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Calculate significance level and power for exponential distribution
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My textbook is unclear about how to do this with exponential distributions (it spends the whole time talking about normal ones), so I'm asking this here to see if I got it right.
The question mentions a statistical test such that:
$H_0 : lambda = 2$
$H_1 : lambda > 2$
$H_0$ is rejected if $X > 3$
(There's only one data point).
a) Calculate the significance level.
Assuming $H_o$ is true, we have that $frac1lambda = 2$ and $lambda = frac12$.
$$P(X > 3mid H_0) = 1 - P(X leq 3)$$
$$1 - e^-lambda (3) = 1 - e^-1.5 = 0.77686983985$$
b) Calculate the power level when $lambda = 2.5$
beginalign
& P(X > 0.7769midlambda = 2.5) \[10pt]
= & 1 - e^-lambda(0.7769) = 1 - e^-2.5(0.7769) = 0.8566
endalign
c) If the actual value of $lambda$ is $2.5$, is this procedure going to make a good decision?
Yes, because the power is high?
Why I think I'm missing something: I haven't used the info $H_1 : lambda > 2$ given in the question
EDIT: I don't know why I said $frac1lambda = 2$. I was thinking about something else. Here's the alternative answer:
Significance level
beginalign
& P(X > 3mid H_0) = 1 - P(X leq 3) \[10pt]
= & 1 - e^-3(2) = 0.9975
endalign
Power
$$P(X > 0.9975midlambda = 2.5)$$
$$1 - e^-2.5(0.9975) = 0.9174$$
probability statistics probability-distributions hypothesis-testing exponential-distribution
edited Jul 30 at 0:50
Michael Hardy
204k23185461
asked Jul 29 at 23:54
Peplm
103
 |Â
show 1 more comment
up vote
1
down vote
favorite
My textbook is unclear about how to do this with exponential distributions (it spends the whole time talking about normal ones), so I'm asking this here to see if I got it right.
The question mentions a statistical test such that:
$H_0 : lambda = 2$
$H_1 : lambda > 2$
$H_0$ is rejected if $X > 3$
(There's only one data point).
a) Calculate the significance level.
Assuming $H_o$ is true, we have that $frac1lambda = 2$ and $lambda = frac12$.
$$P(X > 3mid H_0) = 1 - P(X leq 3)$$
$$1 - e^-lambda (3) = 1 - e^-1.5 = 0.77686983985$$
b) Calculate the power level when $lambda = 2.5$
beginalign
& P(X > 0.7769midlambda = 2.5) \[10pt]
= & 1 - e^-lambda(0.7769) = 1 - e^-2.5(0.7769) = 0.8566
endalign
c) If the actual value of $lambda$ is $2.5$, is this procedure going to make a good decision?
Yes, because the power is high?
Why I think I'm missing something: I haven't used the info $H_1 : lambda > 2$ given in the question
EDIT: I don't know why I said $frac1lambda = 2$. I was thinking about something else. Here's the alternative answer:
Significance level
beginalign
& P(X > 3mid H_0) = 1 - P(X leq 3) \[10pt]
= & 1 - e^-3(2) = 0.9975
endalign
Power
$$P(X > 0.9975midlambda = 2.5)$$
$$1 - e^-2.5(0.9975) = 0.9174$$
probability statistics probability-distributions hypothesis-testing exponential-distribution
edited Jul 30 at 0:50
Michael Hardy
204k23185461
asked Jul 29 at 23:54
Peplm
103
1
Is $2$ the rate or the mean? You say "$H_0 : lambda = 2$" but later say "Assuming $H_o$ is true, we have that $frac1lambda = 2$ and $lambda = frac12$"
– Henry
Jul 30 at 0:05
@Henry , my mistake, I was thinking about something else. I added an alternative answer
– Peplm
Jul 30 at 0:17
You wrote $P(X > 3mid H_0) = 1 - P(X leq 3) = 1 - e^-3(2).$ That is wrong. If the density is $lambda e^-lambda x$ for $xge0,$ then $Pr(X>3mid H_0) = Pr(Xge 3) = e^-lambdacdot3. qquad$
– Michael Hardy
Jul 30 at 0:54
If $lambda$ is the rate, then it doesn't make sense to reject $lambda=2$ in favor of an alternative that $lambda>2$ when $X$ is large. If $lambda$ is the mean, then that would make sense. Hence I am suspicious of this whole thing.
