Power testing and a Poisson Distribution
Clash Royale CLAN TAG#URR8PPP
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Let $X_1, ..., X_n$ be a random sample from a Poisson distribution with parameter $lambda$.
If $X$ is Poisson with parameter $lambda$, then $textE[X] = textVar[X] = lambda$; hence $textE[X] = lambda$ and $textVar[X] = lambda/n$.
(a) Test $H_0 :lambda=4$ versus $H_A : lambda>4$ and that our test procedure is to reject $H_0$ if $x ge k$. Given $n = 100$, what value of $k$ gives a test with significance level approximately equal to $0.05$?
(b) What is the approximate power of the test at $lambda = 5$?
Hi all, I'm new to this so any help is appreciated. For (a), I tried setting up a $95%$ confidence interval because I thought that $95%$ confidence and $5%$ significance were about the same.
statistical-inference hypothesis-testing
edited Jun 30 '14 at 16:12
M. Vinay
6,71522035
asked Jun 30 '14 at 15:44
Alina P
164
 |Â
show 2 more comments
up vote
1
down vote
favorite
Let $X_1, ..., X_n$ be a random sample from a Poisson distribution with parameter $lambda$.
If $X$ is Poisson with parameter $lambda$, then $textE[X] = textVar[X] = lambda$; hence $textE[X] = lambda$ and $textVar[X] = lambda/n$.
(a) Test $H_0 :lambda=4$ versus $H_A : lambda>4$ and that our test procedure is to reject $H_0$ if $x ge k$. Given $n = 100$, what value of $k$ gives a test with significance level approximately equal to $0.05$?
(b) What is the approximate power of the test at $lambda = 5$?
Hi all, I'm new to this so any help is appreciated. For (a), I tried setting up a $95%$ confidence interval because I thought that $95%$ confidence and $5%$ significance were about the same.
statistical-inference hypothesis-testing
edited Jun 30 '14 at 16:12
M. Vinay
6,71522035
asked Jun 30 '14 at 15:44
Alina P
164
Would the downvoter care to explain?
– M. Vinay
Jun 30 '14 at 16:13
When you say x do you mean mean (x bar)
– Kamster
Jun 30 '14 at 16:26
no, not the sample mean. X is a random variable with the characteristics stated above. I am seeking helping with power tests.
– Alina P
Jun 30 '14 at 16:30
Oh I meant what do you mean by $xgeq k$ since you have 100 observations I don't know what x is
– Kamster
Jun 30 '14 at 16:33
From my class notes I'm getting that I want to reject the simple null hypothesis if x, the random variable, is greater than or equal to k. How do I set up a test?
– Alina P
Jun 30 '14 at 16:43
 |Â
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $X_1, ..., X_n$ be a random sample from a Poisson distribution with parameter $lambda$.
If $X$ is Poisson with parameter $lambda$, then $textE[X] = textVar[X] = lambda$; hence $textE[X] = lambda$ and $textVar[X] = lambda/n$.
(a) Test $H_0 :lambda=4$ versus $H_A : lambda>4$ and that our test procedure is to reject $H_0$ if $x ge k$. Given $n = 100$, what value of $k$ gives a test with significance level approximately equal to $0.05$?
(b) What is the approximate power of the test at $lambda = 5$?
Hi all, I'm new to this so any help is appreciated. For (a), I tried setting up a $95%$ confidence interval because I thought that $95%$ confidence and $5%$ significance were about the same.
statistical-inference hypothesis-testing
edited Jun 30 '14 at 16:12
M. Vinay
6,71522035
asked Jun 30 '14 at 15:44
Alina P
164
Let $X_1, ..., X_n$ be a random sample from a Poisson distribution with parameter $lambda$.
If $X$ is Poisson with parameter $lambda$, then $textE[X] = textVar[X] = lambda$; hence $textE[X] = lambda$ and $textVar[X] = lambda/n$.
(a) Test $H_0 :lambda=4$ versus $H_A : lambda>4$ and that our test procedure is to reject $H_0$ if $x ge k$. Given $n = 100$, what value of $k$ gives a test with significance level approximately equal to $0.05$?
(b) What is the approximate power of the test at $lambda = 5$?
