STA258 Lecture 13

• Previously:
  • Hypothesis testing for one sample
• Diet Cola Example:
  • Artificial sweetness index $(0\to 10)$ before during and after storage.
    • Score is Before - After
  • Suppose we have a normal dist with $\sigma =1$
  • $\mu$ is average loss of sweetness.
  • $H_0:\mu =0$
  • $H_a:\mu >0$
  • If colas always lose sweetness, then score is $>0$
  • $\mu \ne 0$ can also be used
  • But ${\mu }_0>0$ is easier and more intuitive to interpret
  • Fallback to two sided if you can't tell.
  • $Z^{\ast }=\frac{\overline{X}-{\mu }_0}{\frac{\sigma }{\sqrt{n}}}=\frac{1.02-0}{\frac{1}{\sqrt{10}}}=3.22552321337175$
  • p-value
    • $P(Z>Z^{\ast })$
      • Follows same inequality direction as $H_a$
    • $P(Z>3.23)=0.0006<0.01$
    • Should we reject $H_0$ at $\alpha =0.01$?
    • Yes, because p-value is less than $\alpha$
    • So we accept the alternative that the average loss of sweetness is greater than 0.
• Deer Example:
  • Find the average weight of deer in and around UTM.
  • $n=36$
  • $\overline{X}=42.53$kg
  • Healthy deer: $45$kg
  • With $\alpha =0.05$
  • Determine whether UTM deer are underweight.
  • $\sigma =5.25$kg
  • $H_0:\mu =45$
  • $H_a:\mu <45$
  • $Z^{\ast }=\frac{42.53-45}{\frac{5.25}{\sqrt{36}}}=-2.82285714285714$
  • p-value
    • $P(Z<-2.82285714285714)$
    • In R
      • pnorm(-2.82285714285714)=0.002379889
    • $0.002379889<0.05$ so we reject $H_0$ and accept $H_a$ that the average weight of deer in and around UTM is less than 45kg.
• Blood Pressure Example:
  • ${\mu }_0=128$
  • $\sigma =15$
  • $\overline{X}=126.07$
  • $n=72$
  • $H_0:\mu =128$
  • $H_a:\mu \ne 128$
  • $Z^{\ast }=\frac{126.07-128}{\frac{15}{\sqrt{72}}}=-1.09177287015203$
    • In R
    • pnorm(-1.09177287015203)=0.1374665
  • p-value
    • $P(Z<-1.09177287015203)+P(Z>1.09177287015203)$
    • $2\cdot P(Z<-1.09177287015203)$
    • $2\cdot 0.1374665=0.274933$
    • $0.274933>0.05$ so we fail to reject $H_0$ and do not have enough evidence to support the claim that the average blood pressure is different from 128.
• Examples of Hypotheses and their P-Values:
  • $H_0:\mu ={\mu }_0$
  • $Z^{\ast }=\frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}$
  • P-Value
    • $H_a:\mu >{\mu }_0$ then $P(Z>Z^{\ast })$
    • $H_a:\mu <{\mu }_0$ then $P(Z<Z^{\ast })$
    • $H_a:\mu \ne {\mu }_0$ then $P(Z>|Z^{\ast }|)$
• Example:
  • $H_0:\mu =20$
  • $H_a:\mu <20$
  • $n=50$
  • $\overline{X}=19.4$
  • $\sigma =2$
  • $Z^{\ast }=\frac{19.4-20}{\frac{2}{\sqrt{50}}}=-2.12132034355965$
  • P-Value
    • $P(Z<-2.12132034355965)$
      • In R
      • pnorm(-2.12132034355965)=0.01694743
    • $\alpha =0.05$
      • $0.01694743<0.05$ so we reject $H_0$ and accept $H_a$ that the average is less than 20.
    • $\alpha =0.01$
      • $0.01694743>0.01$ so we fail to reject $H_0$ and do not have enough evidence to support the claim that the average is less than 20.
    • Time series complexities: STA457.
      • Suppose we get P-Value as $0.009$ but $\alpha =0.01$, should we reject $H_0$?
      • Yes, because $0.009<0.01$.
      • When you reject the first model, then the sequential models can't be accepted.
      • We might need to use subjectivity to accept the p-value to reject $H_0$ or accept it.
• Example:
  • Two sided hypothesis test
  • $H_0:\mu =15$
  • $H_a:\mu \ne 15$
  • $n=50$
  • $\overline{X}=14.15$
  • $\sigma =3$
  • $Z^{\ast }=\frac{14.15-15}{\frac{3}{\sqrt{50}}}=-2.00346921336188$
  • P-Value
    • $P(Z<-2.00346921336188)+P(Z>2.00346921336188)$
    • $2\cdot P(Z<-2.00346921336188)$
    • In R
    • pnorm(-2.00346921336188)=0.02256347
    • $2\cdot 0.02256347=0.04512694$
    • $\alpha =0.05$
      • $0.04512694<0.05$ so we reject $H_0$ and accept $H_a$ that the average is different from 15.
    • $\alpha =0.01$
      • $0.04512694>0.01$ so we fail to reject $H_0$ and do not have enough evidence to support the claim that the average is different from 15.
  • Confidence Intervals
    • $CI=\overline{X}\pm Z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}$
    • $\alpha =0.05$
    • $1-\alpha =0.95$
    • $14.15-1.96\frac{3}{\sqrt{50}}=13.3184424253246$
    • $14.15+1.96\frac{3}{\sqrt{50}}=14.9815575746754$
    • The true mean lies in there. But $15\notin CI$
    • Because our ${\mu }_0\notin CI$ we have to reject the null hypothesis.
    • A Level $\alpha$ two sided hypothesis test rejects $H_0:\mu ={\mu }_0$ exactly when ${\mu }_0$ falls outisde of a level $1-\alpha$ CI for $\mu$.
    • Confidence Intervals don't allow subjectivity, but p-value and hypothesis testing does. We have a little more control.
• Mean variance optimization model for economics. After the internet this model stopped working.
  • There were two other guys who came up with another model which allowed some subjective control. This model is called the Black-Litterman model. It allows you to use your own subjective beliefs to adjust the model. You can use the p-value to reject the null hypothesis and then use the confidence interval to find a range of values for the mean. Then you can use the Black-Litterman model to adjust your portfolio based on your beliefs about the mean.
  • It's not just the math, it's also the subjectivity that allows you to make better decisions. You can use the math to inform your decisions, but you also need to use your judgement and experience to make the final decision.w