STA258 Lecture 14

• Hypothesis Testing based on t distrubtion, one sample, etc.

• Review

  • If we have a normal distribution: $\frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}$
  • If we have a non-normal distribution: $\frac{\overline{X}-\mu }{\frac{S}{\sqrt{n}}}$ where $S$ is the sample standard deviation. This follows a t distribution with $n-1$ degrees of freedom.

• Example:

  • $n=25$
  • Heart rate BPM: $\overline{x}=73.5$
  • $s=6$
  • Restoring heart rate is $71$ BPM
  • Is there evidence that true mean heart rate exceeds $71$ BPM?
  • $\alpha =0.05$
  • Independent random samples.
  • Heart rate has Normal Distribution.
  • $H_0:\mu =71$
  • $H_a:\mu >71$
  • $t^{\ast }=\frac{73.5-71}{\frac{6}{\sqrt{25}}}=2.08333333333333$
  • $t_{0.05,24}$
    • In R: qt(0.95, 24)
    • $1.710882$
  • $p$ value:
    • $P(t_{(24)}>t^{\ast })=P(t_{(24)}>2.08)$
    • In R: pt(2.08333333333333, 24, lower.tail=FALSE)
    • $0.02402205$
  • Since $p$ value is less than $\alpha$, we reject the null hypothesis. There is evidence that the true mean heart rate exceeds $71$ BPM.
    • #tk flashcards on when to reject or accept certain hypotheses based on $p$ value and $\alpha$.
  • Saying that laughter is the best medicine due to increased heart rate doesn't necessarily make sense, but it is a fun example. We would need to do more research to see if there is a causal relationship between laughter and heart rate.
  • Otherwise, we could say that working out increases heart rate and also is the best medicine.
  • $t$ test has a normality assumption

• Example:

  • Average price of a $2$ start motel room have decreased.
  • Last year it was normally distributed with $\mu =89.50$
  • $n=12$
    • $\{85.00,92.05,87.50,89.90,90.00,82.50,87.50,90.00,85.00,89.00,91.50,87.50\}$
  • If the dist of room prices are normal, $\alpha =0.05$. What conclusion should we make?
  • $H_0:\mu =89.50$
  • $H_a:\mu <89.50$
  • $\overline{x}=\frac{1}{12}(85.00+92.05+87.50+89.90+90.00+82.50+87.50+90.00+85.00+89.00+91.50+87.50)=88.1208333333333$
  • $s=\frac{1}{n-1}\sum _{i=1}^n(x_i-\overline{x}{)}^2$
  • $s=\sqrt{\frac{1}{11}((85.00-88.1208333333333{)}^2+(92.05-88.1208333333333{)}^2+(87.50-88.1208333333333{)}^2+(89.90-88.1208333333333{)}^2+(90.00-88.1208333333333{)}^2+(82.50-88.1208333333333{)}^2+(87.50-88.1208333333333{)}^2+(90.00-88.1208333333333{)}^2+(85.00-88.1208333333333{)}^2+(89.00-88.1208333333333{)}^2+(91.50-88.1208333333333{)}^2+(87.50-88.1208333333333{)}^2)}=2.86177458718865$
  • $s=2.86177458718865$
  • $t^{\ast }=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}}$
  • $t^{\ast }=\frac{89.5-88.12083333333333}{\frac{2.86177458718865}{\sqrt{12}}}=1.66944437166086$
  • $t_{0.05,11}$
    • In R: qt(0.95, 11)
    • $1.795885$
  • p-value:
    • $P(t_{(11)}>t^{\ast })=P(t_{(11)}>1.66944437166086)$
    • In R: pt(1.66944437166086, 11, lower.tail=FALSE)
    • $0.06160452$
  • Since $p$ value is greater than $\alpha$, we fail to reject the null hypothesis. There is not enough evidence to conclude that the average price of a $2$ star motel room has decreased.

