4.  More specifically, SE = σ/ ...or in a few minutes... SE = s/  Example: consider a distribution with s = 100; what is our SE if n = 1? If n = 4? If n = 25?

5.  http://blog.minitab.com/blog/michelle- paret/explaining-the-central-limit-theorem-with- bunnies-and-dragons-v2 http://blog.minitab.com/blog/michelle- paret/explaining-the-central-limit-theorem-with- bunnies-and-dragons-v2  http://onlinestatbook.com/stat_sim/sampling_ dist/

8. What would cause the head brewer of the famous Guinness brewery in Dublin, Ireland, not only to use statistics but also to invent new statistical methods? The search for better beer, of course. William S. Gosset joined Guinness as a brewer in 1899. He soon became involved in experiments and in statistics to understand the data from those experiments. What are the best varieties of barley and hops for brewing? How should they be grown, dried, and stored? The results of the field experiments, as you can guess, varied. Gosset faced the problem in using the z test to introduce the reasoning of statistical tests: he didn't know the population standard deviation σ. He observed that just replacing σ by s in the z statistic and calling the result roughly Normal wasn't accurate enough.

9.  After much work, Gosset developed what we now call the t distributions. His new t test identified the best barley variety, and Guinness promptly bought up all the available seed.  Guinness allowed Gosset to publish his discoveries, but not under his own name. He used the name “Student,” so Gosset's t test is sometimes called “Student's t” in his honor.

12.  t-distribution  for means  unknown σ  t  # of standard deviations away from center  farther away from center, the more unlikely the event (example, a female’s height is 5 standard deviations above mean)  less area in that tail of density curve

13.  Confidence Intervals  Primarily 2-sided; middle/center portion of density curve  1-sample or 2-sample  If 2-sample, then dependent (paired data) or independent (2 populations)  Hypothesis Testing  1-sided, 2-sided  1-sample or 2-sample  If 2-sample, then dependent (paired data) or independent (2 populations)

14.  Random sample or random assignment  Normal (or ≈ Normal) distribution.  If population is Normal, then sample is automatically Normal. ✔  If n ≥ 30, then Central Limit Theorem says we have an ≈ Normal distribution. ✔  If n ≤ 30, then need to show that sample is ≈ symmetric with no outliers; then we can proceed in using the t- distribution. ✔  Independence  Population must be at least 10 times sample size.

15.  The mean weight of all 20-year-old women is 128 pounds; and these weights are considered to be Normally distributed. A random sample of 40 vegetarian women who are 20 years old have a sample mean of 122 pounds with a standard deviation of 15 pounds. The women’s measurements were independent of each other.  Some people suspect that vegetarian women’s weights are different than non-vegetarian women. Based on the statistics above, run an appropriate hypothesis test to determine if this is true. Use a significance level of 5%.

16.  n = 40; = 122; s = 15 Ho:μ = 128Ha:μ ≠ 128 where μ is the mean weight of all vegetarian women  Conditions: SRS, Normal, Independent  Input into Stat Crunch; your results should be (≈): t = - 2.529P-value = 0.0156  Ask yourself: Is this unusual? Is this rate?  Reject null hypothesis. With an alpha level of 5% and a p-value of about 1%, there is sufficient evidence to suggest that the weights for all 20-year-old vegetarian women are significantly different than 128 pounds (or non-vegetarian women).

17.  μ = 128; n = 40; = 122; s = 15 Ho:μ = 128Ha:μ ≠ 128  Reject null hypothesis. With an alpha level of 5% and a p-value of about 1%, there is sufficient evidence to suggest that the weights for all 20-year-old vegetarian women are significantly different than 128 pounds.  Let’s calculate a t-interval and see what we get... Does our confidence interval confirm our findings?

18. Stats, t-stats, one sample, with summary, confidence interval  We are 95% confident that the interval from 117.2 to 126.8 captures the true mean weight for all 20- year-old vegetarian women.  Why can we create a confidence interval to confirm our findings in our hypothesis test? Do we need to do this every time? Can we do this with every hypothesis test?

