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This is an example from my book and Iam translating from it so there might be some things that are not all clear. Just let me know so i can edit.

Someone is measuring PH-level in a sea and for that she has taken $\textbf{x} =(x_1,...x_n)$ which is a sample taken from $N(\mu, \sigma^2)$ where $\sigma = 0.2$. The confidence interval is done. A 95% upper confidence level for $\mu$ is: $\bar{\mu} = x + \lambda_{0.05}\frac{\sigma}{\sqrt{n}} = \bar{x} + 1.64499\frac{0.2}{\sqrt{10}} \approx 5.8 +0.10= 5.9$ (observe that $\bar{x} = 5.8$ and that $\lambda_{\alpha}$ is defined as the solution $\Phi(\lambda_{\alpha})= 1- \alpha$. $\Phi$ is the Normal distribution function)

A sea is considered acidified if the pH value is less than $\mu_0 = 6.0$ and we want to test if this limit is undershot(i don't know if undershot is a good transalation ,anyhow) So we test the hypothesis $H_0 : \mu = \mu_0$ against $H_1: \mu < \mu_0$ .

A testvariable should be $\bar{x}$ or equivalently: $T = \dfrac{\bar{x}-\mu_0 }{\frac{\sigma}{ \sqrt{n}}}$. which is a observation from $N(0,1)$. with $5\% $ error risk we get the test: $\textbf{what do they mean by $\bar{x}$ is equivalent to $T$ they are just standardizing the variable? }$

"Discard the hypothesis $H_0$" if $T \leq -\lambda_{0.05}$ ($\textbf{i can't understand what they mean here $T\leq -\lambda_{0.05}$ in terms of seeing that $\mu < \mu_0$ }$)

...this occurs when:

$\bar{x}\leq \mu_{0} -\lambda_{0.05}\frac{\sigma}{\sqrt{n}} $ which is equivalent to:

$\mu_{0} \geq \bar{x} + \lambda_{0.05}\frac{\sigma}{\sqrt{n}}$

.....so we discard $H_0$

TooTone
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Danny
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1 Answers1

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A testvariable should be $\bar{x}$ or equivalently: $T = \dfrac{\bar{x}-\mu_0 }{\frac{\sigma}{ \sqrt{n}}}$. which is a observation from $N(0,1)$. with $5\% $ error risk we get the test: $\textbf{what do they mean by $\bar{x}$ is equivalent to $T$ they are just standardizing the variable? }$

Yes exactly they are just standardizing the random variable $\bar{x}$, the sample mean. If you rearrange to get $\bar{x}$ in terms of $T$ you find $\bar{x} = \frac\sigma{\sqrt{n}}T + \mu_0$. As $T$ is normally distributed with mean $0$ and standard deviation $1$ (under the null hypothesis) then this shows you that $\bar{x}$ must be is normally distributed with mean $\mu_0$ and standard deviation $\frac\sigma{\sqrt{n}}$ (under the null hypothesis) -- although of course we start with the distribution of $\bar{x}$ and work out $T$. This distribution for $\bar{x}$ makes sense because under the null hypothesis the true mean is $\mu_0$ and the population standard deviation is scaled by $1/\sqrt{n}$ to get the sampling distribution of the mean.

"Discard the hypothesis $H_0$" if $T \leq -\lambda_{0.05}$ ($\textbf{i can't understand what they mean here $T\leq -\lambda_{0.05}$ in terms of seeing that $\mu < \mu_0$ }$)

In the earlier part, it seems that they had a 90% confidence interval with the right-hand side given by $\lambda_{0.05} = $ the 95th percentile of the normal distribution $\approx 1.64$. Here they are doing a one-sided test to the left with a significance level of $\alpha=$5% and because of the symmetry of the normal distribution the 5th percentile is exactly as far to the left of the mean (0) as the 95th percentile is to the right. Hence the 5th percentile $=-\lambda_{0.05}$.

Note: usually the point of these examples is to show the equivalence between an $(100\%-\alpha)$ confidence interval and a two-sided hypothesis test with the same $\alpha$. However, here they're showing that a one-sided test of significance $\alpha$ takes its rejection region bound from one side of a $(100\%-\mathbf{2}\alpha)$ confidence interval. Perhaps this is to make the point that using a one-sided test corresponds to a lower-confidence confidence interval than a two-sided test. Usual practice is to use two-sided tests to stop you rigging your test by using a one-sided test and then deciding after collecting the data which side to reject on.

TooTone
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  • thanks for answering, about the second question/statment. when saying $H_1: \mu < \mu_0$ with $5%$ error risk , then they mean, in terms of the test variable $T$, $P( T \leq \mu_0) \geq 5%$. how do i intpret it. @TooTone – Danny Feb 27 '14 at 23:18
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    @Danny I'm not sure about $P(T\le\mu_0)$ because in your question you had "Discard the hypothesis $H_0$" if $T \le -\lambda_{0.05}$. What they are saying is that the region for rejecting $H_0$ in favour of $H_1$ is the region $(-\infty,-\lambda_{0.05})=(-\infty, -1.64)$. Of course, the mean $\bar{x}$ that you sample is random depending on the random data, and so $\alpha=10%$ of the time you will get $\bar{x}$ in $(-\infty,-1.64)$ even when $H_0$ is true (type 1 error). Does this help? – TooTone Feb 27 '14 at 23:29
  • ,ok lets go back one step , that s what was the meaning of standardizing the variable $\bar{x}$ . @TooTone – Danny Feb 28 '14 at 00:16
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    @Danny simply that the test statistic is standardised. It's much easier for someone to understand how extreme something is if it's standardised because you don't have to know the mean or standard deviation. If I say -3 on a standardised distribution you know straightaway that that's extreme without me having to say anything else and without you having to do any more calculations. It's also easier to look up the quantiles of a N(0,1) distribution than an N(6,whatever) distribution but this is less important with modern software. – TooTone Feb 28 '14 at 00:48
  • ok i ask the last question and then i give up(I know Iam really confusing). What i want is something like this: we know that $\bar{x} \sim N(\mu_0 ,\frac{\sigma^2}{n})$ so no we ask ourself what $x$ do we need so $F_{\bar{x}}(x) = 0.05$ , but this is equvivalent of saying $0.05 =\Phi(\frac{x-\mu_0}{\sigma /\sqrt{n}} ) = \Phi(-\lambda_{0.05})$ – Danny Feb 28 '14 at 02:33
  • what i meant above.... $F_{\bar{x}}(x) = 0.05 = P\big(\frac{\bar{x}-\mu_0}{\sigma /\sqrt{n}} \leq (\frac{x-\mu_0}{\sigma /\sqrt{n}} \big) =\Phi(\frac{x-\mu_0}{\sigma /\sqrt{n}} ) = \Phi(-\lambda_{0.05})$ If you can make any sense of this @TooTone – Danny Feb 28 '14 at 02:46
  • so by the chain of equalites we have that $F_{\bar{x}}(x) =0.05$ exactly when $T \leq -\lambda_{0.05}$ – Danny Feb 28 '14 at 02:57
  • what i want is a connection not immediately jump to standardization and $T\leq -\lambda_{0.05}$ – Danny Feb 28 '14 at 02:59
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    Your thinking is sound: that all seems good to me. Because $\bar{x}$ is $N(6,0.2^2/10)$ assuming null hypothesis, then solving $F_{\bar{x}}(x)=0.05$ gives $x=6-1.64\times0.2/\sqrt{10}\approx5.9$. So if $\bar{x}<$ about $5.9$ you reject H0 in favour of H1. – TooTone Feb 28 '14 at 09:22