STA258 Lecture 08

• Review
  • Standard Error
  • Confidence Intervals
• Today we're going to see large sample Confidence Intervals.
• $Z$ Confidence Intervals and $T$ as well.
• #tk Large sample Confidence Intervals are on the test.
• Large Sample Confidence Intervals
  • $Z=\frac{\hat{\theta }-\theta }{{\sigma }_{\hat{\theta }}}\sim N(0,1)$
  • We know ${\sigma }_{\hat{\theta }}=\sqrt{\mathrm{Var}(\hat{\theta })}$
  • If $\hat{\theta }=\overline{X}$
  • Then $\mathrm{Var}(\overline{X})=\frac{{\sigma }^2}{n}=\frac{\sigma }{\sqrt{n}}$
  • So we have $\frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}$
  • Suppose we have $P(a\le Z\le b)=1-\alpha$
  • What should our $a,b$ be?
    • Suppose we have a normal.
    • Our area is $1-\alpha$
    • So our tails are $\frac{\alpha }{2}$
    • $b=Z_{\frac{\alpha }{2}}$
    • Since we have symmetry
    • $a=Z_{\frac{\alpha }{2}}$
  • $P(-Z_{\frac{\alpha }{2}}\le \frac{\hat{\theta }-\theta }{{\sigma }_{\hat{\theta }}}\le Z_{\frac{\alpha }{2}})=1-\alpha$
  • We need to isolate $\theta$
  • $P(-{\sigma }_{\hat{\theta }}{Z}_{\frac{\alpha }{2}}\le \hat{\theta }-\theta \le {\sigma }_{\hat{\theta }}{Z}_{\frac{\alpha }{2}})$
  • $P(\hat{\theta }-Z_{\frac{\alpha }{2}}{\sigma }_{\hat{\theta }}\le \theta \le {\sigma }_{\hat{\theta }}{Z}_{\frac{\alpha }{2}}+\hat{\theta })$
  • So our Confidence Intervals
  • $[\hat{\theta }-Z_{\frac{\alpha }{2}}{\sigma }_{\hat{\theta }},\hat{\theta }+Z_{\frac{\alpha }{2}}{\sigma }_{\hat{\theta }}]=\hat{\theta }\pm Z_{\frac{\alpha }{2}}{\sigma }_{\hat{\theta }}$
  • So our CI for $\theta$ is the point estimator $\pm$ our cutoff times standard error.
• Example:
  • We have the mean annual household income of a set if $119155$
  • Assume this is based on a sample of $80$ households.
  • $\sigma =30000$
  • We have $Y_1,\dots ,Y_n\stackrel{iid}{\sim }N(\mu ,{\sigma }^2)$
  • $\hat{\theta }=\overline{X}$
  • Then $\mathrm{SE}[\overline{X}]=\frac{\sigma }{\sqrt{n}}$
  • Then CI for $\mu$:
    • $\overline{X}\pm Z_{\frac{\alpha }{2}}\cdot \frac{\sigma }{\sqrt{n}}$
  • Compute margin of error.
    • $=Z_{\frac{\alpha }{2}}\cdot \frac{\sigma }{\sqrt{n}}$
    • We want a $90$ CI
    • $1-\alpha =0.9$
    • $\alpha =0.1$
    • $\frac{\alpha }{2}=0.05$
    • Look at $5\mathrm{\%}$ on the table
    • We get $1.64\cdot \frac{\sigma }{\sqrt{n}}$
    • $1.64\cdot \frac{30000}{\sqrt{80}}=5500.72722464948$
    • CI is
      • $\overline{X}\pm Z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}$
      • $119155\pm 5500.7272$
      • $119155+5500.7272=124655.7272$
      • $119155-5500.7272=113654.2728$
  • If we increase the CI. What will happen?
    • Likely we have a larger interval.
    • $1-\alpha =0.95$
    • $\alpha =0.05$
    • $\frac{0.05}{2}=0.025$
    • Look at point on table $=1.96$
    • CI
      • $\overline{X}\pm Z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}$
      • $119155\pm Z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}$
      • $119155\pm (1.96)\frac{30000}{\sqrt{80}}$
      • $119155\pm (1.96)\frac{30000}{\sqrt{80}}$
      • $(1.96)\frac{30000}{\sqrt{80}}=6574.03985384938$
      • See we have a larger interval. Which increases our confidence interval.
      • #tk Can we optimize the tradeoff. Get the most confidence with least interval.
