Mathematics Asked by Hachiman Hikigaya on December 8, 2020

Let $X_{1}, ldots, X_{n}$ be a sample from the distribution $mathrm{N}(mu, 1)$ and consider testing $H_{0}: mu=mu_{0}$ versus $H_{1}: mu=mu_{1}$ where $mu_{0}<mu_{1}$ are known numbers.

(a) For a given level $alpha,$ find the most powerful test.

(b) Calculate the power

(c) For given $mu_{0}, mu_{1}, alpha,$ determine the minimal number of observations needed to have power at least $beta$ (i.e., to reject $H_{0}$ with probability at least $beta$ when $H_{1}$ holds).

Using Neyman Pearson lemma I have found the test statistic to be $bar X$ and reject $H_{0}$ if $bar X > k$, where $k= z_alpha / sqrt{n} + mu_0$ .

And calculated the power to be Power = P{Reject $H_0$} = P{$Z> z_alpha – sqrt{n} (mu-mu_0)$}

But for the Part (c) I don’t know how to find the minimum number of observation for given condition.

I think I have to solve the equation $beta = P_{mu_1}{bar X > z_alpha / sqrt{n} + mu_0 }$ = P{$Z> z_alpha – sqrt{n} (mu_1-mu_0)$}

But I don’t know how to solve this for $n$.

Please help me with this. Thnakyou.

You can state it in terms of the CDF $Phi(z)=P(Z le z)$; I don't think there is a way to avoid it.

begin{align} beta &le 1 - Phi(z_alpha - sqrt{n}(mu_1-mu_0)) \ Phi^{-1}(1-beta) &ge z_alpha - sqrt{n} (mu_1 - mu_0) \ sqrt{n} &ge frac{z_alpha - Phi^{-1}(1-beta)}{mu_1 - mu_0} end{align}

Answered by angryavian on December 8, 2020

As you found, UMP test is given by Neyman Pearson's Lemma with rejection region

$$mathbb {P}[overline{X}_n>k|mu=mu_0]=alpha$$

Now $overline{X}_n>k$ is your decision rule ($k$ now is fixed) and you can calculate the power ( usually indicated with $gamma$ because $beta$ is normally used for type II error)

$$mathbb {P}[overline{X}_n>k|mu=mu_1]=gamma$$

Understood this, finally fix $gamma$ and get $n$

Example

$mu_0=5$

$mu_1=6$

$alpha=5%$

$n=4$

The critical region is

$$(overline {X}_4-5)2=1.64rightarrow overline {X}_4=5.8224$$

Thus your decision rule is

$$overline {X}_4geq 5.8224$$

and you can calculate the power

$$gamma=mathbb {P}[overline {X}_4geq 5.8224|mu=6]=1-Phi(-0.36)approx 64%$$

Now suppose you want a fixed power $gamma geq 90%$, simply re-solve the same inequality in $n$

$$mathbb {P}[overline {X}_ngeq 5.82|mu=6]geq 0.90$$

Getting

$$(5.8224-6)sqrt{n}leq-1.2816$$

That is

$$ngeqBigg lceil Bigg(frac{1.2816}{0.1776}Bigg)^2Biggrceil=53$$

Answered by tommik on December 8, 2020

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