One-sided, one-sample Gaussian test

The one-sided one- sample Gauss test , also known as the one-sided Gauss test , is a special statistical test in test theory , a branch of mathematical statistics . In its basic version, in the case of a normal distribution with an unknown expected value and known variance, it tests whether the expected value is above or below a specified threshold value. If the sample size is large enough and it can be assumed that the central limit theorem applies, the test can also be used as an approximate test . It is then also tested whether the expected value of an unknown distribution is above or below a predetermined threshold value.

Action

If it can be assumed that there is a normal distribution with variance , the test runs according to the following scheme: ${\displaystyle \sigma _{0}^{2}}$

• Choose a threshold for the unknown expected value. The null hypothesis is then of the form ${\displaystyle \mu _{0}}$ ${\displaystyle H_{0}}$

${\displaystyle H_{0}=(-\infty ,\mu _{0}]}$,

the alternative of the form

${\displaystyle H_{1}=(\mu _{0},+\infty )}$

• Choose an error probability that corresponds to type I error . ${\displaystyle \alpha }$
• Calculate the critical value

${\displaystyle k:=\mu _{0}+u_{1-\alpha }\cdot {\sqrt {\frac {\sigma ^{2}}{n}}}}$.

The - is the quantile of the standard normal distribution , which can be looked up in the quantile table of the standard normal distribution. Furthermore, there is the number of elements in the sample and the known variance of the underlying normal distribution. ${\displaystyle u_{1-\alpha }}$ ${\displaystyle (1-\alpha )}$ ${\displaystyle n}$ ${\displaystyle \sigma _{0}^{2}}$

• Find the arithmetic mean

${\displaystyle {\overline {x}}={\tfrac {1}{n}}\sum _{i=1}^{n}x_{i}}$

of the sample elements . ${\displaystyle x_{1},x_{2},\dots ,x_{n}}$

• If the arithmetic mean is greater than the critical value, i.e. if the following applies: reject the null hypothesis. Otherwise, keep the null hypothesis. ${\displaystyle {\overline {x}}>k}$

If you choose the right part of the number line as the null hypothesis , that is ${\displaystyle H_{0}}$

${\displaystyle H_{0}=[\mu _{0},+\infty )}$

and as an alternative accordingly

${\displaystyle H_{1}=(-\infty ,\mu _{0})}$,

so the general procedure remains the same. However, when determining the critical value, the -quantile is replaced by the -quantile . He then calculates to ${\displaystyle (1-\alpha )}$ ${\displaystyle \alpha }$ ${\displaystyle u_{\alpha }}$

${\displaystyle k':=\mu _{0}+u_{\alpha }\cdot {\sqrt {\frac {\sigma ^{2}}{n}}}}$.

The null hypothesis is rejected if the arithmetic mean is smaller than the critical value, i.e. if applies. It is retained accordingly if the arithmetic mean is greater than the critical value. ${\displaystyle {\overline {x}}<k'}$

Optimality

In the case described above, a normal distribution with an unknown expected value and known variance, as well as the hypotheses used above, the one-sided Gaussian test is a uniformly best test . This means that at any given level he always has a smaller Type II error than any other test for that situation. ${\displaystyle \alpha }$

The proof is based on the fact that the normal distribution belongs to the exponential family and has a monotonic density quotient . This allows the Neyman-Pearson lemma for two-element test problems to be extended to the set of normal distributions considered here using monotonic arguments.

Mathematical basis

The underlying statistical model is

${\displaystyle (\mathbb {R} ^{n},{\mathcal {B}}(\mathbb {R} ^{n}),({\mathcal {N}}(\mu ,\sigma _{0}^{2}))_{\mu \in \mathbb {R} })}$

and is also known as the normal distribution model. It is a product model and formalizes the assumption implicitly made above that the sample variables are distributed independently and identically . ${\displaystyle X_{1},X_{2},\dots ,X_{n}}$

This model is the mathematical formalization of the implementation of identical experiments that do not influence each other and the outcome of which is normally distributed with known variance and unknown expected value. ${\displaystyle n}$

The test of the null hypothesis against the alternative is based on the test statistic ${\displaystyle \varphi }$ ${\displaystyle H_{0}=(-\infty ,\mu _{0}]}$ ${\displaystyle H_{1}=(\mu _{0},+\infty )}$

${\displaystyle T(X)={\tfrac {1}{n}}\sum _{i=1}^{n}X_{i}}$,

the so-called sample mean . It is normally distributed to the unknown mean and to the variance , so it applies ${\displaystyle \mu }$ ${\displaystyle {\tfrac {\sigma ^{2}}{n}}}$

${\displaystyle T(X)\sim {\mathcal {N}}(\mu ,{\tfrac {\sigma ^{2}}{n}})}$.