• Analyzing pharma
  • We're looking for some $n$ to achieve a CI with $0.95$ confidence
  • $1-\alpha =0.95$
  • $\alpha =0.05$
  • $\frac{0.05}{2}=0.025$
  • $m=0.005$ as our margin of error
  • We get $Z_{\frac{\alpha }{2}}=Z_{0.025}=1.96$
  • $\sigma =0.0068$
    • How don't we know the mean but we know the $\sigma$
    • From historical data we can estimate $\sigma$
  • Margin of error
  • $Z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}$
  • $m=Z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}$
  • $n={\frac{Z_{\frac{\alpha }{2}}\sigma }{m}}^2$
  • $n={\left(\frac{(1.96)(0.0068)}{(0.005)}\right)}^2=7.10542336$
  • So $n=8$
  • Is it realistic to assume that the population follows a normal?
  • Usually we need that $n=30$ for Central Limit Theorem.
  • There's a danger that our sampling dist is not normal then CLT doesn't hold.
• Example:
  • We have $m=2$
  • and $22.5$ as the value for $\sigma$
  • Find the Sample Size Recommended to Estimate Mean.
    • $1-\alpha =0.9$
    • $\alpha =0.1$
    • $\frac{\alpha }{2}=0.05$
    • $Z_{\frac{\alpha }{2}}=1.65$
    • $n={\left(\frac{Z_{\frac{\alpha }{2}}\sigma }{m}\right)}^2$
    • $n={\left(\frac{(1.65)(22.5)}{(2)}\right)}^2=344.56640625$
    • $n=345$
    • $1-\alpha =0.95$
    • $\alpha =0.05$
    • $\frac{\alpha }{2}=0.025$
    • $Z_{\frac{\alpha }{2}}=1.96$
    • $n={\left(\frac{(1.96)(22.5)}{(2)}\right)}^2=486.2025$
    • $n=487$
  • #tk that's it for test 1.
• Confidence Intervals based on the $t$ dist
  • What if $\sigma$ is not known.
  • Estimate it by $S$
  • $T=\frac{\overline{X}-\mu }{\frac{S}{\sqrt{n}}}\sim t_{(n-1)}$
  • $P(\left|T\right|<m)=1-\alpha$
  • Find two end points of the interval.
  • Our area is $1-\alpha$ our points are $\pm \frac{\alpha }{2}$
  • $P(-t_{\left(\frac{\alpha }{2}\right)}\le T\le t_{\left(\frac{\alpha }{2}\right)})=1-\alpha$
  • Always isolate the unknown parameter.
  • $P(-t_{\left(\frac{\alpha }{2}\right)}\le \frac{\overline{X}-\mu }{\frac{S}{\sqrt{n}}}\le t_{\left(\frac{\alpha }{2}\right)})=1-\alpha$
  • $P(\overline{X}-t_{\left(\frac{\alpha }{2}\right)}\frac{S}{\sqrt{n}}\le \mu \le \overline{X}+t_{\left(\frac{\alpha }{2}\right)}\frac{S}{\sqrt{n}})$
  • $CI=\overline{X}\pm t_{\left(\frac{\alpha }{2}\right)}\frac{S}{\sqrt{n}}$
  • The lower and upper limit are Random Variables.
  • Prior to observation, the CI is random.
  • After an observation, then the CI is deterministic.
  • CI is not the most accurate interpretation that our parameter is in there.
  • If we have 90 percent confidence, 90 out of 100 trials we will have our Estimator correctly Estimate the parameter.
• Normal Population Assumption
  • For small samples of $n<15$. The data should follow a normal dist. If you see outliers or skewness, be cautious.
  • For moderate samples of $15\le n\le 40$. The data should not show strong skewness or outliers. Make a histogram, boxplot or Q-Q plot to check.
  • For large samples of $n>40$. The $t$ procedure is fairly robust to non-normality. Unless the data are extremely skewed or contain outliers. Make a histogram, boxplot or Q-Q plot to check.
    • The reason, as $n\uparrow$, the $t$ dist approaches the normal dist.
• Example:
  • Ancient air.
  • We can examine gas inside ancient amber.
  • Will give us sample of time when amber was formed.
  • We have these observations:
    • $n=9$
    • $\begin{array}{ccccccccc}63.4& 65& 64.4& 63.3& 54.8& 54.5& 50.8& 49.2& 51.0\end{array}$
  • Find a $90\mathrm{\%}$ CI for the mean nitrogen level.
