STA258 Lecture 10

• Estimator
• Confidence Intervals
• Example:
  • ${\sigma }_1^2={\sigma }_2^2={\sigma }^2$
  • $S_p^2=\frac{(n_1-1)S_2+(n_2-1)S_2^2}{n_1+n_2-2}$
  • $CI=({\overline{Y}}_1-{\overline{Y}}_2)\pm t_{(\frac{\alpha }{2},?)}{S}_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$
  • Mutual funds purchases.
  • Can investors do better by buying mutual funds directly rather than through brokers?
  • We have a random sample of net annual returns through brokers and directly.
  • Find a $95\mathrm{\%}$ CI for ${\mu }_1-{\mu }_2$
  • ${\mu }_1$ is direct
  • ${\mu }_2$ is broker
  • Assume ${\sigma }_1^2={\sigma }_2^2$ and annual returns are normal.
  • We get that $n_1=50=n_2$
  • ${\overline{x}}_1=6.63$
  • ${\overline{x}}_2=3.72$
  • $s_1^2=37.49$
  • $s_2^2=43.34$
  • Pooled Variance
    • $S_p^2=\frac{(49)(37.49)+(49)(43.34)}{50+50-2}$
  • $\nu =50+50-2=98$
  • $t_{(\frac{\alpha }{2};98)}=1.984$
  • $CI=(6.63-3.72)\pm 1.984\sqrt{40.42({50}^{-1}+{50}^{-1})}$
  • $CI=2.91\pm 2.52$
  • ${\mu }_1={\mu }_d=$ direct
  • ${\mu }_2={\mu }_b=$ broker
  • Since $0$ is not included, this means the difference is not negligible.
  • Because our difference is positive, it means that ${\mu }_d>{\mu }_b$. So direct is better.
  • If ${\mu }_d={\mu }_b⟹{\mu }_d-{\mu }_b=0$
  • Since ${\mu }_d-{\mu }_b>0⟹{\mu }_d>{\mu }_b$ so that's why it's better.
  • How sure can we be sure that direct is better than broker?
  • $\frac{95}{100}$ times we will have a better return directly. But how strongly can we be confident in this? #tk something about hypothesis testing.
  • Since our interval is almost including $0$. We need to worry about it:
    • If we have $(0.00000001,5.43)$, it's almost $0$
    • If we have $(3,5.43)$, it's pretty far away from $0$
    • $(0.39,5.43)$ isn't as close to $0$ as the first example, but it's almost there.
  • Suppose ${\sigma }_1\ne {\sigma }_2$
    • $CI=({\overline{Y}}_1-{\overline{Y}}_2)\pm t_{(\frac{\alpha }{2};\nu )}\sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2}}$
    • We don't pool it because we need to account for different variance.
    • $\nu =\frac{{(\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2})}^2}{\left(\frac{1}{n_1-1}\right){\left(\frac{S_1^2}{n_1}\right)}^2+\left(\frac{1}{n_2-1}\right){\left(\frac{S_2^2}{n_2}\right)}^2}$
    • This is the Welch method. Easier to do with R.
    • Otherwise, you can do:
      • $min(n_1-1,n_2-1)$
      • This will widen the T Distribution, and will give a conservative answer for your CI.
• Example:
  • A company sells edu materials.
  • A consultant has a $3^{\text{rd}}$ grade glass of $n=21$ for $8$ weeks.
    • This is a treatment group
  • We need a control group, so we have $m=23$ $3^{\text{rd}}$ grade students without the materials.
  • At the end, both groups are given a DRP test to measure.
  • $\begin{array}{ccccccccc}& \text{Treatment}& & & & & \text{Control}& & \\ 24& 61& 59& 46& |& 42& 44& 46& 37\\ \dots \end{array}$
  • Find a $95\mathrm{\%}$ CI for the mean improvement.
  • $CI=({\overline{Y}}_T-{\overline{Y}}_C)\pm t^{\star }\sqrt{\frac{S_T^2}{n_T}+\frac{S_C^2}{n_C}}$
  • Conservative approach $⟹\nu =20$
  • $1-\alpha =0.95⟹\frac{\alpha }{2}=0.025$
  • $t_{(0.025;20)}=2.086$
  • $CI=(51.48-41.52)\pm 2.086(4.31)$
    • $=(0.87,18.95)$
  • The lower bound of mean improvement is barely anything. The higher bound is really effective.
  • But otherwise, $0\notin CI$, so it is effective.
  • Since $0.87$ is pretty close to $0$, how strongly can we know if it would include it or not?