19. Diet colas use artificial sweeteners to avoid sugar. These sweeteners gradually lose their sweetness over time. Manufacturers therefore test new colas for loss of sweetness before marketing them. Trained tasters are randomly assigned to sip the cola along with drinks of standard sweetness and score the cola on a “sweetness scale” of 1 to 10. The cola is then stored for a month at high temperature to imitate the effect of four months' storage at room temperature. Each taster scores the cola again after storage. Our data are the differences (score before storage minus score after storage) in the tasters' scores. The bigger these differences, the bigger the loss of sweetness.

20. Most are positive. That is, most tasters found a loss of sweetness. But the losses are small, and two tasters (the negative scores) thought the cola gained sweetness. Assume that the population is Normally distributed. Are these data good evidence that cola loses sweetness in storage? Or is this variation just due to sampling variability/just by coincidence?

21. Are these data good evidence that the cola loses sweetness over time? Hypotheses: H o : μ before - after = 0 H a : μ before - after > 0 (mean sweetness loss for tasters is +; cola lost sweetness)

22. t-test (because we don’t know σ; we can calculate/we have s, but not σ; and we are dealing with means) SRS:Random assignment Normal/Big Sample: Since population is Normally distributed, our sample is automatically Normal Independence: We must assume the population is at least (10) (10); safe assumption; and must assume one taste tester doesn’t influence any other

23. Let’s use Stat Crunch to run the t-test Input data into Stat Crunch Results (≈): t = 2.6967P-value = 0.0123 Ask yourself: Is this unusual? Is it unlikely? Reject H o. Assuming an α = 0.05, and based on a p- value of ≈ 0.01, there is evidence that supports the alternative hypothesis, that cola does seem to lose sweetness over time.

24.  If there is no α referenced, safe to assume an α = 0.05 (this is the default if none is referenced in problem)  Standard α levels are usually 0.05 or 0.01 (sometimes even 0.10)

25. At α = 0.05, reject H o ; don’t believe μ = 0; but rather do believe μ > 0 (cola lost sweetness) Remember, if we do everything absolutely correct, we can still, sometimes, get an extreme sample (just by chance) which may lead us to an incorrect decision (type I; type II errors) No guarantee we have made the ‘correct’ decision; for the majority of the time, we will never know if we made an error or not. Also... What would happen if our α level was 1%? Would our decision change? Why? Can we create a confidence interval to confirm our findings in our hypothesis test?

26. How well materials conduct heat matters when designing houses. Conductivity is measured in terms of watts of heat power transmitted per square meter of surface per degree Celsius of temperature difference on the two sides of the material. In these units, glass has conductivity about 1. Here is a SRS of 11 measurements of the heat conductivity of a particular type of glass:

27.  Is there evidence that the conductivity of this type of glass is greater than 1? Carry out an appropriate test. We may assume that the population is Normally distributed.  Note: Just looking at the data, all values are greater than 1... But is this just because we happened to get a sample that tested high? Would this happen if we took another sample? That’s why we conduct hypothesis tests.

28. H o :μ = 1H a :μ > 1 where μ = mean heat conductivity transmitted per square meter of surface per degree Celsius difference on the two sides of the glass Appropriate procedure: t-test (why?) Conditions: Random, Normal/Big Sample, Independence

29. H o :μ = 1H a :μ > 1 t = 8.95 and df = 10  p-value = 0.00000216 ≈ 0 Ask yourself: Is this unlikely? It is unusual? Reject H o. Because p-value is less than any reasonable α (say α = 0.05), we can conclude that the mean heat conductivity for this type of glass is greater than 1.

30. H o :μ = 1H a :μ > 1 p-value = 0.00000216 ≈ 0  Reject H o  we conclude that the mean heat conductivity for this type of glass is > 1 How about if α was changed to 10%? 1%?