– Michael Hardy
Jul 30 at 0:59
$ldots,$and I should add that if $lambda$ is the mean rather than the rate, then you have $Pr(Xge 3) = e^-3/lambda. qquad$
– Michael Hardy
Jul 30 at 1:07
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
My textbook is unclear about how to do this with exponential distributions (it spends the whole time talking about normal ones), so I'm asking this here to see if I got it right.
The question mentions a statistical test such that:
$H_0 : lambda = 2$
$H_1 : lambda > 2$
$H_0$ is rejected if $X > 3$
(There's only one data point).
a) Calculate the significance level.
Assuming $H_o$ is true, we have that $frac1lambda = 2$ and $lambda = frac12$.
$$P(X > 3mid H_0) = 1 - P(X leq 3)$$
$$1 - e^-lambda (3) = 1 - e^-1.5 = 0.77686983985$$
b) Calculate the power level when $lambda = 2.5$
beginalign
& P(X > 0.7769midlambda = 2.5) \[10pt]
= & 1 - e^-lambda(0.7769) = 1 - e^-2.5(0.7769) = 0.8566
endalign
c) If the actual value of $lambda$ is $2.5$, is this procedure going to make a good decision?
Yes, because the power is high?
Why I think I'm missing something: I haven't used the info $H_1 : lambda > 2$ given in the question
EDIT: I don't know why I said $frac1lambda = 2$. I was thinking about something else. Here's the alternative answer:
Significance level
beginalign
& P(X > 3mid H_0) = 1 - P(X leq 3) \[10pt]
= & 1 - e^-3(2) = 0.9975
endalign
Power
$$P(X > 0.9975midlambda = 2.5)$$
$$1 - e^-2.5(0.9975) = 0.9174$$
probability statistics probability-distributions hypothesis-testing exponential-distribution
edited Jul 30 at 0:50
Michael Hardy
204k23185461
asked Jul 29 at 23:54
Peplm
103
My textbook is unclear about how to do this with exponential distributions (it spends the whole time talking about normal ones), so I'm asking this here to see if I got it right.
The question mentions a statistical test such that:
$H_0 : lambda = 2$
$H_1 : lambda > 2$
$H_0$ is rejected if $X > 3$
(There's only one data point).
a) Calculate the significance level.
Assuming $H_o$ is true, we have that $frac1lambda = 2$ and $lambda = frac12$.
$$P(X > 3mid H_0) = 1 - P(X leq 3)$$
$$1 - e^-lambda (3) = 1 - e^-1.5 = 0.77686983985$$
b) Calculate the power level when $lambda = 2.5$
beginalign
& P(X > 0.7769midlambda = 2.5) \[10pt]
= & 1 - e^-lambda(0.7769) = 1 - e^-2.5(0.7769) = 0.8566
endalign
c) If the actual value of $lambda$ is $2.5$, is this procedure going to make a good decision?
Yes, because the power is high?
