1. Stat 301 – Day 35 Bootstrapping (4.5) Three handouts…

2. Last time – Inference for Odds Ratios With Case-control studies, would rather look at the odds ratio than the difference in the conditional proportions. Want to estimate , the population odds ratio But the sampling distribution of the sample odds ratio, is not usually well modeled by the normal distribution However, the sampling distribution of the log odds ratio is, with mean zero and a nice formula for the standard error No full night’s sleep in past week At least one full night’s sleep in the past week Sample sizes Case drivers61474535 Control drivers44544588 Total10510181123 About 1% of random samples have OR this large when  =1 What if  =1? OR=1.59

3. Last Time – CI for Odds Ratio Normal based confidence interval  Estimate + z* SE(estimate) So can estimate the population log odds ratio using  sample log odds + z  (1/a+1/b+1/c+1/d) Exponentiate to get endpoints for   Not centered at sample odds  Keep in mind can subtract one and talk about percentage change (1.13, 2.24)  13% to 124% higher odds

4. Note Again, as earlier with categorical data, even though we are estimating the standard deviation, the standard normal curve still provides a reasonable model

5. PP 5.1.3 (p. 426) Sample odds: 2.587 ln(2.587)+ 1.645  (1/65+464+1/30+1/554).950 + 1.645(.230) = (.572, 1.328) (e.572, e 1.328 ) = (1.77, 3.77) I am 90% confident that the odds of having an accident for New Zealand drivers who had less than 5 hours of sleep is between 1.77 and 3.77 times higher than the odds of having a crash for New Zealand drivers who had more than 5 hours of sleep.

6. What we have done so far… Analyzing one sample compared to a claim about the population parameter  Categorical: One-sample z-procedures if sample size is large or binomial  Quantitative: One-sample t-procedures if sample size is large or population is normal

7. What we have done so far… Comparing two proportions (independent samples or randomized experiment)  Categorical: Two-sample z-procedures if sample sizes large or Fisher’s Exact Test (experiment)  Quantitative: Two-sample t-procedures if sample sizes large or normal populations  See Examples 5.1 and 5.2 With quantitative data, “small sample” alternative?

8. Bootstrapping Relatively new approach that allows us to estimate aspects of the sampling distribution of the statistic in cases where the Central Limit Theorem does not apply  Small samples  Statistics other than the mean Previously we considered the randomization distribution  One random sample? Two independent samples?  Confidence interval

9. Investigation 4.5.1 (p. 365) If I take a random sample of 10 words from the population, what do I know about the sampling distribution of the sample mean?

10. Investigation 4.5.1 A bootstrap sample resamples the data from the existing sample, drawing n observations, but with replacement GettysburgSample.mtw  Random sample of 10 words from population  Answer through part (h)

11. Bootstrap Distribution Sampling from existing sample (with replacement, that is, each observed value repeated infinitely many times, demo)demo  Centers around sample statistic  Estimates the standard deviation of the sampling distribution  How does statistic vary around the parameter! If sampling distribution is symmetric, CI could then be statistic + t SE(statistic)  Can do fancier stuff if not symmetric

12. Investigation 4.5.3 (p. 372) Data are truncated so could use median or trimmed mean…

13. Investigation 5.5.1 (p. 454) (d)

14. For Thursday Day 36 handout Quiz 29: Submit question on material, example exam question in Blackboard Review sheet posted  Interest in Monday evening review session?