1. One sample statistical tests, continued…

2. Recall statistics for: Single population mean (known  ) Hypothesis test: Confidence Interval

3. Examples of Sample Statistics: Single population mean (known  ) Single population mean (unknown  ) Single population proportion Difference in means (ttest) Difference in proportions (Z-test) Odds ratio/risk ratio Correlation coefficient Regression coefficient …

4. Sigma is unknown NOTE: if we are actually doing an experiment, we are unlikely to know the standard deviation of the population (  ) ahead of time (unlike with dice, there is no theoretical variance, only a population variance that we can never know exactly without measuring the entire population).  To estimate  :  Estimated standard error of the mean: basically dividing by n twice…

5. Standard error of the mean when true sigma is unknown

6. When  is unknown, use t rather than Z! A t-distribution is like a Z distribution, except has slightly fatter tails to reflect the uncertainty added by estimating . The bigger the sample size (i.e., the bigger the sample size used to estimate  ), then the closer t becomes to Z. If n>100, t approaches Z.

7. Computer simulation of the sampling distribution of the sample mean when standard deviation unknown: 1. Pick any probability distribution and specify a mean and standard deviation. 2. Tell the computer to randomly generate 1000 observations from that probability distributions E.g., the computer is more likely to spit out values with high probabilities 3. Calculate the standard deviation of each sample and calculate 1000 T- statistics: 4. Plot the T-statistics in histograms. 5. Repeat for different sample sizes (n’s).

8. n=2, underlying distribution is normal T-distribution with only 1 degree of freedom.

9. n=5, underlying distribution is normal T-distribution with 4 degrees of freedom.

10. n=10, underlying distribution is normal T-distribution with 9 degrees of freedom.

11. n=30, underlying distribution is normal T-distribution with 29 degrees of freedom.

12. n=100, underlying distribution is normal T-distribution with 99 degrees of freedom. Looks a lot like Z!!

13. Conclusions These simulations show that if we use the sample estimate of variance, we don’t quite get a normal distribution; instead we get a “student’s” t distribution. There is more area in the tails to reflect the added uncertainty from estimating standard deviation. If N is large enough, t and Z are similar.

14. Student’s t Distribution t 0 t (df = 5) t (df = 13) t-distributions are bell- shaped and symmetric, but have ‘fatter’ tails than the normal Standard Normal (t with df =  ) Note: t Z as n increases from “Statistics for Managers” Using Microsoft ® Excel 4 th Edition, Prentice-Hall 2004

15. Student’s t Table Upper Tail Area df.25.10.05 11.0003.0786.314 2 0.8171.886 2.920 30.7651.6382.353 t 0 2.920 The body of the table contains t values, not probabilities Let: n = 3 df = n - 1 = 2  =.10  /2 =.05  /2 =.05 from “Statistics for Managers” Using Microsoft ® Excel 4 th Edition, Prentice-Hall 2004

16. t distribution values With comparison to the Z value Confidence t t t Z Level (10 d.f.) (20 d.f.) (30 d.f.) ____.80 1.372 1.325 1.310 1.28.90 1.812 1.725 1.697 1.64.95 2.228 2.086 2.042 1.96.99 3.169 2.845 2.750 2.58 Note: t Z as n increases from “Statistics for Managers” Using Microsoft ® Excel 4 th Edition, Prentice-Hall 2004

17. The T probability density function What does t look like mathematically? (You may at least recognize some resemblance to the normal distribution function…) Where: v is the degrees of freedom (gamma) is the Gamma function is the constant Pi (3.14...)

18. The t-distribution in SAS Yikes! The t-distribution looks like a mess! Don’t want to integrate! Luckily, there are charts and SAS! MUST SPECIFY DEGREES OF FREEDOM! The t-function in SAS is: probt(t-statistic, df)

19. Ttests have a normality assumption… If the underlying data are not normally distributed, it takes longer for the CLT to kick in and the sample means do not immediately follow a t-distribution… This is the source of the “normality assumption” of the ttest…

20. n=2, underlying distribution is exponential (mean=1, SD=1) This doesn’t yet follow a t-distribution!

