STA260 Lecture 20

• Note
  • If we're asked to find the most powerful ($\alpha$-level) test. We use Neyman-Pearson's Lemma.
• Example on Slide 26:
  • $X_1,\dots ,X_n\stackrel{iid}{\sim }N(\mu ,{\sigma }^2)$
  • ${\sigma }^2$ is known.
  • $H_0:\mu ={\mu }_0$
  • $H_a:\mu >{\mu }_0$
  • $\frac{L({\mu }_0)}{L({\mu }_a)}<k$
    • This means that ${\mu }_a>{\mu }_0$
  • $\frac{\prod _{i=1}^{n}f(x_i|{\mu }_0)}{\prod _{i=1}^{n}f(x_i|{\mu }_a)}<k$
  • $\frac{\prod _{i=1}^n\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{(x_i-{\mu }_0{)}^2}{2{\sigma }^2}}}{\prod _{i=1}^n\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{(x_i-{\mu }_a{)}^2}{2{\sigma }^2}}}<k$
  • $\prod _{i=1}^n{e}^{-\frac{(x_i-{\mu }_0{)}^2}{2{\sigma }^2}+\frac{(x_i-{\mu }_a{)}^2}{2{\sigma }^2}}<k$
  • $e^{\sum _{i=1}^n-\frac{(x_i-{\mu }_0{)}^2}{2{\sigma }^2}+\frac{(x_i-{\mu }_a{)}^2}{2{\sigma }^2}}<k$
  • $\sum _{i=1}^n-\frac{(x_i-{\mu }_0{)}^2}{2{\sigma }^2}+\frac{(x_i-{\mu }_a{)}^2}{2{\sigma }^2}<\ln k$
  • $⋮$
    • The $x_i^2$ terms cancel
    • #tk
  • $\exp (-\frac{n}{2{\sigma }^2})({\mu }_0^2-{\mu }_a^2)\exp (-\frac{1}{{\sigma }^2}({\mu }_a-{\mu }_0)\sum _{i=1}^n{x}_i)<k$
    • The only random component is $\sum _{i=1}^n{x}_i$
    • Everything else is just a constant.
    • For everything to be $<k$, it's equivalent to $\sum _{i=1}^n{x}_i>c$
    • Aside:
      • We want $e^{-x}>1$
      • When $x<0$ we have this.
    • $\sum _{i=1}^n{x}_i>c$ is our simplified criteria
    • As $\sum _{i=1}^n{x}_i\uparrow ⟹\frac{L({\mu }_0)}{L({\mu }_a)}\downarrow$
  • Example a probability of a type 1 error.
    • $P(\text{Type I})=\alpha$
    • $P(\text{Reject }{H}_0|H_0\text{ is true})=\alpha$
    • We know that rejecting $H_0$ is when we have $\sum _{i=1}^n{x}_i>c$
    • $P(\sum _{i=1}^n{x}_i>c|\mu ={\mu }_0)=\alpha$
    • We know that $\overline{X}\sim N(\mu ,\frac{{\sigma }^2}{n})$
    • $P(\frac{1}{n}\sum _{i=1}^n{x}_i>\frac{1}{n}c|\mu ={\mu }_0)=\alpha$
    • $c^{\prime }=\frac{c}{n}$
    • $P(\overline{X}>c^{\prime }|\mu ={\mu }_0)=\alpha$
    • The probability of $P(\overline{X}>c^{\prime })=P(\frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}>\frac{c^{\prime }-\mu }{\frac{\sigma }{\sqrt{n}}})=P(Z>Z_{\alpha })$
    • $c^{″}=\frac{c^{\prime }-\mu }{\frac{\sigma }{\sqrt{n}}}$
    • $c^{″}={\displaystyle \frac{\frac{c}{n}-\mu }{\frac{\sigma }{\sqrt{n}}}}$
    • $c^{″}={\displaystyle \frac{\sqrt{n}(\frac{c}{n}-\mu )}{\sigma }}$
    • We can also call $c^{″}=Z_{\alpha }$
      • $c^{\prime }={\mu }_0+Z_{\alpha }\frac{\sigma }{\sqrt{n}}$ is our threshold for rejecting $H_0$
    • $P(\frac{\overline{X}-{\mu }_0}{\frac{\sigma }{\sqrt{n}}}>\frac{c^{\prime }-{\mu }_0}{\frac{\sigma }{\sqrt{n}}})=\alpha$
    • $P(Z>Z_{\alpha })=\alpha$
    • We want to find $Z_{\alpha }$ such that area to the right is just $\alpha$
    • If we leave it in terms of $c^{\prime }$, then we look at $P(\overline{X}>c^{\prime }|\mu ={\mu }_0)=\alpha$ instead of the original one.
    • So we reject $H_0$ if we have that $\overline{X}>{\mu }_0+Z_{\alpha }\frac{\sigma }{\sqrt{n}}$. This is the most powerful test.
• Exercise:
  • What is the most power level $\alpha$ test for the same $H_0$ and $H_a$ if ${\sigma }^2$ is not known?
  • t-test, we standardize the variable differently, that's it.
    • $P(\frac{1}{n}\sum _{i=1}^n{x}_i>\frac{1}{n}c|\mu ={\mu }_0)=\alpha$
    • $c^{\prime }=\frac{c}{n}$
    • $P(\overline{X}>c^{\prime }|\mu ={\mu }_0)=\alpha$
    • The probability of $P(\overline{X}>c^{\prime })=P(\frac{\overline{X}-{\mu }_0}{\frac{s}{\sqrt{n}}}>\frac{c^{\prime }-{\mu }_0}{\frac{s}{\sqrt{n}}})=\alpha$
    • Then we get $t_{(n-1)}$
    • $P(t_{(n-1)}>t_{(n-1;\alpha )})=\alpha$
• Section 9.3
  • Duality of Confidence Intervals and Hypothesis Tests.