  • We also need to know what the test is, does $0.87$ mean $0.87$ marks, or percent? Could it be rounded down?
• Not caring about domain knowledge is very dangerous:
  • You could end up with a case like the memes about "What they did in the study showing red meat causes cancer" and it's just a video of someone launching cannonballs in a catapult.
  • Outcomes are valid, but interpretations matters in application of it.
• Assumption in the above $3^{\text{rd}}$ grader thing is important too. That the population is normal. We'd need to do a Normal Q-Q Plot. To see if it's normal.
  • We don't have evidence for normality by the chart, but we don't have evidence against it.
• Example:
  • Conservationsist example
  • $12$ unlogged forests
  • $9,8$
  • $\begin{array}{ccccccccccccc}\text{Unlogged:}& 22& 18& 22& 20& 15& 21& 13& 13& 19& 13& 19& 15\end{array}$
  • $\begin{array}{cccccccccc}\text{Logged:}& 17& 4& 18& 13& 18& 15& 15& 10& 12\end{array}$
  • Create a $99\mathrm{\%}$ CI on difference in mean
  • $CI=({\overline{Y}}_U-{\overline{Y}}_L)\pm t_{(\frac{\alpha }{2};\nu )}\sqrt{\frac{S_1^2}{n}+\frac{S_2^2}{m}}$
  • ${\overline{Y}}_1-{\overline{Y}}_2=3.83$
  • $S_1=3.52$
  • $S_2=4.5$
  • $n=n_1=12$
  • $m=n_2=9$
  • $\sqrt{\frac{S_1^2}{n}+\frac{S_2^2}{m}}=1.81$
  • $\nu =min(12-1,9-1)=8$
  • $CI=3.83\pm t_{(\frac{\alpha }{2};\nu )}(1.81)$
  • $t^{\star }=3.35$
  • $CI=3.83\pm (3.35)(1.81)$
  • $=3.83-(3.35)(1.81)=-2.2335$
  • $=3.83+(3.35)(1.81)=9.8935$
  • Does logging have an impact?
    • $0\in CI$ so it doesn't
• Paired Samples
  • Compare two tires
  • $\begin{array}{cccccc}\text{Auto}& 1& 2& 3& 4& 5\\ A& 10.6& 9.8& 12.3& 9.7& 8.8\\ B& 10.2& 9.4& 11.8& 9.1& 8.3\end{array}$
  • Do the data provide evidence in mean wear for tire types $A$ and $B$
  • Mounted on the same car, but when we get the difference, it's independent.
  • First find the difference.
  • $\begin{array}{cccccc}\text{Auto}& 1& 2& 3& 4& 5\\ A& 10.6& 9.8& 12.3& 9.7& 8.8\\ B& 10.2& 9.4& 11.8& 9.1& 8.3\\ & 0.4& 0.4& 0.5& 0.6& 0.5& \overline{X}=0.48\end{array}$
  • $\overline{d}=0.48$
  • $S_d=0.0837$
  • $CI=0.48\pm (2.776)\frac{0.0837}{\sqrt{5}}$
  • $0.48\pm 0.1039$
  • The interval is positive and doesn't contain $0$ so there is a difference between these two tires.
• Comparing two proportions
  • $CI$
    • We use $Z$ critical values
  • $n=1178$
  • Notes whether mother has epidural, do the babies still nurse after $6$ months
  • $p_1=$ those who had an epidural and are nursing at $6$ months
  • $p_2=$ those who didn't who are nursing at $6$ months
  • $\begin{array}{cc}& \text{Nursing at 6 months}\\ \text{Epidural}& Y& N\\ Y& 206& 190\\ N& 498& 284\end{array}$
  • First find the Point Estimators for two proportions.
  • ${\hat{P}}_1=$
    • $=\frac{206}{206+190}=\frac{103}{198}=0.52020202020202$
  • ${\hat{P}}_2=$
    • $=\frac{498}{498+284}=\frac{249}{391}=0.636828644501279$
  • Without doing a confidence interval, it seems we already have a pretty big difference.
  • $\mathrm{SE}[X]=\sqrt{\frac{(0.52)(1-0.52)}{396}+\frac{(0.64)(1-0.64)}{782}}=0.0304126977809148$
  • $CI=(0.52-0.64)\pm (1.96)(0.0304126977809148)$
  • Since $0$ is not in the interval, and we have a negative difference. It means that epidural means the baby will nurse likelier than another.
  • There's an abundance of women who haven't had an epidural. Why are there different sample sizes, make one sample size match the other.