Hi all, I'm new to this so any help is appreciated. For (a), I tried setting up a $95%$ confidence interval because I thought that $95%$ confidence and $5%$ significance were about the same.
statistical-inference hypothesis-testing
statistical-inference hypothesis-testing
edited Jun 30 '14 at 16:12
M. Vinay
6,71522035
asked Jun 30 '14 at 15:44
Alina P
164
edited Jun 30 '14 at 16:12
M. Vinay
6,71522035
asked Jun 30 '14 at 15:44
Alina P
164
edited Jun 30 '14 at 16:12
M. Vinay
6,71522035
edited Jun 30 '14 at 16:12
M. Vinay
6,71522035
edited Jun 30 '14 at 16:12
M. Vinay
6,71522035
6,71522035
asked Jun 30 '14 at 15:44
Alina P
164
asked Jun 30 '14 at 15:44
Alina P
164
asked Jun 30 '14 at 15:44
Alina P
164
164
Would the downvoter care to explain?
– M. Vinay
Jun 30 '14 at 16:13
When you say x do you mean mean (x bar)
– Kamster
Jun 30 '14 at 16:26
no, not the sample mean. X is a random variable with the characteristics stated above. I am seeking helping with power tests.
– Alina P
Jun 30 '14 at 16:30
Oh I meant what do you mean by $xgeq k$ since you have 100 observations I don't know what x is
– Kamster
Jun 30 '14 at 16:33
From my class notes I'm getting that I want to reject the simple null hypothesis if x, the random variable, is greater than or equal to k. How do I set up a test?
– Alina P
Jun 30 '14 at 16:43
 |Â
show 2 more comments
Would the downvoter care to explain?
– M. Vinay
Jun 30 '14 at 16:13
When you say x do you mean mean (x bar)
– Kamster
Jun 30 '14 at 16:26
no, not the sample mean. X is a random variable with the characteristics stated above. I am seeking helping with power tests.
– Alina P
Jun 30 '14 at 16:30
Oh I meant what do you mean by $xgeq k$ since you have 100 observations I don't know what x is
– Kamster
Jun 30 '14 at 16:33
From my class notes I'm getting that I want to reject the simple null hypothesis if x, the random variable, is greater than or equal to k. How do I set up a test?
– Alina P
Jun 30 '14 at 16:43
Would the downvoter care to explain?
– M. Vinay
Jun 30 '14 at 16:13
Would the downvoter care to explain?
– M. Vinay
Jun 30 '14 at 16:13
When you say x do you mean mean (x bar)
– Kamster
Jun 30 '14 at 16:26
When you say x do you mean mean (x bar)
– Kamster
Jun 30 '14 at 16:26
no, not the sample mean. X is a random variable with the characteristics stated above. I am seeking helping with power tests.
– Alina P
Jun 30 '14 at 16:30
no, not the sample mean. X is a random variable with the characteristics stated above. I am seeking helping with power tests.
– Alina P
Jun 30 '14 at 16:30
Oh I meant what do you mean by $xgeq k$ since you have 100 observations I don't know what x is
– Kamster
Jun 30 '14 at 16:33
Oh I meant what do you mean by $xgeq k$ since you have 100 observations I don't know what x is
– Kamster
Jun 30 '14 at 16:33
From my class notes I'm getting that I want to reject the simple null hypothesis if x, the random variable, is greater than or equal to k. How do I set up a test?
– Alina P
Jun 30 '14 at 16:43
From my class notes I'm getting that I want to reject the simple null hypothesis if x, the random variable, is greater than or equal to k. How do I set up a test?
– Alina P
Jun 30 '14 at 16:43
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
0
down vote
So first to find test with significance level of at approximately 0.05 we see that we want
$$P_lambdainTheta_Ho(barXgeq k)=P_lambda=4left(sum Xgeq 100k right) = 1-P_lambda=4left(sum X< 100kright)=0.05$$
Where we see that $sum Xsim operatornamePois(100cdot4)=operatornamePois(400)$ Thus in order to find when this is true it can be a little tedious but you are going to have to calculate a list of possible $100k=t$'s and calculate $1-P(sum X< t)$ (remember for $operatornamePois(400)$ and trying doing on excel) and see which is closest to $0.05$ (not above though). after you get $t^*$ that does this call it $t^*$ then $k=fract^*100$.
