Confidence Intervals

Population Mean

• Confidence interval is a formula that tells us how to use the sample data to calculate an interval that estimates the target parameter
• we always have to add confidence level: $95\%$, $99\%$, etc

Large Samples

• similar to creating the $z$ score
  • we can expect the sample mean $\bar{x}$ is good enough because of Central Limit Theorem $z=\frac{\bar{x}-\mu }{{\sigma }_{\bar{x}}}$ $z\sim N(0,1)$
• region is defined by confidence level as a symmetric range
  • tails share $100\%-\text{confidence level}=1-\alpha$
    • i.e. 95% confidence → tails share 5%
      • each tail has 2.5% $\alpha /2$
    • $-z=\text{2.5 percentile}$
    • $+z=\text{97.5 percentile}$
    • look at any $z$ table to get the results

Alpha Table

$\alpha$	$z_{\alpha /2}$
20%	1.282
15%	1.440
10%	1.645
5%	1.960
1%	2.576

Small Samples

• for small samples the Central Limit Theorem is not holding anymore
  • we need another assumption
• we can use $t$-statistic distributions
  • $t$-statistic distributions have thiccer tails
    • i.e. more extreme events are more likely
  • degree of freedom $df$ defines how exact the $t$-static will match the $z$ value
    • normally $df=(n-1)$ degree of freedom is almost the sample size
    • for large $n$ the $t$ and $z$ value will be ever more similar
      • therefore we don’t need $t$-statistic with larger $n$ sample sizes
• we assume that the population of the sample is normally distributed
  • therefore we can use $s$ for $\sigma$ → normally not allowed $t=\frac{\bar{x}-\mu }{s/\sqrt{n}}$

Large Sample Confidence Intervals

Theory

• when asking a binary question (e.g. Is coffee overpriced?) the result is a binomial
• remember $np(1-p)$ average number of successes from Probability
• $\hat{p}=\frac{\#1}{n}$ … p-hat is number of success over sample size
  • $\hat{p}$ is unbiased, i.e. the expected value = the probability
  • all estimators in QM2 are unbiased
• expressable as $\hat{p}=\frac{1}{n}\ast {\sum }_{i=1}^n\text{Bernoulli}(p)$ where Bournoulli(p) is either 1 or 0
  • summing up all yes (1) and no (0) values and divide by sample size $n$
• after Central Limit Theorem this can be considered the same as sample mean as long as $n$ is “large” enough
  • $\hat{p}=\bar{x}$
  • large is true when
    • $n\ast \hat{p}\ge 15$ and $n\ast (1-\hat{p})\ge 15$
• therefore $\hat{p}\approx N(\text{mean}(\hat{p},\text{sd}(\hat{p}))$
  • therefore $\text{mean}(\hat{p})=p$
  • and $\text{sd}(\hat{p})=\frac{p(1-p)}{n}$
  • results in $\hat{p}\approx N(p,\sqrt{\frac{p(1-p)}{n}})$
• now going for the z-value
  • $z=\frac{\hat{p}-p}{{\sigma }_{\hat{p}}}=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$
  • $z\sim N(0,1)$ … standard normal
• finally for the confidence interval
  • $\hat{p}\pm z_{\alpha /2}\ast {\sigma }_p\approx \hat{p}\pm z_{\alpha /2}\ast \sqrt{\hat{p}(1-\hat{p})/n}$
  • $\hat{p}-z_{\alpha /2}\ast {\sigma }_p\le p\le \hat{p}+z_{\alpha /2}\ast {\sigma }_p$
  • $P(\hat{p})=1-\alpha$ … probability of $\mu$ being in the confidence interval

Example Coffee

• givens:
  • $n=500$
  • $\hat{p}=300/500=0.6$
• asked for: 95% Confidence Interval
• solution:
  • $z_{\alpha /2}(97.5\%)=1.96$
  • ${\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.6\ast 0.4}{500}}=0.022$
  • $0.6-1.96\ast 0.022\le p\le 0.6+1.96\ast 0022$

More Theory - Error

• $S=z_{\alpha /2}\ast \frac{\sigma }{\sqrt{n}}$ is just the half width → careful! sometimes the full width is given/asked for
  • $S$ is the distance from the center to the Confidence Interval edge
• therefore $n=\frac{z_{\alpha /2}^2\ast {\sigma }^2}{S^2}$
  • always round up … $n$ is the minimum, therefore lower would be outside of Confidence Interval