Why I think I'm missing something: I haven't used the info $H_1 : lambda > 2$ given in the question
EDIT: I don't know why I said $frac1lambda = 2$. I was thinking about something else. Here's the alternative answer:
Significance level
beginalign
& P(X > 3mid H_0) = 1 - P(X leq 3) \[10pt]
= & 1 - e^-3(2) = 0.9975
endalign
Power
$$P(X > 0.9975midlambda = 2.5)$$
$$1 - e^-2.5(0.9975) = 0.9174$$
probability statistics probability-distributions hypothesis-testing exponential-distribution
edited Jul 30 at 0:50
Michael Hardy
204k23185461
asked Jul 29 at 23:54
Peplm
103
edited Jul 30 at 0:50
Michael Hardy
204k23185461
edited Jul 30 at 0:50
Michael Hardy
204k23185461
edited Jul 30 at 0:50
Michael Hardy
204k23185461
204k23185461
asked Jul 29 at 23:54
Peplm
103
asked Jul 29 at 23:54
Peplm
103
asked Jul 29 at 23:54
Peplm
103
103
1
Is $2$ the rate or the mean? You say "$H_0 : lambda = 2$" but later say "Assuming $H_o$ is true, we have that $frac1lambda = 2$ and $lambda = frac12$"
– Henry
Jul 30 at 0:05
@Henry , my mistake, I was thinking about something else. I added an alternative answer
– Peplm
Jul 30 at 0:17
You wrote $P(X > 3mid H_0) = 1 - P(X leq 3) = 1 - e^-3(2).$ That is wrong. If the density is $lambda e^-lambda x$ for $xge0,$ then $Pr(X>3mid H_0) = Pr(Xge 3) = e^-lambdacdot3. qquad$
– Michael Hardy
Jul 30 at 0:54
If $lambda$ is the rate, then it doesn't make sense to reject $lambda=2$ in favor of an alternative that $lambda>2$ when $X$ is large. If $lambda$ is the mean, then that would make sense. Hence I am suspicious of this whole thing.
– Michael Hardy
Jul 30 at 0:59
$ldots,$and I should add that if $lambda$ is the mean rather than the rate, then you have $Pr(Xge 3) = e^-3/lambda. qquad$
– Michael Hardy
Jul 30 at 1:07
 |Â
show 1 more comment
1
Is $2$ the rate or the mean? You say "$H_0 : lambda = 2$" but later say "Assuming $H_o$ is true, we have that $frac1lambda = 2$ and $lambda = frac12$"
– Henry
Jul 30 at 0:05
@Henry , my mistake, I was thinking about something else. I added an alternative answer
– Peplm
Jul 30 at 0:17
You wrote $P(X > 3mid H_0) = 1 - P(X leq 3) = 1 - e^-3(2).$ That is wrong. If the density is $lambda e^-lambda x$ for $xge0,$ then $Pr(X>3mid H_0) = Pr(Xge 3) = e^-lambdacdot3. qquad$
– Michael Hardy
Jul 30 at 0:54
If $lambda$ is the rate, then it doesn't make sense to reject $lambda=2$ in favor of an alternative that $lambda>2$ when $X$ is large. If $lambda$ is the mean, then that would make sense. Hence I am suspicious of this whole thing.
– Michael Hardy
Jul 30 at 0:59
$ldots,$and I should add that if $lambda$ is the mean rather than the rate, then you have $Pr(Xge 3) = e^-3/lambda. qquad$
– Michael Hardy
Jul 30 at 1:07
1
1
Is $2$ the rate or the mean? You say "$H_0 : lambda = 2$" but later say "Assuming $H_o$ is true, we have that $frac1lambda = 2$ and $lambda = frac12$"
– Henry
Jul 30 at 0:05
Is $2$ the rate or the mean? You say "$H_0 : lambda = 2$" but later say "Assuming $H_o$ is true, we have that $frac1lambda = 2$ and $lambda = frac12$"
– Henry
Jul 30 at 0:05
@Henry , my mistake, I was thinking about something else. I added an alternative answer
– Peplm
Jul 30 at 0:17
@Henry , my mistake, I was thinking about something else. I added an alternative answer
– Peplm
Jul 30 at 0:17
You wrote $P(X > 3mid H_0) = 1 - P(X leq 3) = 1 - e^-3(2).$ That is wrong. If the density is $lambda e^-lambda x$ for $xge0,$ then $Pr(X>3mid H_0) = Pr(Xge 3) = e^-lambdacdot3. qquad$
– Michael Hardy
Jul 30 at 0:54
You wrote $P(X > 3mid H_0) = 1 - P(X leq 3) = 1 - e^-3(2).$ That is wrong. If the density is $lambda e^-lambda x$ for $xge0,$ then $Pr(X>3mid H_0) = Pr(Xge 3) = e^-lambdacdot3. qquad$
– Michael Hardy
Jul 30 at 0:54
If $lambda$ is the rate, then it doesn't make sense to reject $lambda=2$ in favor of an alternative that $lambda>2$ when $X$ is large. If $lambda$ is the mean, then that would make sense. Hence I am suspicious of this whole thing.