31. Some claim that listening to Mozart improves students' performance on tests. Perhaps pleasant scents have a similar effect. To test this idea, 21 subjects worked a paper-and-pencil maze while wearing a mask. The mask was either unscented or carried a floral scent. The response variable is their average time on three trials. Each subject worked the maze with both masks, in a random order. The randomization is important because subjects tend to improve their times as they work a maze repeatedly. The following gives the subjects' average times with both masks.

32. The 21 differences form a single sample. They appear in the “Difference” columns. The first subject, for example, was 7.37 seconds slower wearing the scented mask, so the difference is negative. Positive differences show that the subject did better/faster when wearing the scented mask.

33. H 0 : μ ( unscented – scented) = 0H a : μ (unscented – scented) > 0 Where μ is the mean difference in the population from which the subjects were drawn. The null hypothesis says that no improvement occurs, and H a says that unscented times are longer than scented times on average. Assume a Normal distribution.

34. H 0 : μ ( unscented – scented) = 0H a : μ (unscented – scented) > 0 Perform a t test SRS, Normal, Independence Input values from the ‘difference’ column into Stat Crunch & perform the t-test Based on the p-value, is this unusual? Is it unlikely?

35. H 0 : μ ( unscented – scented) = 0H a : μ (unscented – scented) > 0 Interpretation: Fail to reject H o (don’t believe H a ). Since P-value of 0.3652 is larger than any reasonable α (say α = 0.05), data doesn’t support that floral scent improve all test performances. (outside rejection zone)

36. H 0 : μ ( unscented – scented) = 0H a : μ (unscented – scented) > 0 Possible errors in context: Reject H o when H o is really true. We would determine that floral scent improves test performance when it really doesn’t. Consequences? Fail to reject H o, when H a is really true. We would determine that floral scent does not improve test performance when it really does. Consequences?

37. H 0 : μ ( unscented – scented) = 0H a : μ (unscented – scented) > 0 What if alpha level was something other than 5%? Would our decision change? Why or why not?

38. The design of controls and instruments affects how easily people can use them. A student project investigated this effect by asking a SRS of 25 right-handed students to turn a knob (with their right hands) that moved an indicator. There were two identical instruments, one with a right-hand thread (the knob turns clockwise) and the other with a left-hand thread (the knob turns counterclockwise). The following table gives the times in seconds each subject took to move the indicator a fixed distance. The project designers hoped to show that right-handed people find right- hand threads easier to use. Carry out a significance test at the 5% significance level to investigate this claim.

39. μ (right – left)

40.  Right threat sample mean = 104.12  Right threat sample standard deviation = 15.80  Left threat sample mean = 117.44  Left threat sample standard deviation = 27.26

41. H o :μ (right – left) = 0 H a :μ (right – left) < 0 We want to assess if right-handed people find right-hand threads easier (faster) to use.

42.  1-sample t-procedure  SRS – Stated in problem.  Normal/Big Sample – n = 25, so CLT; so distribution/sample is ≈ Normal  Independence – We must assume the population of students is at least (10) (25). This is reasonable to assume. Also one subject does not influence the other.

43.  Use Stat Crunch to run the test  Right threat sample mean = 104.12  Right threat sample standard deviation = 15.80  Left threat sample mean = 117.44  Left threat sample standard deviation = 27.26  Is our result unlikely? Is it unusual?

44. H o :μ (right – left) = 0H a :μ (right – left) < 0 Reject H o. With a p-value of 0.0039 (well below any reasonable α of, say, 0.05), data does supports H a, right-ies find right-handed threads easier & faster Possible errors? What if we changed our alpha level?