21. n=5, underlying distribution is exponential (mean=1, SD=1) This doesn’t yet follow a t-distribution!

22. n=10, underlying distribution is exponential (mean=1, SD=1) This doesn’t yet follow a t-distribution!

23. n=30, underlying distribution is exponential (mean=1, SD=1) Still not quite a t-distribution! Note the left skew.

24. n=100, underlying distribution is exponential (mean=1, SD=1) Now, pretty close to a T-distribution!

25. Conclusions If the underlying data are not normally distributed AND n is small**, the means do not follow a t-distribution (so using a ttest will result in erroneous inferences). Non-parametric tests should be used instead. **How small is too small? No hard and fast rule—depends on the true shape of the underlying distribution. Here N>30 (closer to 100) is needed.

26. Practice Problem: A manufacturer of light bulbs claims that its light bulbs have a mean life of 1520 hours with an unknown standard deviation. A random sample of 40 such bulbs is selected for testing. If the sample produces a mean value of 1505 hours and a sample standard deviation of 86, is there sufficient evidence to claim that the mean life is significantly less than the manufacturer claimed? Assume that light bulb lifetimes are roughly normally distributed.

27. Answer 1. What is your null hypothesis? Null hypothesis: mean life = 1520 hours Alternative hypothesis: mean life < 1520 hours 2. What is your null distribution? Since we have to estimate the standard deviation, we need to make inferences from a T-curve with 39 degrees of freedom. 3. Empirical evidence: 1 random sample of 40 has a mean of 1498.3 hours 5. Probably not sufficient evidence to reject the null.  We cannot sue the light bulb manufacturer for false advertising! Notice that using t-distribution to calculate the p-value didn’t change much! With n>30, might as well use Z table.

28. Practice problem You want to estimate the average ages of kids that ride a particular kid’s ride at Disneyland. You take a random sample of 8 kids exiting the ride, and find that their ages are: 2,3,4,5,6,6,7,7. Assume that ages are roughly normally distributed. a. Calculate the sample mean. b. Calculate the sample standard deviation. c. Calculate the standard error of the mean. d. Calculate the 99% confidence interval.

29. Answer (a,b) a. Calculate the sample mean. b. Calculate the sample standard deviation.

30. Answer (c) c. Calculate the standard error of the mean.

31. Answer (d) d. Calculate the 99% confidence interval. t 7,.005 =3.5

32. Example problem, class data: A two-tailed hypothesis test: A researcher claims that Stanford affiliates eat fewer than the recommended intake of 5 fruits and vegetables per week. We have data to address this claim: 22 people in the class provided data on their daily fruit and vegetable intake. Do we have evidence to dispute her claim?

33. Histogram fruit and veggie intake… Mean=4.0 servings Median=4.5 servings Mode=5.0 servings Std Dev=1.8 servings

34. Answer 1. Define your hypotheses (null, alternative) H 0 : P(average servings)=5.0 Ha: P(average servings)≠5.0 hours (two-sided) 2. Specify your null distribution We do not know the true standard deviation of homework times, so we must use a T-distribution to make inferences, rather than a Z- distribution.

35. Answer, continued 5. Reject or fail to reject (~accept) the null hypothesis Reject! Stanford affiliates eat significantly fewer than the recommended servings of fruits and veggies. T 21 critical value for p<.05, two tailed = 2.08 3. Do an experiment observed mean in our experiment = 4.0 servings 4. Calculate the p-value of what you observed p-value <.05;

36. 95% Confidence Interval H 0 : P(average servings)=5.0 The 95% CI excludes 5, so p-value <.05

37. Paired data (repeated measures) PatientBP Before (diastolic)BP After 110092 28984 38380 49893 510898 69590 What about these data? How do you analyze these?

