STA258 Lecture 09

• Answer to bonus question:
  • #tk
  • Song: Purple Rain by Prince
• Confidence Intervals.
• Review
  • We have a large sample CI for $\mu$ mean, $p$ proportion, or differences ${\mu }_1-{\mu }_2$ or $p_1-p_2$.
• If we have $Y$ and $X$
  • $Y:n_1,{\mu }_1,{\sigma }_1^2$
  • $X=n_2,{\mu }_2,{\sigma }_2^2$
  • $\mathrm{\Delta }={\mu }_1-{\mu }_2$ unknown, so estimate.
  • $\hat{\mathrm{\Delta }}=\overline{Y}-\overline{X}$
    • Good estimate because it's unbiased.
  • $\mathrm{E}[\hat{\mathrm{\Delta }}]=\mathrm{E}[\overline{Y}-\overline{X}]=\mathrm{E}[\overline{Y}]-\mathrm{E}[\overline{X}]$
    • $={\mu }_1-{\mu }_2=\mathrm{\Delta }$ which is the original.
  • So this is an unbiased estimator
  • $\frac{\hat{\mathrm{\Delta }}-\mathrm{\Delta }}{{\sigma }_{\hat{\mathrm{\Delta }}}}\sim N(0,1)$ by Central Limit Theorem
  • $\mathrm{Var}(\hat{\mathrm{\Delta }})=\mathrm{Var}(\overline{Y})+\mathrm{Var}(\overline{X})$
    • $=\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}$
    • ${\sigma }_{\hat{\mathrm{\Delta }}}=\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}$
  • $Z={\displaystyle \frac{\hat{\mathrm{\Delta }}-\mathrm{\Delta }}{\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}}}\sim N(0,1)$
  • To construct the Confidence Intervals. You need to set it between two critical values, and say that the probability of that is $1-\alpha$
  • $CI=\hat{\mathrm{\Delta }}\pm Z_{\frac{\alpha }{2}}\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}$
  • If the variances are equal from the two populations.
    • ${\sigma }_1^2={\sigma }_2^2$
    • $CI=\hat{\mathrm{\Delta }}\pm Z_{\frac{\alpha }{2}}\sigma \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$
    • If we know $\sigma$, no issues.
    • If we don't know $\sigma$, we can use the plug-in principle and use an estimator.
      • #tk is this note accurate?
    • Example:
      • Two STA258 Courses
      • $n_1=10$
      • $n_2=100$
      • Each has it's own variance.
      • How do we find the popl. variance?
      • We can do the average of the two variances.
        • $\frac{1}{2}{S}_1^2+\frac{1}{2}{s}_2^2$
      • But how do account the sample sizes?
      • Do a weighted average.
      • $\frac{n_1-1}{n_1+n_2-2}{S}_1^2+\frac{n_2-1}{n_1+n_2-2}{S}_2^2$
      • Ex to show it adds up to $1$
        • $\frac{n_1-1}{n_1+n_2-2}\alpha +\frac{n_2-1}{n_1+n_2-2}\alpha =\alpha$
      • This is a convex combination.
      • $w_1{x}_1+w_2{x}_2$
      • $w_1+w_2=1$
      • Then $0<w_1{w}_2<1$
    • This is the pooled estimate.
      • $S_p^2=\frac{n_1-1}{n_1+n_2-2}{S}_1^2+\frac{n_2-1}{n_1+n_2-2}{S}_2^2$
    • Is the pooled estimate an unbiased estimator?
      • We still have ${\sigma }_1^2={\sigma }_2^2={\sigma }^2$
      • $\mathrm{E}[S_p^2]$
      • $\mathrm{E}[\frac{n_1-1}{n_1+n_2-2}{S}_1^2+\frac{n_2-1}{n_1+n_2-2}{S}_2^2]$
      • $\frac{n_1-1}{n_1+n_2-2}\mathrm{E}[S_1^2]+\frac{n_2-1}{n_1+n_2-2}\mathrm{E}[S_2^2]$
      • $\frac{n_1-1}{n_1+n_2-2}\mathrm{E}[{\cancel{S_1^2}}^{{\sigma }^2}]+\frac{n_2-1}{n_1+n_2-2}\mathrm{E}[{\cancel{S_2^2}}^{{\sigma }^2}]$
      • $=\sigma$
      • So it's unbiased
        • #tk he skipped too many steps
    • We know:
      • $\frac{(n_1-1)S^2}{{\sigma }^2}\sim {\chi }_{(n_1-1)}^2$
      • $\frac{(n_2-1)S^2}{{\sigma }^2}\sim {\chi }_{(n_2-1)}^2$
      • $S_1^2\perp S_2^2$
      • $\frac{(n_1-1)S^2}{{\sigma }^2}+\frac{(n_2-1)S^2}{{\sigma }^2}\sim {\chi }_{(n_1-1+n_2-1)}^2$
      • $\frac{(n_1-1)S^2}{{\sigma }^2}+\frac{(n_2-1)S^2}{{\sigma }^2}\sim {\chi }_{(n_1+n_2-2)}^2$
    • We also know:
      • Sample Mean and Sample Variance are two independent RVs
      • $\overline{Y}\perp \perp S_1^2$
      • $\overline{X}\perp \perp S_2^2$
      • $S_p^2$ is independent from $\frac{(\overline{Y}-\overline{X})-({\mu }_1-{\mu }_2)}{\sigma \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$
    • So $T={\displaystyle \frac{{\displaystyle \frac{(\overline{Y}-\overline{X})-({\mu }_1-{\mu }_2)}{\sigma \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}}}{{\displaystyle \frac{\sqrt{{\displaystyle \frac{(n_1+n_2-2)S_p^2}{{\sigma }^2}}}}{n_1+n_2-2}}}}\sim t_{(n_1+n_2-2)}$
      • We have a normal over a chi over df.