This follows from the calculation rules for normally distributed random variables . The actual statistical test is then given by

${\displaystyle \varphi (X)={\begin{cases}1&{\text{ falls }}\quad T(X)>k\\0&{\text{ falls }}\quad T(X)\leq k\end{cases}}}$

Here is the critical value . It determines the level (and thus the first type of error). If the level is as specified, the critical value is calculated as above ${\displaystyle k}$ ${\displaystyle \alpha }$

${\displaystyle k:=\mu _{0}+u_{1-\alpha }\cdot {\sqrt {\frac {\sigma ^{2}}{n}}}}$

with the quantile of the standard normal distribution. ${\displaystyle 1-\alpha }$ ${\displaystyle u_{1-\alpha }}$

The rejection area of the test is thus ${\displaystyle A}$

${\displaystyle A=\{x\in \mathbb {R} ^{n}\mid {\overline {x}}>k\}}$,

what in vector notation

${\displaystyle A=\{x\in \mathbb {R} ^{n}\mid \mathbf {1} ^{T}x>nk\}}$

corresponds. Here is the one vector and . The rejection area thus forms a half-space . ${\displaystyle \mathbf {1} }$ ${\displaystyle x=(x_{1},x_{2},\dots ,x_{n})}$

The mathematical basis for the analog test of the null hypothesis against the alternative is the same, only the inequality signs in the test are reversed and the critical value is determined as indicated in the section above. ${\displaystyle H_{0}=[\mu _{0},+\infty )}$ ${\displaystyle H_{1}=(-\infty ,\mu _{0})}$

Different definitions

Both the definition of the test and

${\displaystyle \varphi (X)={\begin{cases}1&{\text{ falls }}\quad T(X)>k\\0&{\text{ falls }}\quad T(X)\leq k\end{cases}}}$

as well as

${\displaystyle \varphi (X)={\begin{cases}1&{\text{ falls }}\quad T(X)\geq k\\0&{\text{ falls }}\quad T(X)<k\end{cases}}}$

However, the different inequality signs have no influence on the result of the test, which always applies. The probability that the test statistic assumes exactly the critical value and that the two definitions thus assume different values ​​is zero and can be neglected. ${\displaystyle P(T(X)=k)=0}$

Derived confidence interval

The confidence interval for the confidence level derived from the one-sample Gaussian test is ${\displaystyle 1-\alpha }$

${\displaystyle C_{1}(X)=\left[{\overline {X}}-u_{1-\alpha }\cdot {\sqrt {\tfrac {\sigma _{0}^{2}}{n}}};\infty \right)}$.

respectively

${\displaystyle C_{2}(X)=\left(-\infty ;{\overline {X}}+u_{1-\alpha }\cdot {\sqrt {\tfrac {\sigma _{0}^{2}}{n}}}\right]}$.

This follows directly from the duality of tests and confidence areas . The confidence intervals are thus equally the best confidence intervals .

Individual evidence

1. ^ ^{A b} Ludger Rüschendorf: Mathematical Statistics . Springer Verlag, Berlin Heidelberg 2014, ISBN 978-3-642-41996-6 , p. 195 , doi : 10.1007 / 978-3-642-41997-3 .
2. ↑ ^{a b c} Hans-Otto Georgii: Stochastics . Introduction to probability theory and statistics. 4th edition. Walter de Gruyter, Berlin 2009, ISBN 978-3-11-021526-7 , p. 276 , doi : 10.1515 / 9783110215274 .
3. ^ Hans-Otto Georgii: Stochastics . Introduction to probability theory and statistics. 4th edition. Walter de Gruyter, Berlin 2009, ISBN 978-3-11-021526-7 , p. 265-266 , doi : 10.1515 / 9783110215274 .
4. ↑ Claudia Czado, Thorsten Schmidt: Mathematical Statistics . Springer-Verlag, Berlin Heidelberg 2011, ISBN 978-3-642-17260-1 , p. 153 , doi : 10.1007 / 978-3-642-17261-8 .