• Proportions

  • $H_0:P_0$
  • $Z^{\ast }=\frac{\hat{P}-P_0}{\sqrt{\frac{P_0(1-P_0)}{n}}}\sim N(0,1)$
    • $np>10$
    • $n(1-p)>10$
    • These are the conditions for CLT to kick in.
      • #tk flashcards of this.
  • $H_a:P>P_0$
    • $P(Z>Z^{\ast })$
  • $H_a:P<P_0$
    • $P(Z<Z^{\ast })$
  • $H_a:P\ne P_0$
    • $2P(Z>\left|Z^{\ast }\right|)$

• Example:

  • Tim Hortons roll up the rim.
  • 100 empty cups
  • $12$ winning cups
  • $p=\frac{1}{6}$
  • Is there enough evidence to suggest that the probability of winning is less than $\frac{1}{6}$?
  • $H_0:p=\frac{1}{6}$
  • $H_a:p<\frac{1}{6}$
  • $Z^{\ast }=\frac{\frac{12}{100}-\frac{1}{6}}{\sqrt{\frac{\frac{1}{6}(1-\frac{1}{6})}{100}}}=-\frac{14\sqrt{5}}{25}=-1.25219806739988$
  • $p$ value:
    • $P(Z<Z^{\ast })=P(Z\le -1.25219806739988)$
    • In R: pnorm(-1.25219806739988, lower.tail=TRUE)
    • $0.1054618$
  • Since $p$ value is greater than $\alpha$, we fail to reject the null hypothesis. There is not enough evidence to suggest that the probability of winning is less than $\frac{1}{6}$.
  • Since sample was $100$, it's a strong result.

• Example:

  • Congenital abnormality occurs in $5\mathrm{\%}$
  • We have a study that $\frac{46}{384}$ have abnormality
  • Is there evidence that the true proportion of congenital abnormality is greater than $5\mathrm{\%}$?
  • $H_0:p=0.05$
  • $H_a:p>0.05$
  • $Z^{\ast }=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$
  • $Z^{\ast }=\frac{\frac{46}{384}-0.05}{\sqrt{\frac{(0.05)(1-0.05)}{384}}}=6.27512493759735$
  • $p$ value:
    • $P(Z>Z^{\ast })=P(Z>6.27512493759735)$
    • In R: pnorm(6.27512493759735, lower.tail=FALSE)
    • $1.746768e-10$
    • Very small. This is a very strong result.
    • Since $p$ value is less than $\alpha$, we reject the null hypothesis. There is evidence that the true proportion of congenital abnormality is greater than $5\mathrm{\%}$.
    • If we assume $\alpha =0.001$ for medical study
    • We still have a very small $p$ value, so we would still reject the null hypothesis. There is still evidence that the true proportion of congenital abnormality is greater than $5\mathrm{\%}$.

• Example:

  • $12.5\mathrm{\%}$ of us workers were unionized in 2005.
  • Suppose we have a sample of $400=n$ in $2006$
  • Have union efforts increased membership?
  • $H_0:p=12.5$
  • $H_a:p>12.5$
  • In 2006 $52$ were unionized, so $\hat{p}=\frac{52}{400}=0.13$.
  • $Z^{\ast }=\frac{0.13-0.125}{\sqrt{\frac{(0.125)(1-0.125)}{400}}}=0.302371578407382$
  • $P(Z>Z^{\ast })=P(Z>0.302371578407382)$
  • In R: pnorm(0.302371578407382, lower.tail=FALSE)
  • $0.3811844$
  • With $\alpha =0.05$
  • $p>\alpha$ so we fail to reject the null hypothesis. There is not enough evidence to suggest that union efforts have increased membership.
  • By visual inspection, we can see that $13\mathrm{\%}\sim 12.5\mathrm{\%}$

• Example:

  • $0.64$ of supermarket shoppers, believe no-name have the same quality as name brand.
  • $H_0:p=0.64$
  • $H_a:p\ne 0.64$
  • $n=100$ and $52$ believe that no-name have the same quality as name brand, so $\hat{p}=\frac{52}{100}=0.52$.
  • $Z^{\ast }=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$
  • $Z^{\ast }=\frac{0.52-0.64}{\sqrt{\frac{(0.64)(1-0.64)}{100}}}=-2.5$
  • $2P(Z>Z^{\ast })=2P(Z>2.5)$
    • In R: pnorm(2.5, lower.tail=FALSE)*2
    • $0.01241933$
  • With $\alpha =0.05$ $p<\alpha$ so we reject the null hypothesis. There is evidence to suggest that the proportion of supermarket shoppers who believe no-name have the same quality as name brand is different from $0.64$.