Now to calculate Power we just calculate
$$operatornamepower(lambda=5)=P_lambda=5left(sum Xgeq t^*right) = 1-P_lambda=5left(sum X< t^*right)$$
where now $sum Xsim operatornamePois(100cdot5)=operatornamePois(500)$
also note $sum X=sum_i=1^100 X_i$
Also to your question about confidence interval, it is true you could calculate a $95%$ confidence interval and that would correspond to $5%$ significance level but the problem with trying to do this is that I don't really know what CI interval is for mean in poisson and also even if you could derive it, CI intervals still rely on some critical value and since you can't really standardize or have table to look at for that it becomes situation above where you just have calculate probabilities for list of values and see which ones closets to significance level
edited May 17 at 17:41
Michael Hardy
205k23187466
answered Jun 30 '14 at 16:59
Kamster
1,687917
I've never used excel to do calculations before, so could you help me with what I need to input? I entered poisson distribution with mean 4, x=100 and cumulative 0. My answer was 3.154 and I don't think that's right.
– Alina P
Jun 30 '14 at 20:42
Ok so on excel when you have cumulative=TRUE that means it calculated Pr(X<=x) thus to get => you must subrtact that one from 1 then and Pr(X=x)
– Kamster
Jun 30 '14 at 23:53
You can write $100times4$ or $100cdot4.$ Using an asterisk for that purpose is a workaround for occasions when you are limited to the symbols on the keyboard.
– Michael Hardy
May 17 at 17:41
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
So first to find test with significance level of at approximately 0.05 we see that we want
$$P_lambdainTheta_Ho(barXgeq k)=P_lambda=4left(sum Xgeq 100k right) = 1-P_lambda=4left(sum X< 100kright)=0.05$$
Where we see that $sum Xsim operatornamePois(100cdot4)=operatornamePois(400)$ Thus in order to find when this is true it can be a little tedious but you are going to have to calculate a list of possible $100k=t$'s and calculate $1-P(sum X< t)$ (remember for $operatornamePois(400)$ and trying doing on excel) and see which is closest to $0.05$ (not above though). after you get $t^*$ that does this call it $t^*$ then $k=fract^*100$.
Now to calculate Power we just calculate
$$operatornamepower(lambda=5)=P_lambda=5left(sum Xgeq t^*right) = 1-P_lambda=5left(sum X< t^*right)$$
where now $sum Xsim operatornamePois(100cdot5)=operatornamePois(500)$
also note $sum X=sum_i=1^100 X_i$
Also to your question about confidence interval, it is true you could calculate a $95%$ confidence interval and that would correspond to $5%$ significance level but the problem with trying to do this is that I don't really know what CI interval is for mean in poisson and also even if you could derive it, CI intervals still rely on some critical value and since you can't really standardize or have table to look at for that it becomes situation above where you just have calculate probabilities for list of values and see which ones closets to significance level
edited May 17 at 17:41
Michael Hardy
205k23187466
answered Jun 30 '14 at 16:59
Kamster
1,687917
I've never used excel to do calculations before, so could you help me with what I need to input? I entered poisson distribution with mean 4, x=100 and cumulative 0. My answer was 3.154 and I don't think that's right.
– Alina P
Jun 30 '14 at 20:42
Ok so on excel when you have cumulative=TRUE that means it calculated Pr(X<=x) thus to get => you must subrtact that one from 1 then and Pr(X=x)
– Kamster
Jun 30 '14 at 23:53
You can write $100times4$ or $100cdot4.$ Using an asterisk for that purpose is a workaround for occasions when you are limited to the symbols on the keyboard.
– Michael Hardy
May 17 at 17:41
add a comment |Â
up vote
0
down vote
So first to find test with significance level of at approximately 0.05 we see that we want
$$P_lambdainTheta_Ho(barXgeq k)=P_lambda=4left(sum Xgeq 100k right) = 1-P_lambda=4left(sum X< 100kright)=0.05$$
Where we see that $sum Xsim operatornamePois(100cdot4)=operatornamePois(400)$ Thus in order to find when this is true it can be a little tedious but you are going to have to calculate a list of possible $100k=t$'s and calculate $1-P(sum X< t)$ (remember for $operatornamePois(400)$ and trying doing on excel) and see which is closest to $0.05$ (not above though). after you get $t^*$ that does this call it $t^*$ then $k=fract^*100$.
