Exercise 6. Let X1, …, Xn iid∼ N(μ, σ^2) where σ^2 is known....

Exercise 6. Let $X_1, \dots, X_n \stackrel{iid}{\sim} N(\mu, \sigma^2)$ where $\sigma^2$ is known. Consider testing $H_0: \mu = \mu_0$ versus $H_A: \mu > \mu_0$ (a) Determine the power function of the test procedure $\delta$ that rejects $H_0$ if $\bar{X} > \mu_0 + z_\alpha \frac{\sigma}{\sqrt{n}}$. (b) Determine the sample size required to detect the value $\mu = \mu_1 > \mu_0$ at level $\alpha$ with power $\gamma$.

Transcript

00:01 Hello student for the given problem.

00:02 We have to perform the likelihood ratio test for the hypothesis not equal to theta equal to theta naught versus alternative hypothesis theta is not equal to for that we will first formulate the likelihood and under the null hypothesis.

00:18 So the likelihood function become l of theta naught given x is equal to f of x 1 upon theta naught into f of x 2 upon theta so on up to f of x n upon theta where xi given theta naught is the probability density function of normal distribution with mean mu naught and variance.

00:41 Therefore f of xi upon theta naught is equal to 1 upon under root of 2 pi theta naught multiplied by e to the power minus of xi minus mu naught the whole square upon theta.

00:59 Under the alternative hypothesis the likelihood function becomes l of theta upon x equal to f of x 1 upon theta into f of x 2 upon theta so on up to f of x n upon theta where f of xi upon theta is the probability density function of normal distribution with mean mu naught and variance theta.

01:29 Therefore f of xi upon theta become 1 upon the root of pi theta into e to the power minus of xi minus mu naught the whole square upon theta.

01:41 Next we need to calculate the likelihood ratio based statistic which is denoted by w which is equal to l of theta naught upon x given l of theta upon theta.

01:53 Now we can calculate the value of w using likelihood function from the above step.

02:00 Therefore w becomes f of x 1 upon theta naught into f of x 2 upon theta naught so on up to f of x 1 upon theta divided by f of x 1 theta into f of x 2 theta so on up to f of x n upon theta.

02:25 Next to determine the null distribution of w to determine the null distribution of w we need to use the fact that under the null hypothesis the likelihood ratio based up follows chi -square distribution with degrees of freedom equal to the difference in the number of parameters.

02:54 So in this case the null hypothesis has one parameter theta naught an alternative hypothesis has also one parameter therefore degrees of freedom become 1...

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