  • Our present atmosphere is $78.% $ nitrogen
  • $\overline{X}=59.58$
  • $S^2=\frac{\sum (X_i-\overline{X}{)}^2}{n-1}=6.2552$
  • $1-\alpha =0.9$
  • $\alpha =0.1$
  • $\frac{\alpha }{2}=0.05$
  • $t_{\left(\frac{\alpha }{2}\right)}=t_{0.05;8}=1.860$
  • $P(\overline{X}-t_{\left(\frac{\alpha }{2}\right)}\frac{S}{\sqrt{n}}\le \mu \le \overline{X}+t_{\left(\frac{\alpha }{2}\right)}\frac{S}{\sqrt{n}})=1-\alpha$
  • $P(59.58-(1.860)\frac{6.2552}{\sqrt{9}}\le \mu \le 59.58+(1.860)\frac{6.2552}{\sqrt{9}})=0.9$
  • $CI=59.58\pm (1.860)\frac{6.2552}{\sqrt{9}}$
  • $=59.58\pm 3.872$
  • $=(55.708,63.452)$
• Example:
  • A film-processing company want to know how many pictures were stored on computers.
  • Random sample of 10 digital camera owners.
  • Estimate with $95\mathrm{\%}$ confidence the mean number of pictures stored.
  • Data:
    • $\begin{array}{cccccccccc}25& 6& 22& 26& 31& 18& 13& 20& 13& 2\end{array}$
  • $n=10$
  • $\overline{X}=17.7$
  • $S=9.08$
  • $1-\alpha =0.95$
  • $\alpha =0.05$
  • $\frac{\alpha }{2}=0.025$
  • $t_{\left(\frac{\alpha }{2}\right)}=t_{0.025;9}=6.49$
  • Margin of error:
  • $=t_{(0.025;9)}\frac{S}{\sqrt{n}}$
  • $2.262\left(\frac{9.08}{\sqrt{10}}\right)=6.49498943710919$
  • CI:
    • $17.7\pm 6.49498943710919$
  • Assumption of normality is critical here since $n$ is small.
  • #tk learn q-q-plots Normal Q-Q Plot.
• Example:
  • Suppose we have a random sample:
  • $Y_1,\dots ,Y_n\stackrel{iid}{\sim }\mathrm{Bernoulli}(p)$
  • How can we estimate $p$?
  • Flip a coin $100$ times, how can you estimate $p$?
  • It's the number of successes over number of trials.
  • $S=\sum _{i=1}^n{X}_i\sim \mathrm{Bin}(n,p)$
  • $S$ is total number of successes.
  • $\hat{p}=\frac{S}{n}$
  • $\hat{p}$ is our proportion of successes. An estimator for $p$.
  • We need to standardize.
  • $\mathrm{E}[\hat{p}]=\mathrm{E}\left[\frac{S}{n}\right]=\frac{1}{n}\mathrm{E}[S]=\frac{1}{n}np=p$
  • $\mathrm{Var}(\hat{p})=\mathrm{Var}\left(\frac{S}{n}\right)=\frac{1}{n^2}\mathrm{Var}(S)=\frac{1}{n^2}np(1-p)=\frac{p(1-p)}{n}$
  • $\mathrm{SE}[\hat{p}]=\sqrt{\mathrm{Var}(\hat{p})}=\sqrt{\frac{p(1-p)}{n}}$
  • Based on Central Limit Theorem we have:
    • $\hat{P}\approx N(p,\frac{p(1-p)}{n})⟹\frac{\hat{P}-P}{\sqrt{\frac{p(1-p)}{n}}}$
    • CI for $p$:
  • $\hat{p}\pm Z_{\frac{\alpha }{2}}\sqrt{\frac{p(1-p)}{n}}$
• Example:
  • In a poll of $800$ adults
  • $45\mathrm{\%}$ indicated that movies are getting better.
  • $43\mathrm{\%}$ indicated that movies are getting worse.
  • Find a $98\mathrm{\%}$ CI for the proportion of all adults who think movies are getting better.
  • $\hat{p}=0.45$
  • $1-a=0.98$
  • $\alpha =0.02$
  • $\frac{\alpha }{2}=0.01$
  • ${\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$
  • ${\sigma }_{\hat{p}}=\sqrt{\frac{0.45(1-0.45)}{800}}=0.0175890590993379$
  • CI for $p$:
  • $0.45\pm Z_{\frac{\alpha }{2}}\sqrt{\frac{p(1-p)}{n}}\approx Z_{\frac{\alpha }{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$
    • We need to use the plug-in estimate since we don't know $p$.
    • But we want a result in terms of known quantities.