Now to calculate Power we just calculate
$$operatornamepower(lambda=5)=P_lambda=5left(sum Xgeq t^*right) = 1-P_lambda=5left(sum X< t^*right)$$
where now $sum Xsim operatornamePois(100cdot5)=operatornamePois(500)$
also note $sum X=sum_i=1^100 X_i$
Also to your question about confidence interval, it is true you could calculate a $95%$ confidence interval and that would correspond to $5%$ significance level but the problem with trying to do this is that I don't really know what CI interval is for mean in poisson and also even if you could derive it, CI intervals still rely on some critical value and since you can't really standardize or have table to look at for that it becomes situation above where you just have calculate probabilities for list of values and see which ones closets to significance level
edited May 17 at 17:41
Michael Hardy
205k23187466
answered Jun 30 '14 at 16:59
Kamster
1,687917
I've never used excel to do calculations before, so could you help me with what I need to input? I entered poisson distribution with mean 4, x=100 and cumulative 0. My answer was 3.154 and I don't think that's right.
– Alina P
Jun 30 '14 at 20:42
Ok so on excel when you have cumulative=TRUE that means it calculated Pr(X<=x) thus to get => you must subrtact that one from 1 then and Pr(X=x)
– Kamster
Jun 30 '14 at 23:53
You can write $100times4$ or $100cdot4.$ Using an asterisk for that purpose is a workaround for occasions when you are limited to the symbols on the keyboard.
– Michael Hardy
May 17 at 17:41
add a comment |Â
up vote
0
down vote
up vote
0
down vote
So first to find test with significance level of at approximately 0.05 we see that we want
$$P_lambdainTheta_Ho(barXgeq k)=P_lambda=4left(sum Xgeq 100k right) = 1-P_lambda=4left(sum X< 100kright)=0.05$$
Where we see that $sum Xsim operatornamePois(100cdot4)=operatornamePois(400)$ Thus in order to find when this is true it can be a little tedious but you are going to have to calculate a list of possible $100k=t$'s and calculate $1-P(sum X< t)$ (remember for $operatornamePois(400)$ and trying doing on excel) and see which is closest to $0.05$ (not above though). after you get $t^*$ that does this call it $t^*$ then $k=fract^*100$.
Now to calculate Power we just calculate
$$operatornamepower(lambda=5)=P_lambda=5left(sum Xgeq t^*right) = 1-P_lambda=5left(sum X< t^*right)$$
where now $sum Xsim operatornamePois(100cdot5)=operatornamePois(500)$
also note $sum X=sum_i=1^100 X_i$
Also to your question about confidence interval, it is true you could calculate a $95%$ confidence interval and that would correspond to $5%$ significance level but the problem with trying to do this is that I don't really know what CI interval is for mean in poisson and also even if you could derive it, CI intervals still rely on some critical value and since you can't really standardize or have table to look at for that it becomes situation above where you just have calculate probabilities for list of values and see which ones closets to significance level
edited May 17 at 17:41
Michael Hardy
205k23187466
answered Jun 30 '14 at 16:59
Kamster
1,687917
So first to find test with significance level of at approximately 0.05 we see that we want
$$P_lambdainTheta_Ho(barXgeq k)=P_lambda=4left(sum Xgeq 100k right) = 1-P_lambda=4left(sum X< 100kright)=0.05$$
Where we see that $sum Xsim operatornamePois(100cdot4)=operatornamePois(400)$ Thus in order to find when this is true it can be a little tedious but you are going to have to calculate a list of possible $100k=t$'s and calculate $1-P(sum X< t)$ (remember for $operatornamePois(400)$ and trying doing on excel) and see which is closest to $0.05$ (not above though). after you get $t^*$ that does this call it $t^*$ then $k=fract^*100$.