46.  A large pet store buys the identical species of adult tropical fish from two different suppliers — Buy-Rite Pets and Fish Friends. Several of the managers at the pet store suspect that the lengths of the fish from Fish Friends are consistently greater than the lengths of the fish from Buy-Rite Pets.  Random samples of 8 adult fish of the species from Buy-Rite Pets and 10 adult fish of the same species from Fish Friends were selected and the lengths of the fish, in inches, were recorded, as shown in the following table.  It is known that the lengths of both fish populations is considered Normal.

47. Do the data provide convincing evidence that the mean length of the adult fish of the species from Fish Friends is greater than the mean length of the adult fish of the same species from Buy-Rite Pets?

48. Ho: μ BRPets = μ FF Ha: μ BRPets < μ FF OR Ho: μ FF = μ BRPets Ha: μ FF > μ BRPets Where μ BRPets is the true unknown population parameter (mean length in inches) for all fish of this type from Buy- Rite Pets; and μ FF is the true unknown population parameter (mean length in inches) for all fish of this type from Fish Friends. FYI: Order doesn’t matter; just get your inequality sign correct and be consistent.

49. Ho: μ BRPets = μ FF Ha: μ BRPets < μ FF OR Ho: μ FF = μ BRPets Ha: μ FF > μ BRPets Random, Normal, Independent: Let’s discuss. Run the procedure in Stat Crunch. Stat, T-Stat, Two Sample, With Summary

50. Ho: μ BRPets = μ FF Ha: μ BRPets < μ FF OR Ho: μ FF = μ BRPets Ha: μ FF > μ BRPets

51. Ho: μ BRPets = μ FF Ha: μ BRPets < μ FF  Fail to reject Ho. With a p-value of about 0.40 and an alpha level of 5%, we do not have sufficient evidence to conclude that the mean length of all fish like these from Buy-Rite Pets is less than the mean length of all fish like these from Fish Friends.

53. H o : μ Ambulance = μ Self or H o : μ Ambulance - μ Self = 0 H a : μ Ambulance ≠ μ Self or H o : μ Ambulance - μ Self ≠ 0 Where μ Ambulance is the true unknown population parameter (mean wait time in minutes) for all ambulance patients and μ Self is the true unknown population parameter (mean wait time in minutes) for all self–transported patients.

54. H o : μ Ambulance = μ Self or H o : μ Ambulance - μ Self = 0 H a : μ Ambulance ≠ μ Self or H o : μ Ambulance - μ Self ≠ 0 Conditions: Random, Normal, Independence (check each conditions) Stat Crunch: Stat, T-Stat, Two Sample, With Summary

55. H o : μ Ambulance = μ Self or H o : μ Ambulance - μ Self = 0 H a : μ Ambulance ≠ μ Self or H o : μ Ambulance - μ Self ≠ 0

56. H o : μ Ambulance = μ Self or H o : μ Ambulance - μ Self = 0 H a : μ Ambulance ≠ μ Self or H o : μ Ambulance - μ Self ≠ 0 Is this rare? Is it unusual? Unexpected? Reject Ho. With a p-value of less than 1% and an alpha level of 5%, we have sufficient evidence to conclude that the wait times for all patients like these transported via ambulance is not the same as the wait times for all patients like these who were self-transported.

57. H o : μ Ambulance - μ Self = 0 H o : μ Ambulance - μ Self ≠ 0 Reject Ho. With a p-value of less than 1% and an alpha level of 5%, we have sufficient evidence to conclude that the wait times for all patients like these transported via ambulance is not the same as the wait times for all patients like these who were self- transported. Let’s confirm our findings with a confidence interval. Go to Stat Crunch, stat, t-stat, two sample, with summary & interpret what this means

58. Does increasing the amount of calcium in our diet reduce blood pressure? Examination of a large sample of people revealed a relationship between calcium intake and blood pressure. The relationship was strongest for black men. Such observational studies do not establish causation. Researchers therefore designed a randomized comparative experiment to determine if calcium in the diet reduces blood pressure. The subjects in part of the experiment were 21 healthy black men. A randomly assigned group of 10 of the men received a calcium supplement for 12 weeks. The control group of 11 men received a placebo pill that looked identical. The experiment was double-blind. The response variable is the decrease in systolic (top number) blood pressure for a subject after 12 weeks, in millimeters of mercury. An increase appears as a negative response (because the researchers set it up as ‘before’ minus ‘after’. Assume the population from which these samples were randomly taken is Normal.

