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My textbook just says that the following test statistic is normal without actually going through the derivation. Here is the problem:

Suppose that $X_1...X_n$ are iid RV with each being $N(\mu,\sigma^2)$ where $\sigma^2$ is known. We want to test:

$$H_0:\mu=0$$ $$H_1:\mu>0$$

We then take the following test statistic:

$$\tau_n=\frac{\sqrt{n}\bar{X_n}}{\sigma}$$

Now what I don't understand is why did the book choose this test statistic? It cannot be arbitrary can it? But more importantly, the book says that this test statistic is standard normal, can anyone please show me mathematically why this is true?

When I take the expectation and variance of $\tau_n$ I do not get $0$ for the expectation and $1$ for the variance so how can that be?

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This is a classic Z-test. To calculate a Z-score you simply take your observed sample mean, $\bar{X}_n$ minus your hypothesized value under $H_0: \mu=0$, and divide the result by the standard deviation of the the sample mean. To calculate the sample mean, we first obtain the standard deviation:

$Var(\bar{X}_n)$ = $Var(\sum_{i=1}^nX_i)\over{n^{2}}$ = $n\sigma^2\over{n^2}$ = $\sigma^2\over{n}$.

Taking the square root of the variance gives the standard deviation of the mean:

$\sqrt{Var(\bar{X}_n)}=\sqrt{\sigma^2\over{n}} =$ $\sigma\over{\sqrt{n}}$.

Now, we obtain the Z-score:

$Z$=${\bar{X}_n-\mu_0}\over{\sigma\over{\sqrt{n}}}$ = ${\bar{X}_n-0}\over{\sigma\over{\sqrt{n}}}$=$\sqrt{n}\bar{X}\over{\sigma}$

You then compare the Z score with percentiles of the standard normal distribution.

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Since this is a self study, let me give you a few pointes. One useful property of normals is that a sum of normals is normal. Believe that or look for a proof. Also, for independent random variables, $var(\sum X_i) = \sum var(X_i)$. Try to prove that. Yet another good property is $var(\lambda X) = \lambda^2 var(X)$. As for the mean of $\tau_n$, it is not zero. However, under $H_0$, it equals zero. What did you get for the variance?

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