– Michael Hardy
Jul 30 at 0:59
If $lambda$ is the rate, then it doesn't make sense to reject $lambda=2$ in favor of an alternative that $lambda>2$ when $X$ is large. If $lambda$ is the mean, then that would make sense. Hence I am suspicious of this whole thing.
– Michael Hardy
Jul 30 at 0:59
$ldots,$and I should add that if $lambda$ is the mean rather than the rate, then you have $Pr(Xge 3) = e^-3/lambda. qquad$
– Michael Hardy
Jul 30 at 1:07
$ldots,$and I should add that if $lambda$ is the mean rather than the rate, then you have $Pr(Xge 3) = e^-3/lambda. qquad$
– Michael Hardy
Jul 30 at 1:07
 |Â
show 1 more comment
1 Answer
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up vote
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As stated the problem makes no sense, and not surprisingly, you have made
some mistakes in dealing with exponential probabilities. I will do my best
to write something helpful.
Suppose $H_0: mu = 2$ and $H_a: mu > 2.$ Then it makes sense to reject
if $X > 3.$
Because large values of $X$ would be associated with large parameter $mu$ and small values of rate $lambda = 1/mu,$ it would make no sense to
reject for large $X$ if the null and and alternative hypotheses were
$H_0: lambda = 2$ and $H_a: lambda > 2.$
Then the significance level is
$$P(textRej,|,mu = 2) = P(X > 3,|,mu = 2) = e^-lambda(3) =
e^-1.5 = 0.2231.$$
In R statistical software, the exponential functions use the rate $lambda = 1/mu$ as parameter and pexp is an exponential CDF. So the computation in R is:
1 - pexp(3, rate=1/2)
[1] 0.2231302
exp(-1.5)
[1] 0.2231302
The power against the alternative that $mu = 2.5$ is
$$P(textRej,|,mu = 2.5) = P(X > 3,|, mu= 2.5)
= e^-lambda(3) = e^-(1/2.5)(3) = e^-1.2 = 0.3012.$$
1 - pexp(3, rate=0.4)
[1] 0.3011942
exp(-1.2)
[1] 0.3011942
The figure below shows the null distribution with $mu = 2,,$ $(lambda = 0.5)$ (in blue)
and the particular alternative distribution with $mu = 2.5,,$ $(lambda = 0.4)$
(in red). The rejection region is the interval $(3, infty).$ Rejection
probabilities are areas under the curves to the right of the vertical dotted
line. The probability of rejection is a little higher under the alternative. This is a difficult testing situation because there is so little difference
between the null and alternative distributions.
Note: Tests based on only one observation are seldom of practical use.
Instead, suppose $n = 100,,$ $H_0: mu = 2,,$ $H_a: mu > 2.$ Then it makes sense
to reject $H_0$ when $bar X > 2.34,$ which gives significance level about
$alpha = 0.05.$
Using moment generating functions, one can show that
$bar X sim mathsfGamma(shape = 100, scale = nlambda),$ where $lambda = 1/mu.$ Then $E(bar X) = mu$ and $SD(bar X) = mu/sqrtn = mu/10.$
The significance level is 5% and the power against the alternative $mu = 2.5,,
lambda = 0.4$ is about 73%.
1 - pgamma(2.34, 100, 100*.5)
[1] 0.04997338
1 - pgamma(2.34, 100, 100*.4)
[1] 0.7325898
A plot of density curves of the distribution of $bar X$ for $mu = .5$ (null hypothesis) and $mu = .4$ (specific alternative) are much different, so that
a practical test is possible.
answered Jul 30 at 3:17
BruceET
33.1k61440
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
accepted
As stated the problem makes no sense, and not surprisingly, you have made
some mistakes in dealing with exponential probabilities. I will do my best
to write something helpful.