• Variance

  • Suppose we want to test on variance
  • $H_0:{\sigma }^2={\sigma }_0^2$
    • We know $\frac{(n-1)S^2}{{\sigma }^2}\sim {\chi }_{(n-1)}^2={\chi }_{\star }^2$
    • $H_a:{\sigma }^2={\sigma }_0^2$
    • Reject $H_0$ if ${\chi }_{\star }^2>{\chi }_{(n-1;\frac{\alpha }{2})}^2$ or ${\chi }_{\star }^2<{\chi }_{(n-1;\frac{\alpha }{2})}^2$
  • $H_a:{\sigma }^2>{\sigma }_0^2$
    • Reject if ${\chi }_{\star }^2>{\chi }_{(n-1;\alpha )}^2$ or $P({\chi }_{(n-1)}^2>{\chi }_{\star }^2)$ is too small
  • $H_a:{\sigma }^2<{\sigma }_0^2$
    • Reject if ${\chi }_{\star }^2<{\chi }_{(n-1;1-\alpha )}^2$ or $P({\chi }_{(n-1)}^2<{\chi }_{\star }^2)$ is too small

• example:

  • A compnay produces standard metal pipes with length $\sigma =1.2$
  • A client took $n=25$
  • $s=1.5$
  • Does this undermine the comany's claim?
  • $\alpha =0.05$
  • Length is normally distributed
  • $H_0:{\sigma }^2\le {1.2}^2$
  • $H_a{\sigma }^2\ge {1.2}^2$
  • ${\chi }_{\star }^2=\frac{(24)(1.5{)}^2}{(1.2{)}^2}=37.5$
  • p-value
    • $P({\chi }_{(24)}^2>{\chi }_{\star }^2)$
    • $P({\chi }_{(24)}^2>37.5)$
    • In R: pchisq(37.5, 24, lower.tail=FALSE)
    • $0.0389818$
  • Since $p$ value is less than $\alpha$, we reject the null hypothesis. There is evidence to suggest that the variance of the length of the pipes is greater than ${1.2}^2$.

• Example:

  • Car batteries life normally distributed with $\sigma =0.9$
  • $n=10$
  • $s=1.2$
  • Is $\sigma >0.9$
  • $\alpha =0.05$
  • ${\chi }_{\star }^2=\frac{(9)(1.2{)}^2}{(0.9{)}^2}$
  • $\frac{(9)(1.2{)}^2}{(0.9{)}^2}=16.0$
  • p-value
    • $P({\chi }_{(9)}^2>{\chi }_{\star }^2)$
    • $P({\chi }_{(9)}^2>16)$
    • In R: pchisq(16, 9, lower.tail=FALSE)
    • $0.06688159$
  • Since $p$ value is greater than $\alpha$, we fail to reject the null hypothesis. There is not enough evidence to suggest that the variance of the life of the car batteries is greater than ${0.9}^2$.

• Example:

  • Cola
  • Sweetness loss
  • #tk make flashcards on type 1 and 2 errors.
  • Sweetness loss is too much if $\mu =1.1$
  • $\alpha =0.05$
  • $H_0:\mu =0$
  • $H_a:\mu >0$
  • $n=10$
  • We want the power of the test against the alternative.
  • $\sigma =1$
  • Reject $H_0$ at $\alpha =0.05$ if:
    • $Z=\frac{\overline{x}-0}{\frac{1}{\sqrt{10}}}>1.645$
    • $\overline{x}>\frac{1.645}{\sqrt{10}}$
  • Reject $H_0$ if $\overline{x}>0.52$
  • $\text{power}=P(\overline{X}>0.52|\mu =1.1)$
    • $P(\frac{\overline{X}-1.1}{\frac{1}{\sqrt{10}}}>\frac{0.52-1.1}{\frac{1}{\sqrt{10}}})$
    • $\frac{0.52-1.1}{\frac{1}{\sqrt{10}}}=-1.83412104289766$
    • $P(\frac{\overline{X}-1.1}{\frac{1}{\sqrt{10}}}>-1.83412104289766)=0.9644$
    • Type 1 error is $1-\text{power}=1-0.9644=0.0356$
    • #tk true?

