Determining the most powerful test for a simple hypothesis
Let $X_1,...,X_n$ be $N(mu,sigma^2)$ where $mu$ is unknown and $sigma$ is known. Let the null hypothesis $H_0$ state that $mu=mu_0$ and the alternate be that $mu=mu_1$ where $mu_1>|mu_0$.
Using the Neyman-Pearson Lemma, I plan to reject the null hypothesis for a sufficiently large value of $frac{L(x|mu_1)}{L(x|mu_0}$, the likelihood function under the alternate divided by the likelihood function under the null. The question specifies that my response should be in “simplest implementable form” which I will get to in a second.
After having simplified my ratio of likelihood functions, I ultimately determined that the ratio is positively related to $sum_{I=1}^nX_i$, so we will reject the null if there is a sufficiently large value for $sum_{I=1}^nX_i$.
For my answer, I use the fact that $sum_{I=1}^nX_i$ is normally distributed with mean $nmu$ and variance $nsigma^2$, so the most powerful test tin simplest implementable form would be as follows:
Reject the null hypothesis if $z_{alpha}<phi(frac{sum_{I=1}^nX_i-nmu_0}{sqrt{n}sigma})$, where $z_{alpha}$ is the upper $100alpha$ percentile of the standard normal distribution.
I’m not sure at all if this is correct, so I would appreciate some feedback.
statistics statistical-inference hypothesis-testing
asked Nov 26 at 23:06
DavidS
337111
add a comment |
Let $X_1,...,X_n$ be $N(mu,sigma^2)$ where $mu$ is unknown and $sigma$ is known. Let the null hypothesis $H_0$ state that $mu=mu_0$ and the alternate be that $mu=mu_1$ where $mu_1>|mu_0$.
Using the Neyman-Pearson Lemma, I plan to reject the null hypothesis for a sufficiently large value of $frac{L(x|mu_1)}{L(x|mu_0}$, the likelihood function under the alternate divided by the likelihood function under the null. The question specifies that my response should be in “simplest implementable form” which I will get to in a second.
After having simplified my ratio of likelihood functions, I ultimately determined that the ratio is positively related to $sum_{I=1}^nX_i$, so we will reject the null if there is a sufficiently large value for $sum_{I=1}^nX_i$.
For my answer, I use the fact that $sum_{I=1}^nX_i$ is normally distributed with mean $nmu$ and variance $nsigma^2$, so the most powerful test tin simplest implementable form would be as follows:
Reject the null hypothesis if $z_{alpha}<phi(frac{sum_{I=1}^nX_i-nmu_0}{sqrt{n}sigma})$, where $z_{alpha}$ is the upper $100alpha$ percentile of the standard normal distribution.
I’m not sure at all if this is correct, so I would appreciate some feedback.
statistics statistical-inference hypothesis-testing
asked Nov 26 at 23:06
DavidS
337111
add a comment |
Let $X_1,...,X_n$ be $N(mu,sigma^2)$ where $mu$ is unknown and $sigma$ is known. Let the null hypothesis $H_0$ state that $mu=mu_0$ and the alternate be that $mu=mu_1$ where $mu_1>|mu_0$.
Using the Neyman-Pearson Lemma, I plan to reject the null hypothesis for a sufficiently large value of $frac{L(x|mu_1)}{L(x|mu_0}$, the likelihood function under the alternate divided by the likelihood function under the null. The question specifies that my response should be in “simplest implementable form” which I will get to in a second.
After having simplified my ratio of likelihood functions, I ultimately determined that the ratio is positively related to $sum_{I=1}^nX_i$, so we will reject the null if there is a sufficiently large value for $sum_{I=1}^nX_i$.
For my answer, I use the fact that $sum_{I=1}^nX_i$ is normally distributed with mean $nmu$ and variance $nsigma^2$, so the most powerful test tin simplest implementable form would be as follows:
Reject the null hypothesis if $z_{alpha}<phi(frac{sum_{I=1}^nX_i-nmu_0}{sqrt{n}sigma})$, where $z_{alpha}$ is the upper $100alpha$ percentile of the standard normal distribution.
I’m not sure at all if this is correct, so I would appreciate some feedback.
statistics statistical-inference hypothesis-testing
asked Nov 26 at 23:06
DavidS
337111
Let $X_1,...,X_n$ be $N(mu,sigma^2)$ where $mu$ is unknown and $sigma$ is known. Let the null hypothesis $H_0$ state that $mu=mu_0$ and the alternate be that $mu=mu_1$ where $mu_1>|mu_0$.
Using the Neyman-Pearson Lemma, I plan to reject the null hypothesis for a sufficiently large value of $frac{L(x|mu_1)}{L(x|mu_0}$, the likelihood function under the alternate divided by the likelihood function under the null. The question specifies that my response should be in “simplest implementable form” which I will get to in a second.
After having simplified my ratio of likelihood functions, I ultimately determined that the ratio is positively related to $sum_{I=1}^nX_i$, so we will reject the null if there is a sufficiently large value for $sum_{I=1}^nX_i$.
For my answer, I use the fact that $sum_{I=1}^nX_i$ is normally distributed with mean $nmu$ and variance $nsigma^2$, so the most powerful test tin simplest implementable form would be as follows:
Reject the null hypothesis if $z_{alpha}<phi(frac{sum_{I=1}^nX_i-nmu_0}{sqrt{n}sigma})$, where $z_{alpha}$ is the upper $100alpha$ percentile of the standard normal distribution.
I’m not sure at all if this is correct, so I would appreciate some feedback.
statistics statistical-inference hypothesis-testing
statistics statistical-inference hypothesis-testing
asked Nov 26 at 23:06
DavidS
337111
asked Nov 26 at 23:06
DavidS
337111
asked Nov 26 at 23:06
DavidS
337111
asked Nov 26 at 23:06
DavidS
337111
asked Nov 26 at 23:06
DavidS
337111
337111
add a comment |
add a comment |
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