59.  Here is the data from group 1, n = 10 (calcium group; before - after):  Here is data from group 2, n = 11 (placebo group; before - after):  Remember, an increase in blood pressure appears as a negative response.

60. An increase in blood pressure appears as a negative response so... +  decrease in blood pressure -  increase in blood pressure. Group 1 received calcium; group 2 received placebo. H o :μ 1 = μ 2 ORμ 1 – μ 2 = 0 H a :μ 1 > μ 2 ORμ 1 – μ 2 > 0

61.  2-sample t-procedure  SRS – Subjects were randomly assigned (good)  Normality/Big Sample – Generally we must look at n 1 & n 2 separately; stated both were from a Normal population  Independence – Blood pressure for one subject/group should not influence the blood pressure for other subjects/group; also assume each population at least 10 times each sample size

62.  Use Stat Crunch; 2-sample t-procedure  Don’t ever pool (don’t ever assume equal variances)  t = 1.60 P-Value = 0.0644 df = 15

63.  Fail to reject H o. With a P-value of 0.0644, df = 15, and at any reasonable α level (5% or 1%), there is little evidence to show that calcium reduces blood pressure.  WHAT IF.... Our α = 0.10. Would this change our decision?? Why?  Possible errors? Could we calculate a confidence interval to confirm our findings?

64.  t-distributions/procedures (1-sample, matched pair, 2-sample have the quality of being considered “robust”  “Robust”... think of the energizer bunny... it can take a lot of abuse (i.e., not all conditions have to be met exactly) and still be accurate, precise, produce good numbers; just need be careful about outliers and extreme skewness  2-sample t-distributions are especially robust when the 2 samples have similar shapes and n for both samples are equal (n 1 = n 2 )

65.  Observational studies suggest that moderate use of alcohol reduces heart attacks, and that red wine may have special benefits. One reason may be that red wine contains polyphenols, substances that do good things to cholesterol in the blood and so may reduce the risk of heart attacks.  In an experiment, healthy men were assigned at random to drink half a bottle of either red or white wine each day for two weeks. The level of polyphenols in their blood was measured before and after the two- week period. Here are the percent changes in level for the subjects in both groups:

66.  Is there good evidence that red wine drinkers’ mean polyphenol levels were different from white wine drinkers’ mean polyphenol levels? Assume both populations are approximately Normal.  We want to test: H o :μ R = μ W orμ R – μ W = 0 H a :μ R ≠ μ W orμ R – μ W ≠ 0 where μ R & μ W are the mean percent change in polyphenols for men who drink red and white wine, respectfully.

67. H o :μ R = μ W orμ R – μ W = 0 H a :μ R ≠ μ W orμ R – μ W ≠ 0 Conditions: SRS – random assignment stated in the problem Normality – Both distributions assumed Normal (what if they hadn’t told us that?) Independence - Independence – We must assume that EACH population (red and white wine subjects) are at least (10) times the sample size; and that one group does not influence the other.

68. H o :μ R = μ W orμ R – μ W = 0 H a :μ R ≠ μ W orμ R – μ W ≠ 0 Calculations: 2-sample t-test (do not assume variances =) t = 3.81 p = 0.0019 df = 14

69. Interpretation: Reject the null hypothesis. With t = 3.81, p ≈ 0, & df = 14, we have strong evidence at any reasonable α level that red wine drinkers’ polyphenol levels are different from white wine drinkers’ polyphenol levels.

