STA258 Lecture 15

• Example:
  • Two sample t test for difference
  • $Y_1,\dots ,Y_n\stackrel{iid}{\sim }N(\mu ,{\sigma }^2)$
  • $X_1,\dots ,X_n\stackrel{iid}{\sim }N(\mu ,{\sigma }^2)$
  • ${\sigma }_1^2={\sigma }_2^2={\sigma }^2$
  • $S_p^2=\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}$
  • $\frac{(\overline{Y}-\overline{X})-({\mu }_1-{\mu }_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\sim t_{(n_1+n_2-2)}$
  • $H_0:{\mu }_1={\mu }_2$
  • $H_a:{\mu }_1\ne {\mu }_2$
• Example:
  • Two groups
  • Managerial success index
  • Higher is better
  • Group 1:
    • Interaction with outsiders
  • Group 2:
    • Fewer interactions
  • $n_1=12$
  • ${\overline{X}}_1=65.33$
  • $S_1^2={6.61}^2$
  • $n_2=15$
  • ${\overline{X}}_2=49.67$
  • $S_2^2={9.33}^2$
  • See whether assumptions of problems are satisfied. See if both are normal then.
  • Use Boxplots and Q-Q plots to check for normality.
  • They should also be independent.
  • Spread of Group 1 is almost same as Group 2, so we can assume equal variances.
  • We can assume variances are the same if two samples come from the same population.
  • We can also say:
    • ${\sigma }_1^2\approx {\sigma }_2^2⟺\frac{max(S_1^2,S_2^2)}{min(S_1^2,S_2^2)}<2$
  • Compute pooled variance:
    • $S_p^2=\frac{(12-1)(6.61{)}^2+(15-1)(9.33{)}^2}{(12+15-2)}=67.971908$
    • DF: $n_1+n_2-2=12+15-2=25$
  • $t^{\ast }=\frac{{\overline{X}}_1-{\overline{X}}_2}{\sqrt{\frac{S_p^2}{n_1}+\frac{S_p^2}{n_2}}}=\frac{65.33-49.67}{\sqrt{\frac{67.97}{12}+\frac{67.97}{15}}}=4.97$
    • $\frac{65.33-49.67}{\sqrt{\frac{67.97}{12}+\frac{67.97}{15}}}=4.90441817549598$
      • #tk why the discrepancy?
  • p-value:
    • $2P(t_{(25)}>4.97)=0.00<0.01$ so we reject $H_0$ and conclude that there is a significant difference between the two groups.
  • We have strong evidence to suggest that there is a difference in managerial success index between the two groups.
• What if we have difference variances?
  • We can't use the pooled variance anymore.
  • $t^{\ast }=\frac{{\overline{X}}_1-{\overline{X}}_2}{\sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2}}}$
  • This is still a t dist, however df:
    • $\frac{{(\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2})}^2}{\left(\frac{1}{n_1-1}\right){\left(\frac{S_1^2}{n_1}\right)}^2+\left(\frac{1}{n_2-1}\right){\left(\frac{S_2^2}{n_2}\right)}^2}$
    • Or $min(n_1-1,n_2-1)$
• Example:
  • Conservationsists want to see destructino of a rain forest over time
  • ${\mu }_1=$ mean unlogged forests
  • ${\mu }_2=$ mean of logged forests
  • $H_0:{\mu }_1={\mu }_2$
  • $H_a:{\mu }_1<{\mu }_2$
  • $n_1=12$
  • ${\overline{X}}_1=17.5$
  • $S_1^2=3.52$
  • $n_2=9$
  • ${\overline{X}}_2=13.6$
  • $S_2^2=4.5$
  • $t^{\ast }=\frac{{\overline{X}}_1-{\overline{X}}_2}{\sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2}}}=2.11$
  • df: $n_2-1=8$
  • p-value: $P(t_{(8)}>t^{\ast })=P(t_{(8)}>2.11)<0.05$
  • Reject $H_0$ and conclude that there is a significant difference between the two groups. We have evidence to suggest that unlogged forests have a higher mean than logged forests.
    • #tk flashcard on what $\alpha$ really means, when to reject what, as well as type 1 and 2 errors, $\alpha$ and $\beta$
  • There is strong evidence that the mean number of species in unlogged plots is greater than the mean number of species in logged plots.
• Example:
  • A company sells educational materials.
  • They post studies that their directed reading activities improve reading comprehension.
  • 3rd grade class of $21$ students take part.
  • Control class of $23$ students.
  • Degree of Reading Power test
  • ${\mu }_1$: treatment group
  • ${\mu }_2$: control group
  • $H_0:{\mu }_1={\mu }_2$
  • $H_a:{\mu }_1>{\mu }_2$
  • You can only prove something by rejecting the null hypothesis. You can never prove something by accepting the null hypothesis.
  • Q-Q plot for treatment group is normal. Q-Q plot for control group is normal. So we can assume normality.
  • ${\overline{X}}_1=51.48$
  • ${\overline{X}}_2=41.52$
  • $n_1=21$
  • $n_2=23$
  • $S_1^2=11.01$
  • $S_2^2=17.15$
  • $t^{\ast }=\frac{{\overline{X}}_1-{\overline{X}}_1}{\sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2}}}=2.31$
  • $=\frac{51.48-41.52}{\sqrt{\frac{11.01}{21}+\frac{17.15}{23}}}=8.83828696316849$
    • #tk why the discrepancy?
  • df = $20$
  • $P(t_{(20)}>2.31)<0.05$ so we can reject $H_0$ and conclude that there is a significant difference between the two groups. We have evidence to suggest that the treatment group has a higher mean than the control group.
• Paired Data
  • Condition:
    • Data must be paired and Quantitative Data
  • Example:
    • Students have improved scores from test 1 to 2.
    • Student Alpha is not independent from himself. So we pair his test 1 and 2.
    • Student Alpha might be independent from Student Beta though.
    • $D_i=Y_i-X_i$
  • To compute the responses to the two treatments in a matched pair design. We need to find the difference between the responses to the two treatments for each respondent. Then we can perform a one sample t test on the differences.
• Example:
  • Tires.
  • Type a and b
  • One of each are randomly mounted on rear wheels of each car.
  • Amount of wear was noted:
  • $\begin{array}{cccccc}\text{Auto}& 1& 2& 3& 4& 5\\ A& 10.6& 9.8& 12.3& 9.7& 8.8\\ B& 10.2& 9.4& 11.8& 9.1& 8.3\\ \mathrm{\Delta }& 0.4& 0.4& 0.5& 0.6& 0.5\end{array}$
  • $\overline{\mathrm{\Delta }}=0.48$
  • $S_{\mathrm{\Delta }}=0.08$
  • $H_0:{\mu }_{\mathrm{\Delta }}=0$
  • $H_a:{\mu }_{\mathrm{\Delta }}\ne 0$
  • $t^{\ast }=\frac{\overline{\mathrm{\Delta }}-0}{\frac{S_{\mathrm{\Delta }}}{\sqrt{n}}}=\frac{0.48}{\frac{0.08}{\sqrt{5}}}=12.8$
  • p-value:
    • $2P(t_{(4)}>12.8)<0.05$
    • We can reject $H_0$ and conclude that there is a significant difference between the two types of tires. We have evidence to suggest that type A has more wear than type B.