1. Stat 512 – Day 8 Tests of Significance (Ch. 6)

2. Last Time Use random sampling to eliminate sampling errors Use caution to reduce nonsampling errors Use probability theory to estimate the size of random sampling errors  Normal distribution is often useful

3. Last Time – Central Limit Theorem Case 1: Variable of interest is quantitative  Eg.,date of pennies, length of words  Parameters: population mean, , population standard deviation,   Sampling distribution for sample means will be centered at  with standard deviation (assuming random sample)  Sampling distribution will follow a normal distribution if population is normal otherwise sampling distribution will be approximately normal if n is “large” (n > 30)

4. Last Time – Central Limit Theorem Case 2: Variable of interest is qualitative (binary)  Eg., orange or not orange candy, did heart transplant patient survive  Parameters: population proportion/probability,   Sampling distribution for sample proportions will be centered at  with standard deviation (assuming random sample)  Sampling distribution will approximately follow a normal distribution if if n is “large” (n  > 10; n(1-  )>10)

5. Population samples pop proportion Sampling distribution Is the sample proportion that I observed surprising?  Sample proportions 

6. Last Time – Central Limit Theorem Moral: Don’t have to keep running these simulations with means and proportions, know what the answer is going to be!

7. Therefore Knowledge of the sampling distribution, allows us to make judgments of whether an observation is surprising  Instead of running simulations, have the Central Limit Theorem to tell us about the behavior of many (hypothetical) samples

8. Technical Conditions Keep in mind that we can’t always apply the Central Limit Theorem  Means: random sample, n > 30 or normal population  Proportions: random sample, n  >10 and n(1-  )>10 Consider both sample size and value of  Recall Gettysburg Address

9. Practice Problem 2 Hospital under suspicion for high mortality rate from heart transplantations  Last ten cases has 80% mortality rate Researchers determined that 15% was a reasonable benchmark for transplantation. Parameter of interest:  = the overall mortality rate for heart transplants at this hospital Give the hospital the benefit of the doubt and assume  is equal to.15 See if that is consistent with the data at the hospital

10. Practice Problem 2 Observed: 19.7% of the 361 heart transplant patients between 1986 and 2000 died at this hospital Can this be explained “by chance”? Central Limit Theorem If  =.15, we would see a sample proportion at least as large as.197 in only.62% of random samples. Either really unlucky coincidence or  =.15 belief is wrong

11. More Formal Structure 1. Assumptions  Review data collection plan Simple random sample?  Quantitative or qualitative data?  Are the conditions for the Central Limit Theorem met? Check technical conditions

12. More Formal Structure 2. State competing conjectures about the parameter of interest   =.15 (equals national rate) Null hypothesis, H 0   >.15 (what actually suspect) Alternative hypothesis, H a

13. More Formal Structure 3. Measure the discrepancy between what observed and what hypothesized “test statistic”

14. More Formal Structure 4. How unusual is this discrepancy, assuming the null hypothesis is true  Probability >.197 = probability Z > 2.50 =.0062 “p-value” assuming  =.15  The smaller the p-value, the stronger the evidence against the null hypothesis Would get such an extreme sample proportion in only.6% of random samples if  =.15. Matches direction of alternative hypothesis

15. More Formal Structure 5. Draw a conclusion  Use same cut-off guidelines E.g., p-value <.05  statistically significant evidence against null hypothesis  “Reject the null hypothesis” If p-value is considered large, we “fail to reject the null hypothesis”

16. Test of Significance 0. Define parameter of interest 1. Check technical conditions 2. State null and alternative hypotheses about the population parameter of interest (in symbols and in words) 3. Calculate test statistic 4. Determine p-value Be able to interpret 5. State conclusion (about the null hypothesis) Sketch a picture of the sampling distribution! In English!

17. Example 1: Cohen v. Brown University Observational units = Brown university athletes; variable = gender Population = process of males and females becoming athletes at Brown Parameter of interest = probability of a Brown University athlete being female,  H 0 : .51 (proportional to student body) H a : .51 (smaller prob of an athlete being F)

18. Example 1: Cohen v. Brown Univ Give the university the benefit of doubt Does the CLT apply?  897(.51) = 457.5 > 10 and 897(.49)=439.5 > 10  Consider this a “random sample” from this process Test statistic: z = -7.79 P-value <.000000287 (from Table A) Reject H 0, very unlikely to have only 38% females in sample if randomly selected from population w/ 51% Strong evidence that  <.51

19. Example 2: Kissing the Right Way “Human Behavior: Adult Persistence of Head- Turning Asymmetry,”  Güntürkün, O.  Nature, 421: 771, 2003.

20. Example 2: Kissing the Right Way Let  represent the probability of a kissing couple turning to the right H 0 :  =.5 (equally likely left or right) H a :  >.5 (majority of couples turn right) CLT?.5(124) = 62 > 10, assuming it’s a representative sample Test statistic: z = 3.23 P-value =.000619 <.05 Reject H 0 Conclude a majority of couples to the right as long as believe this is a representative sample

21. Example 3: Body Temperatures Observational units: healthy adults Variable: temperature (quantitative) Parameter: let  represent the average body temperature of a healthy adult Sample size is large (n = 130 > 30) and we are assuming this is a representative sample H 0 :  = 98.6 o FH a …. Test statistic: - 6.22…

22. Behavior of t distribution, Table B

23. Minitab/applet (p. 5)

24. Example 4: Golden Rectangles Sample is skewed to the right  Evidence that population is skewed to the right as well Population is probably not normal so need large n. Sample size is 20, which isn’t all that large.  Proceed with caution

25. Example 4: Golden Rectangles Let  represent the ratio used by American Indians H 0 :  =.618 (same ratio on average) H a :  ≠.618 (ratio used by American Indians differs) t = 2.05 with df = 19 P-value =.054 Weak evidence against H 0 Not overwhelmingly convincing that the mean ratio used by American Indians differs from.618.

26. For Tuesday Read Ch. 5 (confidence intervals) Complete PP 9 Turn in second project report (see syllabus for details) Start HW 5 Online anonymous Blackboard survey by Thursday