Now to calculate Power we just calculate
$$operatornamepower(lambda=5)=P_lambda=5left(sum Xgeq t^*right) = 1-P_lambda=5left(sum X< t^*right)$$
where now $sum Xsim operatornamePois(100cdot5)=operatornamePois(500)$
also note $sum X=sum_i=1^100 X_i$
Also to your question about confidence interval, it is true you could calculate a $95%$ confidence interval and that would correspond to $5%$ significance level but the problem with trying to do this is that I don't really know what CI interval is for mean in poisson and also even if you could derive it, CI intervals still rely on some critical value and since you can't really standardize or have table to look at for that it becomes situation above where you just have calculate probabilities for list of values and see which ones closets to significance level
edited May 17 at 17:41
Michael Hardy
205k23187466
answered Jun 30 '14 at 16:59
Kamster
1,687917
edited May 17 at 17:41
Michael Hardy
205k23187466
edited May 17 at 17:41
Michael Hardy
205k23187466
edited May 17 at 17:41
Michael Hardy
205k23187466
205k23187466
answered Jun 30 '14 at 16:59
Kamster
1,687917
answered Jun 30 '14 at 16:59
Kamster
1,687917
answered Jun 30 '14 at 16:59
Kamster
1,687917
1,687917
I've never used excel to do calculations before, so could you help me with what I need to input? I entered poisson distribution with mean 4, x=100 and cumulative 0. My answer was 3.154 and I don't think that's right.
– Alina P
Jun 30 '14 at 20:42
Ok so on excel when you have cumulative=TRUE that means it calculated Pr(X<=x) thus to get => you must subrtact that one from 1 then and Pr(X=x)
– Kamster
Jun 30 '14 at 23:53
You can write $100times4$ or $100cdot4.$ Using an asterisk for that purpose is a workaround for occasions when you are limited to the symbols on the keyboard.
– Michael Hardy
May 17 at 17:41
add a comment |Â
I've never used excel to do calculations before, so could you help me with what I need to input? I entered poisson distribution with mean 4, x=100 and cumulative 0. My answer was 3.154 and I don't think that's right.
– Alina P
Jun 30 '14 at 20:42
Ok so on excel when you have cumulative=TRUE that means it calculated Pr(X<=x) thus to get => you must subrtact that one from 1 then and Pr(X=x)
– Kamster
Jun 30 '14 at 23:53
You can write $100times4$ or $100cdot4.$ Using an asterisk for that purpose is a workaround for occasions when you are limited to the symbols on the keyboard.
– Michael Hardy
May 17 at 17:41
I've never used excel to do calculations before, so could you help me with what I need to input? I entered poisson distribution with mean 4, x=100 and cumulative 0. My answer was 3.154 and I don't think that's right.
– Alina P
Jun 30 '14 at 20:42
I've never used excel to do calculations before, so could you help me with what I need to input? I entered poisson distribution with mean 4, x=100 and cumulative 0. My answer was 3.154 and I don't think that's right.
– Alina P
Jun 30 '14 at 20:42
Ok so on excel when you have cumulative=TRUE that means it calculated Pr(X<=x) thus to get => you must subrtact that one from 1 then and Pr(X=x)
– Kamster
Jun 30 '14 at 23:53
Ok so on excel when you have cumulative=TRUE that means it calculated Pr(X<=x) thus to get => you must subrtact that one from 1 then and Pr(X=x)
– Kamster
Jun 30 '14 at 23:53
You can write $100times4$ or $100cdot4.$ Using an asterisk for that purpose is a workaround for occasions when you are limited to the symbols on the keyboard.
– Michael Hardy
May 17 at 17:41
You can write $100times4$ or $100cdot4.$ Using an asterisk for that purpose is a workaround for occasions when you are limited to the symbols on the keyboard.
– Michael Hardy
May 17 at 17:41
add a comment |Â
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Would the downvoter care to explain?
– M. Vinay
Jun 30 '14 at 16:13
When you say x do you mean mean (x bar)
– Kamster
Jun 30 '14 at 16:26
no, not the sample mean. X is a random variable with the characteristics stated above. I am seeking helping with power tests.
– Alina P
Jun 30 '14 at 16:30
Oh I meant what do you mean by $xgeq k$ since you have 100 observations I don't know what x is
– Kamster
Jun 30 '14 at 16:33
From my class notes I'm getting that I want to reject the simple null hypothesis if x, the random variable, is greater than or equal to k. How do I set up a test?
– Alina P
Jun 30 '14 at 16:43