  • The impact is $n_2$
  • The numbers aren't balanced. They're getting weighed by the sample sizes.
  • Extreme sample:
    • Control group is 1000 people
    • What would happen to the Confidence Intervals?
    • It would be smaller, but we don't have balance in the dataset regarding those who had an epidural.
  • We are $95\mathrm{\%}$ confident that percent of nursing is mothers at 6 months for those who had an epidural are between $0\mathrm{\%}$ and $17\mathrm{\%}$ less than those who didn't.
• Example:
  • Smoking
  • We have $244$ smokers who want to stop to receive nicotine patches
  • $245$ receive both a path and the anti-depressant
  • After a year $40$ in the nicotine group had abstained
  • $87$ in the patch plus drug group have.
  • Give a $99\mathrm{\%}$ CI for the difference, $t-c$ in the proportion of smokers who quit.
  • Find the Point Estimator for proportions
  • ${\hat{P}}_1=$
    • $=\frac{40}{244}=0.163934426229508$
  • ${\hat{P}}_2=$
    • $=\frac{87}{245}=\frac{87}{245}=0.355102040816327$
  • Already we see that there are a lot more who quit in the treatment group.
  • $1-\alpha =0.99$
  • $\alpha =0.01$
  • $\frac{\alpha }{2}=\frac{0.01}{2}=0.005$
  • $\mathrm{SE}[X]=\sqrt{\frac{(0.16)(1-0.16)}{244}+\frac{(0.36)(1-0.36)}{245}}=0.0386164192461763$
  • $Z_{\frac{\alpha }{2}}=2.575829$
  • $(0.36-0.16)+(2.575829)(0.0386164192461763)=0.299469292570459$
  • $(0.36-0.16)-(2.575829)(0.0386164192461763)=0.100530707429541$
  • There is a benefit. Because $0\notin C$, if $0\in C$ then the difference is $0$ and there aren't benefits.
  • You don't remove someone from another group if someone withdrew from the first.
  • If you remove someone, then you need to start the experiment again from day 0.
    • You don't extend it to day 60 for the initial
    • You don't have the effects through 10 days with the new guy if you add one.
• Comparing two population variances
  • $\frac{\frac{S_1^2}{{\sigma }_1^2}}{\frac{S_2^2}{{\sigma }_2^2}}\sim F_{(n_1-1,n_2-2)}$
  • $P(F_{(n_1-1,n_2-1;1-\frac{\alpha }{2})}\le \frac{{\sigma }_1^2{S}_1^2}{{\sigma }_2^2{S}_2^2}\le F_{(n_1-1,n_2-2;\alpha )})=1-\alpha$
  • Compare the ratio of $\frac{{\sigma }_1^2}{{\sigma }_2^2}$
  • $P(\frac{S_1^2}{S_2^2}\frac{1}{F_{(n_1-1,n_2-1;\frac{\alpha }{2})}}\le \frac{{\sigma }_1^2}{{\sigma }_2^2}\le \frac{S_1^2}{S_2^2}\frac{1}{F_{(n_1-1,n_2-1;1-\frac{\alpha }{2})}})$
  • We know that $F_{(n_1,n_2)}=\frac{1}{F_{(n_2,n_1)}}$
  • $P(\frac{S_1^2}{S_2^2}\frac{1}{F_{(n_2-1,n_1-1;1-\frac{\alpha }{2})}}\le \frac{{\sigma }_1^2}{{\sigma }_2^2}\le \frac{S_1^2}{S_2^2}{F}_{(n_2-1,n_1-1;\frac{\alpha }{2})})=1-\alpha$
  • Example:
    • Managerial success
    • Index based on length of time in the org, and level within the term.
    • Compare group 1, high volume of interactions with people.
    • Group 2, rarely interact with people outside their work unit.
    • ${\mu }_{G_1}=65.3$
    • ${\mu }_{G_2}=49.5$
    • Is there a significant difference?
      • Already, yes.
      • However, we don't know about the index, so we don't know if $15$ points is or isn't really significant.
    • $(\frac{{6.61}^2}{{9.33}^2(3.09)},\frac{{6.61}^2}{{9.33}^2(3.29)})$
    • $(0.16,1.68)$
    • There is not a significant variance based on our $CI$. Because $1\in CI$
    • No evidence to show that the population variances of managerial success index is different.
    • $\frac{{\sigma }_1^2}{{\sigma }_2^2}=1⟺{\sigma }_1^2={\sigma }_2^2$