  • $0.45\pm 2.33\cdot 0.0175890590993379$
  • $0.45\pm 0.0409757533251168$
  • $=(0.409024246674883,0.490975753325117)$
  • Since we have less than $50\mathrm{\%}$ for our CI.
  • Then even in the best case for our $98\mathrm{\%}$ confidence interval, less than half of adults think movies are getting better.
• Example:
  • Utility of mobile devices raises questions on intrusion of work into personal life.
  • $\frac{158}{473}$ of employees took work with them on vacation.
  • a:
    • What is the point estimate of the population proportion of all employees who take work with them on vacation?
    • $\hat{p}=\frac{158}{473}=0.334$
  • b:
    • At $0.9$ confidence, what is the margin of error for the population proportion of all employees who take work with them on vacation?
    • $=Z_{\frac{\alpha }{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=0.035$
    • $1-\alpha =0.9$
    • $\alpha =0.1$
    • $\frac{\alpha }{2}=0.05$
    • $Z_{\frac{\alpha }{2}}=1.65$
    • $\sqrt{\frac{(0.334)(1-0.334)}{473}}=0.0216860161877937$
    • $1.65\cdot 0.0216860161877937=0.035$
    • CI:
      • $0.334\pm 0.035=(0.299,0.369)$
• Example:
  • Aisha Shariff and Yvette Ye are candidates for mayor.
  • You are planning a small survey to determine the percent of voters to vote for shariff.
  • $p$ is population proportion of voters who will vote for Shariff.
  • You want to be $95\mathrm{\%}$ confident that your estimate of $p$ is within $0.03$ of the true value.
  • How large a sample should you take?
  • $m=0.03$
  • $1-\alpha =0.95$
  • $\alpha =0.05$
  • $\frac{\alpha }{2}=0.025$
  • $m=Z_{\frac{\alpha }{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$
  • What is $p$?
    • When we don't know that value of $\hat{p}$
    • We can use $0.5$
    • If you sketch the plot of $p(1-p)$
    • At $50\mathrm{\%}$ it's maximized.
  • ${\left(\frac{1.96}{0.03}\right)}^2(0.5)(1-0.5)=1067.11111111111$
  • $n=1068$
• Sample size for an interval estimate of a population proportion:
  • $n={\frac{Z_{\ast }}{E}}^2{p}^{\ast }(1-p^{\ast })$
  • Planning values $p^{\ast }$ can be chosen by:
    • Sample proportion from a previous sample of the same or similar units
    • Use a planning value of $0.5$ to maximize the required sample size when no reasonable estimate of $p$ is available.
• $10\mathrm{\%}$ condition
  • If less than $10\mathrm{\%}$ of the population is sampled, the sample observations can be assumed to be independent.
  • We also consider $n\ge 30$ in this fashion, $n\hat{p}\ge 10$ and $n(1-\hat{p})\ge 10$ instead of the $30$ rules for CLT.
• Confidence Intervals for Variance
  • You know $\frac{(n-1)S^2}{{\sigma }^2}\sim {\chi }_{(n-1)}^2$ as long as we have a normal population from which we draw our sample.
  • We want $P(?\le \frac{(n-1)S^2}{{\sigma }^2}\le ?)=1-\alpha$
  • We need to find two points on the chi-square dist.
    • Top value is:
      • ${\chi }_{\left(\frac{\sigma }{2}\right)}^2$
    • Bottom value is:
      • ${\chi }_{(1-\frac{\alpha }{2})}^2$
  • $P(\frac{1}{{\chi }_{(1-\frac{\alpha }{2})}^2}\ge \frac{{\sigma }^2}{(n-1)S^2}\ge \frac{1}{{\chi }_{\left(\frac{\alpha }{2}\right)}^2})=1-\alpha$
  • $P(\frac{(n-1)S^2}{{\chi }_{(1-\frac{\alpha }{2})}^2}\ge {\sigma }^2\ge \frac{(n-1)S^2}{{\chi }_{\left(\frac{\alpha }{2}\right)}^2})$
  • CI:
    • $[\frac{(n-1)S^2}{{\chi }_{\left(\frac{\alpha }{2}\right)}^2},\frac{(n-1)S^2}{{\chi }_{(1-\frac{\alpha }{2})}^2}]$
  • Estimate ${\sigma }^2$ with CC $0.9$
  • $n=3$ means $\mathrm{d}\mathrm{f}=2$
  • $\overline{Y}=$
  • $S^2=10.57$
  • $[3.53,205.25]$
    • Crazy wide interval.
    • Maybe because Skewness as well as low samples.