38. Example problem: paired ttest PatientDiastolic BP BeforeD. BP AfterChange 110092-8 28984-5 38380-3 49893-5 510898-10 69590-5 Null Hypothesis: Average Change = 0

39. Example problem: paired ttest Change -8 -5 -3 -5 -10 -5 With 5 df, T>2.571 corresponds to p<.05 (two-sided test) Null Hypothesis: Average Change = 0

40. Example problem: paired ttest Change -8 -5 -3 -5 -10 -5 Note: does not include 0.

41. Summary: Single population mean (unknown  ) Hypothesis test: Confidence Interval

42. Summary: paired ttest Hypothesis test: Confidence Interval Where d=change over time or difference within a pair.

43. Examples of Sample Statistics: Single population mean (known  ) Single population mean (unknown  ) Single population proportion Difference in means (ttest) Difference in proportions (Z-test) Odds ratio/risk ratio Correlation coefficient Regression coefficient …

44. Recall: normal approximation to the binomial… Statistics for proportions are based on a normal distribution, because the binomial can be approximated as normal if np>5

45. Recall: stats for proportions For binomial: For proportion: P-hat stands for “sample proportion.” Differs by a factor of n.

46. Sampling distribution of a sample proportion Always a normal distribution! p=true population proportion. BUT… if you knew p you wouldn’t be doing the experiment!

47. Practice Problem A fellow researcher claims that at least 15% of smokers fail to eat any fruits and vegetables at least 3 days a week. You find this hard to believe and decide to check the validity of this statistic by taking a random (representative) sample of smokers. Do you have sufficient evidence to reject your colleague’s claim if you discover that 17 of the 200 smokers in your sample eat no fruits and vegetables at least 3 days a week?

48. Answer 1. What is your null hypothesis? Null hypothesis: p=proportion of smokers who skip fruits and veggies frequently =.15 Alternative hypothesis: p <.15 2. What is your null distribution? Var( ) =.15*.85/200 =.00064  SD( ) =.025 ~ N (.15,.025) 3. Empirical evidence: 1 random sample: = 17/200 =.085 4. Z = (.085-.15)/.025 = -2.6 p-value = P(Z<-2.6) =.0047 5. Sufficient evidence to reject the claim.

49. OR, use computer simulation… 1. Have SAS randomly pick 200 observations from a binomial distribution with p=.15 (the null). 2. Divide the resulting count by 200 to get the observed sample proportion. 3. Repeat this 1000 times (or some arbitrarily large number of times). 4. Plot the resulting distribution of sample proportions in a histogram:

50. How often did we get observed values of 0.085 or lower when true p=.15? Only 4/1000 times! Emprical p-value=.004

51. Practice Problem In Saturday’s newspaper, in a story about poll results from Ohio, the article said that 625 people in Ohio were sampled and claimed that the margin of error in the results was 4%. Can you explain where that 4% margin of error came from?

52. Answer

53. Paired data proportions test… Analogous to paired ttest… Also takes on a slightly different form known as McNemar’s test (we’ll see lots more on this next term…)

54. Paired data proportions test… 1000 subjects were treated with antidepressants for 6 months and with placebo for 6 months (order of tx was randomly assigned) Question: do suicide attempts (yes/no) differ depending on whether a subject is on antidepressants or on placebo?

55. Paired data proportions test… Data: 15 subjects attempted suicide in both conditions (non- informative) 10 subjects attempted suicide in the antidepressant condition but not the placebo condition 5 subjects attempted suicide in the placebo condition but not the antidepressant condition 970 did not attempt suicide in either condition (non- informative) Data boils down to 15 observations… In 10/15 cases (66.6%), antidepressant>placebo.

56. Paired proportions test… Single proportions test: Under the null hypothesis, antidepressants and placebo work equally well. So, H o : among discordant cases, p (antidepressant>placebo) = 0.5 Observed p =.666 Not enough evidence to reject the null!

57. Key one-sample Hypothesis Tests… Test for H o : μ = μ 0 (σ 2 unknown): Test for H o : p = p o :

58. Corresponding confidence intervals… For a mean (σ 2 unknown): For a proportion:

59. Symbol overload! n: Sample size Z: Z-statistic (standard normal) t df: T-statistic (t-distribution with df degrees of freedom) p: (“p-hat”): sample proportion X: (“X-bar”): sample mean s: Sample standard deviation p 0: Null hypothesis proportion  0: Null hypothesis mean