Suppose $H_0: mu = 2$ and $H_a: mu > 2.$ Then it makes sense to reject
if $X > 3.$
Because large values of $X$ would be associated with large parameter $mu$ and small values of rate $lambda = 1/mu,$ it would make no sense to
reject for large $X$ if the null and and alternative hypotheses were
$H_0: lambda = 2$ and $H_a: lambda > 2.$
Then the significance level is
$$P(textRej,|,mu = 2) = P(X > 3,|,mu = 2) = e^-lambda(3) =
e^-1.5 = 0.2231.$$
In R statistical software, the exponential functions use the rate $lambda = 1/mu$ as parameter and pexp is an exponential CDF. So the computation in R is:
1 - pexp(3, rate=1/2)
[1] 0.2231302
exp(-1.5)
[1] 0.2231302
The power against the alternative that $mu = 2.5$ is
$$P(textRej,|,mu = 2.5) = P(X > 3,|, mu= 2.5)
= e^-lambda(3) = e^-(1/2.5)(3) = e^-1.2 = 0.3012.$$
1 - pexp(3, rate=0.4)
[1] 0.3011942
exp(-1.2)
[1] 0.3011942
The figure below shows the null distribution with $mu = 2,,$ $(lambda = 0.5)$ (in blue)
and the particular alternative distribution with $mu = 2.5,,$ $(lambda = 0.4)$
(in red). The rejection region is the interval $(3, infty).$ Rejection
probabilities are areas under the curves to the right of the vertical dotted
line. The probability of rejection is a little higher under the alternative. This is a difficult testing situation because there is so little difference
between the null and alternative distributions.
Note: Tests based on only one observation are seldom of practical use.
Instead, suppose $n = 100,,$ $H_0: mu = 2,,$ $H_a: mu > 2.$ Then it makes sense
to reject $H_0$ when $bar X > 2.34,$ which gives significance level about
$alpha = 0.05.$
Using moment generating functions, one can show that
$bar X sim mathsfGamma(shape = 100, scale = nlambda),$ where $lambda = 1/mu.$ Then $E(bar X) = mu$ and $SD(bar X) = mu/sqrtn = mu/10.$
The significance level is 5% and the power against the alternative $mu = 2.5,,
lambda = 0.4$ is about 73%.
1 - pgamma(2.34, 100, 100*.5)
[1] 0.04997338
1 - pgamma(2.34, 100, 100*.4)
[1] 0.7325898
A plot of density curves of the distribution of $bar X$ for $mu = .5$ (null hypothesis) and $mu = .4$ (specific alternative) are much different, so that
a practical test is possible.
answered Jul 30 at 3:17
BruceET
33.1k61440
add a comment |Â
up vote
0
down vote
accepted
As stated the problem makes no sense, and not surprisingly, you have made
some mistakes in dealing with exponential probabilities. I will do my best
to write something helpful.
Suppose $H_0: mu = 2$ and $H_a: mu > 2.$ Then it makes sense to reject
if $X > 3.$
Because large values of $X$ would be associated with large parameter $mu$ and small values of rate $lambda = 1/mu,$ it would make no sense to
reject for large $X$ if the null and and alternative hypotheses were
$H_0: lambda = 2$ and $H_a: lambda > 2.$
Then the significance level is
$$P(textRej,|,mu = 2) = P(X > 3,|,mu = 2) = e^-lambda(3) =
e^-1.5 = 0.2231.$$
In R statistical software, the exponential functions use the rate $lambda = 1/mu$ as parameter and pexp is an exponential CDF. So the computation in R is:
1 - pexp(3, rate=1/2)
[1] 0.2231302
exp(-1.5)
[1] 0.2231302
The power against the alternative that $mu = 2.5$ is
$$P(textRej,|,mu = 2.5) = P(X > 3,|, mu= 2.5)
= e^-lambda(3) = e^-(1/2.5)(3) = e^-1.2 = 0.3012.$$
1 - pexp(3, rate=0.4)
[1] 0.3011942
exp(-1.2)
[1] 0.3011942
The figure below shows the null distribution with $mu = 2,,$ $(lambda = 0.5)$ (in blue)
and the particular alternative distribution with $mu = 2.5,,$ $(lambda = 0.4)$
(in red). The rejection region is the interval $(3, infty).$ Rejection
probabilities are areas under the curves to the right of the vertical dotted
line. The probability of rejection is a little higher under the alternative. This is a difficult testing situation because there is so little difference
between the null and alternative distributions.
