares $2^2$ and $3^2$. Then, you get the correct results.
Maybe you may draw the region $T<2$ on the square with side length $3$ to get a visual grip of what you are calculating.
answered Jan 9 at 12:42
trancelocationtrancelocation
14.2k1829
edited Jan 9 at 13:24
answered Jan 9 at 12:42
trancelocationtrancelocation
14.2k1829
answered Jan 9 at 12:42
trancelocationtrancelocation
14.2k1829
answered Jan 9 at 12:42
trancelocationtrancelocation
14.2k1829
14.2k1829
$begingroup$
Thanks for the Hint, so I have to compute P(T<2|$mu$=3) right? My problem is that I don't know how to include the real value of $mu$ that is equal to 3 and the $mu=2.5$. I think that the result should be $frac{2-0}{3-0}=0.66$ but the correct answer is 0.44, where I'm wrong? maybe I should consider also the value 2.5
$endgroup$
– FTAC
Jan 9 at 13:15
1
$begingroup$
You do not need the value $2.5$ here as we are dealing with a type II error: accepting $H_0$ although $H_1: mu = 3 >2.5$ is true. Since for each $mu > 2.5$ the probability of the type II error changes, the specific $mu = 3$ is indicated.
$endgroup$
– trancelocation
Jan 9 at 13:20
1
$begingroup$
@FabioTaccaliti : I added some more info for clarification. Hope this helps.
$endgroup$
– trancelocation
Jan 9 at 13:25
$begingroup$
Thank you a lot! Now is clear!
$endgroup$
– FTAC
Jan 9 at 13:48
add a comment |
$begingroup$
Thanks for the Hint, so I have to compute P(T<2|$mu$=3) right? My problem is that I don't know how to include the real value of $mu$ that is equal to 3 and the $mu=2.5$. I think that the result should be $frac{2-0}{3-0}=0.66$ but the correct answer is 0.44, where I'm wrong? maybe I should consider also the value 2.5
$endgroup$
– FTAC
Jan 9 at 13:15
1
$begingroup$
You do not need the value $2.5$ here as we are dealing with a type II error: accepting $H_0$ although $H_1: mu = 3 >2.5$ is true. Since for each $mu > 2.5$ the probability of the type II error changes, the specific $mu = 3$ is indicated.
$endgroup$
– trancelocation
Jan 9 at 13:20
1
$begingroup$
@FabioTaccaliti : I added some more info for clarification. Hope this helps.
$endgroup$
– trancelocation
Jan 9 at 13:25
$begingroup$
Thank you a lot! Now is clear!
$endgroup$
– FTAC
Jan 9 at 13:48
$begingroup$
Thanks for the Hint, so I have to compute P(T<2|$mu$=3) right? My problem is that I don't know how to include the real value of $mu$ that is equal to 3 and the $mu=2.5$. I think that the result should be $frac{2-0}{3-0}=0.66$ but the correct answer is 0.44, where I'm wrong? maybe I should consider also the value 2.5
$endgroup$
– FTAC
Jan 9 at 13:15
$begingroup$
Thanks for the Hint, so I have to compute P(T<2|$mu$=3) right? My problem is that I don't know how to include the real value of $mu$ that is equal to 3 and the $mu=2.5$. I think that the result should be $frac{2-0}{3-0}=0.66$ but the correct answer is 0.44, where I'm wrong? maybe I should consider also the value 2.5
$endgroup$
– FTAC
Jan 9 at 13:15
1
1
$begingroup$
You do not need the value $2.5$ here as we are dealing with a type II error: accepting $H_0$ although $H_1: mu = 3 >2.5$ is true. Since for each $mu > 2.5$ the probability of the type II error changes, the specific $mu = 3$ is indicated.
$endgroup$
– trancelocation
Jan 9 at 13:20
$begingroup$
You do not need the value $2.5$ here as we are dealing with a type II error: accepting $H_0$ although $H_1: mu = 3 >2.5$ is true. Since for each $mu > 2.5$ the probability of the type II error changes, the specific $mu = 3$ is indicated.
$endgroup$
– trancelocation
Jan 9 at 13:20
1
1
$begingroup$
@FabioTaccaliti : I added some more info for clarification. Hope this helps.
$endgroup$
– trancelocation
Jan 9 at 13:25
$begingroup$
@FabioTaccaliti : I added some more info for clarification. Hope this helps.
$endgroup$
– trancelocation
Jan 9 at 13:25
$begingroup$
Thank you a lot! Now is clear!
$endgroup$
– FTAC
Jan 9 at 13:48
$begingroup$
Thank you a lot! Now is clear!
$endgroup$
– FTAC
Jan 9 at 13:48
add a comment |
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