    • $CI=(\overline{Y}-\overline{X})+t_{(\frac{\alpha }{2};n_1+n_2-2)}{S}_p\sqrt{n_1^{-1}+n_2^{-1}}$
    • Example:
      • We have scores for two sets of students.
      • $\begin{array}{lll}\times & \text{Verbal}& \text{Math}\\ \text{Eng}& \overline{y}=446,s=42& \overline{y}=548,s=57\\ \text{Lit}& \overline{y}=534,s=45& \overline{y}=517,s=62\end{array}$
      • Construct a $95\mathrm{\%}CI$, for the difference in means of verbal scores for students in Eng and Lit.
      • Interpret the results.
      • $n_1=n_2=15$
      • Compute the pooled $S_p^2$
        • $S_p^2=\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{(n_1+n_2-2)}=\frac{(15-1)(42{)}^2+(15-1)(45{)}^2}{(15+15-2)}$
          • $=\frac{(15-1)(42{)}^2+(15-1)(45{)}^2}{(15+15-2)}=1894.5$
      • $1-\alpha =0.95$
        • $\alpha =0.05$
        • $\frac{0.05}{2}=0.025$
        • $t_{(0.025;28)}=2.048$
      • Now we just need to use the formula:
      • $CI=(\overline{Y}-\overline{X})+t_{(\frac{\alpha }{2};n_1+n_2-2)}{S}_p\sqrt{n_1^{-1}+n_2^{-1}}$
      • $CI=[-120.55,-55.45]$
      • Since we don't have $0$ in the interval. it means that in no way, do we have the chance of these two sets of students being equal.
      • So there is a difference.
      • The hypothesis that ${\mu }_1={\mu }_2$ is rejected at $95\mathrm{\%}$ level of confidence.
        • #tk Is this a case where we can jack up the sample numbers to achieve a better CI interval?
        • I'm thinking not really.
      • Since avg of lit students in verbal is higher, also based on CI we have negative numbers.
      • So ${\mu }_1-{\mu }_2<0$ so ${\mu }_1$ is lesser than ${\mu }_2$.
    • Example:
      • We have $n_1=4$
      • ${\overline{y}}_1=0.22$
      • $S_1^2=0.001$
      • $n_2=5$
      • ${\overline{y}}_2=0.17$
      • $S_2^2=0.002$
      • We want to construct a $95\mathrm{\%}$ confidence interval for difference in means.
      • ${\mu }_1-{\mu }_2$
      • Compute the pooled $S_p^2$
        • $S_p^2=\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{(n_1+n_2-2)}=\frac{(4-1)(0.001)+(5-1)(0.002)}{(4+5-2)}$
          • $=\frac{(4-1)(0.001)+(5-1)(0.002)}{(4+5-2)}=0.00157142857142857$
        • $S_p$
          • $=\sqrt{\frac{(4-1)(0.001)+(5-1)(0.002)}{(4+5-2)}}=0.0396412483586046$
      • $df=n_1+n_2-2=7$
      • $CI=({\overline{y}}_{1-{\overline{y}}_2})\pm t_{(\frac{\alpha }{2};df)}{S}_p\sqrt{n_1^{-1}+n_2^{-1}}$
        • $t_{(0.025;7)}=2.365$
      • $CI=(0.22-0.17)\pm (2.365)(0.0396412483586046)\sqrt{5^{-1}+4^{-1}}$
      • $(0.22-0.17)+(2.365)(0.0396412483586046)\sqrt{5^{-1}+4^{-1}}=0.112890453227361$
      • $(0.22-0.17)-(2.365)(0.0396412483586046)\sqrt{5^{-1}+4^{-1}}=-0.0128904532273608$
      • Is there significant evidence against the difference of the means?
        • $0\in CI⟹$ there is no significant evidence for difference in means.