Note: Tests based on only one observation are seldom of practical use.
Instead, suppose $n = 100,,$ $H_0: mu = 2,,$ $H_a: mu > 2.$ Then it makes sense
to reject $H_0$ when $bar X > 2.34,$ which gives significance level about
$alpha = 0.05.$
Using moment generating functions, one can show that
$bar X sim mathsfGamma(shape = 100, scale = nlambda),$ where $lambda = 1/mu.$ Then $E(bar X) = mu$ and $SD(bar X) = mu/sqrtn = mu/10.$
The significance level is 5% and the power against the alternative $mu = 2.5,,
lambda = 0.4$ is about 73%.
1 - pgamma(2.34, 100, 100*.5)
[1] 0.04997338
1 - pgamma(2.34, 100, 100*.4)
[1] 0.7325898
A plot of density curves of the distribution of $bar X$ for $mu = .5$ (null hypothesis) and $mu = .4$ (specific alternative) are much different, so that
a practical test is possible.
answered Jul 30 at 3:17
BruceET
33.1k61440
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
As stated the problem makes no sense, and not surprisingly, you have made
some mistakes in dealing with exponential probabilities. I will do my best
to write something helpful.
Suppose $H_0: mu = 2$ and $H_a: mu > 2.$ Then it makes sense to reject
if $X > 3.$
Because large values of $X$ would be associated with large parameter $mu$ and small values of rate $lambda = 1/mu,$ it would make no sense to
reject for large $X$ if the null and and alternative hypotheses were
$H_0: lambda = 2$ and $H_a: lambda > 2.$
Then the significance level is
$$P(textRej,|,mu = 2) = P(X > 3,|,mu = 2) = e^-lambda(3) =
e^-1.5 = 0.2231.$$
In R statistical software, the exponential functions use the rate $lambda = 1/mu$ as parameter and pexp is an exponential CDF. So the computation in R is:
1 - pexp(3, rate=1/2)
[1] 0.2231302
exp(-1.5)
[1] 0.2231302
The power against the alternative that $mu = 2.5$ is
$$P(textRej,|,mu = 2.5) = P(X > 3,|, mu= 2.5)
= e^-lambda(3) = e^-(1/2.5)(3) = e^-1.2 = 0.3012.$$
1 - pexp(3, rate=0.4)
[1] 0.3011942
exp(-1.2)
[1] 0.3011942
The figure below shows the null distribution with $mu = 2,,$ $(lambda = 0.5)$ (in blue)
and the particular alternative distribution with $mu = 2.5,,$ $(lambda = 0.4)$
(in red). The rejection region is the interval $(3, infty).$ Rejection
probabilities are areas under the curves to the right of the vertical dotted
line. The probability of rejection is a little higher under the alternative. This is a difficult testing situation because there is so little difference
between the null and alternative distributions.
Note: Tests based on only one observation are seldom of practical use.
Instead, suppose $n = 100,,$ $H_0: mu = 2,,$ $H_a: mu > 2.$ Then it makes sense
to reject $H_0$ when $bar X > 2.34,$ which gives significance level about
$alpha = 0.05.$
Using moment generating functions, one can show that
$bar X sim mathsfGamma(shape = 100, scale = nlambda),$ where $lambda = 1/mu.$ Then $E(bar X) = mu$ and $SD(bar X) = mu/sqrtn = mu/10.$
The significance level is 5% and the power against the alternative $mu = 2.5,,
lambda = 0.4$ is about 73%.
1 - pgamma(2.34, 100, 100*.5)
[1] 0.04997338
1 - pgamma(2.34, 100, 100*.4)
[1] 0.7325898
A plot of density curves of the distribution of $bar X$ for $mu = .5$ (null hypothesis) and $mu = .4$ (specific alternative) are much different, so that
a practical test is possible.
answered Jul 30 at 3:17
BruceET
33.1k61440
As stated the problem makes no sense, and not surprisingly, you have made
some mistakes in dealing with exponential probabilities. I will do my best
to write something helpful.