70. Let’s look at our 95% (why 95%?) confidence interval and confirm our hypothesis decision from our confidence interval. Why can we do this? We are 95% confident that the interval from 2.32% to 8.21% capture the true mean percentage difference between all red-wine drinkers and all white wine drinkers’ levels of polyphenol. Or, in other words, on average, all red-wine drinkers have from 2.32% to 8.21% higher polyphenol levels than all white wine drinkers.

71. ... and think about ‘what if’ we changed our alpha level?... and what types of errors could we make... even if we did everything perfectly correct?

72. How badly does logging damage tropical rain forests? One study compared 2 forest plots in Borneo that had never been logged with similar plots nearby that had been logged 8 years earlier. Plots can be considered randomly assigned. Here are the data on the number of tree species in 12 unlogged plots and 9 logged plots:

73. Does logging significantly change the mean number of species in a plot after 8 years? Give appropriate statistical evidence to support your conclusion. Assume both populations are Normally distributed. We want to test H o : μ U = μ L ORμ U – μ L = 0 H a : μ U ≠ μ L ORμ U – μ L ≠ 0 where μ U & μ L are the mean number of species in unlogged and logged plots, respectfully

74. H o : μ U = μ L ORμ U – μ L = 0 H a : μ U ≠ μ L ORμ U – μ L ≠ 0 Conditions: SRS; and both unlogged and logged plots were randomly assigned Normal/Big Sample: Stated in problem Independence – we must assume each population (unlogged and logged) is at least (10) times each sample size; and that one area does not effect the other.

75. H o : μ U = μ L ORμ U – μ L = 0 H a : μ U ≠ μ L ORμ U – μ L ≠ 0 Use Stat Crunch to run your 2-sample t procedure t = 2.114P-value = 0.0519

76. Interpretation: Fail to reject null hypothesis. Assuming an α = 0.05, and t = 2.11, and p = 0.0519 (df = 14), there is evidence to determine that logging does not significantly changed the number of tree species in a plot of land. What do you think about this decision?

77. Let’s look at our 95% confidence interval from this data. What additional information does the confidence interval give us? We are 95% confident that the interval from - 0.036 to 7.703 captures the true population difference of all tree species between all unlogged and logged plots of land in Borneo after 8 years.

78. A “subliminal” message is below our threshold of awareness but may nonetheless influence us. Can subliminal messages help students learn math? A group of students who had failed the mathematics part of the City University of New York Skills Assessment Test agreed to participate in a study to find out. All received a daily subliminal message, flashed on a screen too rapidly to be consciously read. The treatment group of 10 students (assigned at random) was exposed to “Each day I am getting better in math.” The control group of 8 students was exposed to a neutral message, “People are walking on the street.” All students participated in a summer program designed to raise their math skills, and all took the assessment test again at the end of the program.

79. Is there good evidence at the 5% level that the treatment brought about a greater improvement in math scores than the neutral message? Assume all conditions have been checked and met.

80. sample mean improvement/change for treatment group = 11.4 sample standard deviation treatment group = 3.17 sample size n = 10 for treatment group sample mean improvement/change for control group =8.25 sample standard deviation for control group = 3.69 sample size n = 8 for control group

81. Randomly assigned pairs of two... Turn into me in 15 minutes. Null & Alternative Hypotheses Check Conditions Do Calculations Interpretation

82.  According to the National Institute of Health, the average resting heart rate for humans 10 years old or older is approximately 80 beats per minute.  Gather our data; assume population is Normally distributed  Perform an hypothesis test (state null, alternative, define parameters, check conditions, etc.; entire process); you choose your alpha level.  What type of error could we have possibly committed and what would that mean? Describe the error in context.  Calculate and interpret a confidence interval (that would make sense with the hypothesis test you just did).

83.  Do you think our temperatures are higher or lower than the mean healthy body temperature of adults in US?  Class activity: Let’s gather our data and perform a hypothesis test.

84.  Chapter 9 Concepts/HW quiz....  Exam 3 will cover chapters 7, 8, & 9...