  • Theorem 1:
    • $X$: is our observed data.
    • ${\theta }_0$: parameter of interest.
    • $\mathrm{\Theta }$: set of all possible values of ${\theta }_0$.
    • ${\theta }_0\in \mathrm{\Theta }$: where $\mathrm{\Theta }$ is the parameter space.
    • $A({\theta }_0)$: Acceptance region for testing $H_0:\theta ={\theta }_0$.
      • Do not reject null if $X\in A({\theta }_0)$.
    • $C(\mathbf{X})=\{\theta \in \mathrm{\Theta }|X\in A(\theta )\}$
      • Confidence interval for $\theta$.
    • $C(\mathbf{X})$: is the set of parameter values of $\theta$, which the test does not reject $H_0:\theta ={\theta }_0$ at $\alpha$
    • Suppose for every ${\theta }_0$ in $\mathrm{\Theta }$ ($\mathrm{\Theta }$ is a set) there is a test at level $\alpha$ where $H_0:\theta ={\theta }_0$. The acceptance region is $A({\theta }_0)$.
    • This means that given some acceptance region, $C(\mathbf{X})$ is the $100(1-\alpha )\mathrm{\%}$ confidence region for $\theta$.
    • Moving from a hypothesis test to a confidence interval
  • Theorem 2:
    • The other way around from Theorem 1.
    • Moving from a confidence interval to a hypothesis test.
    • $C(\mathbf{X})$: is a $100(1-\alpha )\mathrm{\%}$ confidence region for $\theta$.
    • For a fixed ${\theta }_0$: $P(\theta \in C(\mathbf{X})|\theta ={\theta }_0)=1-\alpha$
    • The acceptance region:
      • $A({\theta }_0)=\{X|{\theta }_0\in C(\mathbf{X})\}$
    • For testing the null hypothesis $H_0:\theta ={\theta }_0$, we reject $H_0$ if $X\notin A({\theta }_0)$.
  • This means you can move between the two, via these theorems.
  • Example:
    • $X_1,\dots ,X_n\stackrel{iid}{\sim }N(\mu ,{\sigma }^2)$
    • ${\sigma }^2$ is known.
      • $H_0:\mu ={\mu }_0$
      • $H_a:\mu \ne {\mu }_0$
    • Before we had $H_a:\mu >{\mu }_0$, but now we have $H_a:\mu \ne {\mu }_0$, so it's a two-sided test.
      • $\overline{X}-{\mu }_0>Z_{\alpha }\frac{\sigma }{\sqrt{n}}$
    • Now we have:
      • $\overline{X}-{\mu }_0>Z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}$ or $\overline{X}-{\mu }_0<-Z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}$
      • So $|\overline{X}-{\mu }_0|>Z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}$
    • For acceptance, the test fails to reject $H_0$ when $|\overline{X}-{\mu }_0|\le Z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}$.
    • So $-Z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}\le \overline{X}-{\mu }_0\le Z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}$
    • By theorem 1, the set of values of ${\mu }_0$ where $H_0$ is not rejected is:
      • $C(\mathbf{X})=\{{\mu }_0\in \mathbb{R}|-Z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}\le \overline{X}-{\mu }_0\le Z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}\}$
    • Application of Theorem 2:
      • #tk
• Generalized Likelihood Ratio Test:
  • $H_0:\theta \in {\mathrm{\Theta }}_0$
  • $H_a:\theta \in {\mathrm{\Theta }}_a$
  • The rejection region is $\lambda \le k$
  • $\lambda =\frac{\underset{\theta \in {\mathrm{\Theta }}_0}{max}L(\theta )}{\underset{\theta \in {\mathrm{\Theta }}_a}{max}L(\theta )}\le k$
    • Likelihood ratio test statistic.
  • ${\mathrm{\Theta }}_0$: parameter space under $H_0$
  • ${\mathrm{\Theta }}_a$: parameter space under $H_a$
  • $\mathrm{\Theta }$: parameter space for $\theta$, so for null and alternative, so ${\mathrm{\Theta }}_0\cup {\mathrm{\Theta }}_a=\mathrm{\Theta }$.
  • Test rejects the null for $\lambda \le k$. $k$ is our threshold.
  • Example:
    • $X_1,\dots ,X_n\stackrel{iid}{\sim }N(\mu ,{\sigma }^2)$
    • $H_0:\mu ={\mu }_0$
    • $H_a:\mu \ne {\mu }_0$
    • Find the appropriate likelihood ratio test.
    • Identify ${\mathrm{\Omega }}_0,{\mathrm{\Omega }}_a,\mathrm{\Omega }$
    • $\mathrm{\Omega }=\{(\mu ,{\sigma }^2)|\mu ,{\sigma }^2\in \mathbb{R},{\sigma }^2>0\}$
    • Under $H_0$:
      • ${\mathrm{\Omega }}_0=\{(\mu ,{\sigma }^2)|\mu ={\mu }_0,{\sigma }^2\in \mathbb{R},{\sigma }^2>0\}$
    • Under $H_a$:
      • ${\mathrm{\Omega }}_a=\{(\mu ,{\sigma }^2)|\mu \ne {\mu }_0,{\sigma }^2\in \mathbb{R},{\sigma }^2>0\}$