• Example:

  • Suppose we wish to test
  • $H_0:\mu =100$
  • $H_a:\mu >100$
  • $\alpha =0.05$
  • $1-\beta =0.6$
  • when $\mu =103$
  • What's the smallest sample size that will achieve that objective?
  • Var measured is normally distributed with $\sigma =14$
  • $\alpha =P(\text{rejecting }{h}_0|H_0\text{ is true})$
    • $=P(\overline{X}>{\overline{x}}^{\ast }|\mu ={\mu }_0)$
      • ${\mu }_0$ is when $H_0$ is true
      • Here ${\mu }_0=100$
    • $=P(\frac{\overline{X}-100}{\frac{14}{\sqrt{n}}}>\frac{{\overline{x}}^{\ast }-100}{\frac{14}{\sqrt{n}}})$
    • $=P(Z>\frac{{\overline{x}}^{\ast }-100}{\frac{14}{\sqrt{n}}})=\underset{\alpha }{\underbrace{0.05}}$
    • Critical value to achieve $0.05$ is $1.645$
    • $\overline{x}=100+1.645\frac{14}{\sqrt{n}}$
    • $\text{power}=1-\beta =P(\overline{X}>{\overline{x}}^{\ast }|\mu =103)$
      • $=P(\frac{\overline{X}-103}{\frac{14}{\sqrt{n}}}>\frac{{\overline{x}}^{\ast }-103}{\frac{14}{\sqrt{n}}})=0.6$
      • $Z^{\ast }=-0.25$
    • $\overline{x}=103-0.25\frac{14}{\sqrt{n}}$
      • #tk why do we do this, instead of $-0.25=\frac{{\overline{x}}^{\ast }-103}{\frac{14}{\sqrt{n}}}$
    • $100+1.645\frac{14}{\sqrt{n}}=103-0.25\frac{14}{\sqrt{n}}$
    • $n=78.2045444444445$
    • $n\approx 79$

• example:

  • Vending machine
  • $225mL$ of coffee
  • $\sigma =7mL$
  • $H_0:\mu =225$
  • $H_a:\mu <225$
  • $n=40$
  • If true mean is $223$, what is the power of your test at $\alpha =0.05$
  • How many coffee cups should you sample if you want to raise the power to $0.8$
  • $\alpha =P(\text{rejecting }{H}_0|H_0\text{ is true})$
    • $P(\overline{X}<{\overline{X}}^{\ast }|\mu =225)$
    • $P(\frac{\overline{X}-225}{\frac{7}{\sqrt{n}}}<\frac{{\overline{X}}^{\ast }-225}{\frac{7}{\sqrt{n}}})$
    • $P(Z<\frac{{\overline{X}}^{\ast }-225}{\frac{7}{\sqrt{n}}})=0.05$
    • In R: qnorm(0.05, lower.tail=TRUE)
    • $-1.644854\approx -1.645$
    • $\frac{{\overline{X}}^{\ast }-225}{\frac{7}{\sqrt{n}}}=-1.645$
    • ${\overline{X}}^{\ast }=225-1.644854\frac{7}{\sqrt{n}}$
  • $\text{power}=1-\beta =P(\overline{X}<{\overline{X}}^{\ast }|\mu =223)$
    • $P(\frac{\overline{X}-223}{\frac{7}{\sqrt{n}}}<\frac{{\overline{X}}^{\ast }-223}{\frac{7}{\sqrt{n}}})$
    • $P(Z<\frac{{\overline{X}}^{\ast }-223}{\frac{7}{\sqrt{n}}})=0.8$
    • In R: qnorm(0.8, lower.tail=TRUE)
    • $Z^{\ast }=0.8416212$
    • $\frac{{\overline{X}}^{\ast }-223}{\frac{7}{\sqrt{n}}}=0.8416212$
    • $225-1.644854\frac{7}{\sqrt{n}}=223+0.8416212\frac{7}{\sqrt{n}}$
    • ${\overline{X}}^{\ast }=223+0.8416212\frac{7}{\sqrt{n}}$
    • $n=75.7363467726342$
    • $n\approx 76$