Suppose $H_0: mu = 2$ and $H_a: mu > 2.$ Then it makes sense to reject
if $X > 3.$
Because large values of $X$ would be associated with large parameter $mu$ and small values of rate $lambda = 1/mu,$ it would make no sense to
reject for large $X$ if the null and and alternative hypotheses were
$H_0: lambda = 2$ and $H_a: lambda > 2.$
Then the significance level is
$$P(textRej,|,mu = 2) = P(X > 3,|,mu = 2) = e^-lambda(3) =
e^-1.5 = 0.2231.$$
In R statistical software, the exponential functions use the rate $lambda = 1/mu$ as parameter and pexp is an exponential CDF. So the computation in R is:
1 - pexp(3, rate=1/2)
[1] 0.2231302
exp(-1.5)
[1] 0.2231302
The power against the alternative that $mu = 2.5$ is
$$P(textRej,|,mu = 2.5) = P(X > 3,|, mu= 2.5)
= e^-lambda(3) = e^-(1/2.5)(3) = e^-1.2 = 0.3012.$$
1 - pexp(3, rate=0.4)
[1] 0.3011942
exp(-1.2)
[1] 0.3011942
The figure below shows the null distribution with $mu = 2,,$ $(lambda = 0.5)$ (in blue)
and the particular alternative distribution with $mu = 2.5,,$ $(lambda = 0.4)$
(in red). The rejection region is the interval $(3, infty).$ Rejection
probabilities are areas under the curves to the right of the vertical dotted
line. The probability of rejection is a little higher under the alternative. This is a difficult testing situation because there is so little difference
between the null and alternative distributions.
Note: Tests based on only one observation are seldom of practical use.
Instead, suppose $n = 100,,$ $H_0: mu = 2,,$ $H_a: mu > 2.$ Then it makes sense
to reject $H_0$ when $bar X > 2.34,$ which gives significance level about
$alpha = 0.05.$
Using moment generating functions, one can show that
$bar X sim mathsfGamma(shape = 100, scale = nlambda),$ where $lambda = 1/mu.$ Then $E(bar X) = mu$ and $SD(bar X) = mu/sqrtn = mu/10.$
The significance level is 5% and the power against the alternative $mu = 2.5,,
lambda = 0.4$ is about 73%.
1 - pgamma(2.34, 100, 100*.5)
[1] 0.04997338
1 - pgamma(2.34, 100, 100*.4)
[1] 0.7325898
A plot of density curves of the distribution of $bar X$ for $mu = .5$ (null hypothesis) and $mu = .4$ (specific alternative) are much different, so that
a practical test is possible.
answered Jul 30 at 3:17
BruceET
33.1k61440
answered Jul 30 at 3:17
BruceET
33.1k61440
answered Jul 30 at 3:17
BruceET
33.1k61440
answered Jul 30 at 3:17
BruceET
33.1k61440
33.1k61440
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1
Is $2$ the rate or the mean? You say "$H_0 : lambda = 2$" but later say "Assuming $H_o$ is true, we have that $frac1lambda = 2$ and $lambda = frac12$"
– Henry
Jul 30 at 0:05
@Henry , my mistake, I was thinking about something else. I added an alternative answer
– Peplm
Jul 30 at 0:17
You wrote $P(X > 3mid H_0) = 1 - P(X leq 3) = 1 - e^-3(2).$ That is wrong. If the density is $lambda e^-lambda x$ for $xge0,$ then $Pr(X>3mid H_0) = Pr(Xge 3) = e^-lambdacdot3. qquad$
– Michael Hardy
Jul 30 at 0:54
If $lambda$ is the rate, then it doesn't make sense to reject $lambda=2$ in favor of an alternative that $lambda>2$ when $X$ is large. If $lambda$ is the mean, then that would make sense. Hence I am suspicious of this whole thing.
– Michael Hardy
Jul 30 at 0:59
$ldots,$and I should add that if $lambda$ is the mean rather than the rate, then you have $Pr(Xge 3) = e^-3/lambda. qquad$
– Michael Hardy
Jul 30 at 1:07