 More specifically, SE = σ/ ...or in a few minutes... SE = s/  Example: consider a distribution with s = 13; what is our SE if n = 10? If n = 55? If n = 100?

 paret/explaining-the-central-limit-theorem-with- bunnies-and-dragons-v2 paret/explaining-the-central-limit-theorem-with- bunnies-and-dragons-v2  dist/

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 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 we noted 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.

 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.

 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

How much time do students spend on the internet? We collect a SRS of data from the 320 students and calculate the mean amount of time that each student spent on the internet yesterday. Let’s practice checking conditions too...

The stem plot below shows the force required to pull apart 20 pieces of randomly selected Douglas fir.

Discuss whether it would be appropriate to construct a t interval to estimate the population mean for: We want to estimate the average age at which U.S. presidents have died. So we obtain a list of all U.S. presidents who have died and their ages at death.

 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)

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.

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?

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)

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

Let’s use Minitab to run the t-test Input data into Minitab

 Reject H o. Assuming an α = 0.05, and based on a p-value of ≈ 0.02, there is evidence that supports the alternative hypothesis, that cola does lose sweetness over time.

 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)

p-value ≈ 0.02 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 which may lead us to an incorrect decision (type I; type II errors) No guarantee we have made the ‘correct’ decision Also... What would happen if our α level was 1%? Would our decision change? Why?

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:

 Is there evidence that the conductivity of this type of glass is greater than 1? Carry out an appropriate test.  (Assume the population is Normally distributed)

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 Conditions: Randomization, Normality/Big Sample, Independence

H o :μ = 1H a :μ > 1 t = 8.95 and df = 10  p-value = ≈ 0 Interpretation: 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.

H o :μ = 1H a :μ > 1 p-value = ≈ 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%?

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.

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.

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.

H 0 : μ ( unscented – scented) = 0H a : μ (unscented – scented) > 0 Conditions (we do not know the standard deviation of the population of differences, so we should perform a paired t test)  SRS The data come from a randomized, matched pairs experiment.  Normality/Big Sample: Problem stated we may assume a Normal distribution.  Independence We are willing to assume that the 21 subjects' differences in average maze completion times with the two types of masks are independent observations. Also we assume at least (10) (21) in our population.

H 0 : μ ( unscented – scented) = 0H a : μ (unscented – scented) > 0 Run the test in Minitab...

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 is larger than any reasonable α (say α = 0.05), data doesn’t support that floral scent improves test performance. (outside rejection zone)

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?

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?

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.

μ (right – left)

 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.

 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.

 Use Minitab to run the test

H o :μ (right – left) = 0H a :μ (right – left) < 0 Reject H o. With a p-value of (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?

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.

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

Remember, the response variable is the decrease in systolic (top number) blood pressure for a subject after 12 weeks, in millimeters of mercury. An increase in blood pressure appears as a negative response So +  decrease in blood pressure; -  increase in blood pressure.

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

 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

 Use Minitab; 2-sample t-procedure  Use summary statistics or raw data (either one)  Don’t ever pool (don’t ever assume equal variances)  t = 1.60 p = df = 15

 Fail to reject H o. With a p-value of , df = 15, and at any reasonable α level (5% or 1%), there is little evidence that calcium reduces blood pressure.  WHAT IF.... Our α = Would this change our decision?? Why?  Possible errors?

 Let’s go back into Minitab and look at some of our calculations...

 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  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 )

 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:

 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.

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.

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 = df = 14

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.

Let’s look at our 95% Confidence Interval and confirm our hypothesis decision from our confidence interval. Why/how does this CI confirm our findings?

... 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?

How badly does logging damage tropical rain forests? One study compared 2 SRS’s of 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:

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

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.

H o : μ U = μ L ORμ U – μ L = 0 H a : μ U ≠ μ L ORμ U – μ L ≠ 0 Use Minitab to run your 2-sample t procedure

Interpretation: Fail to reject null hypothesis. Assuming an α = 0.05, and t = 2.11, and p = (df = 14), there is sufficient 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?

Let’s look at our 95% confidence interval from this data. What additional information does the CI give us?

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.

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.

t = 1.91p-value = 0.038df = 14 Reject null hypothesis. At the 5% level, and a p- value of 0.038, there is good evidence that the positive subliminal message brought about greater improvement in math scores than the control.

Do piano lessons change (improve or worsen) the spatial-temporal reasoning of preschool children? A study designed to test this measured the spatial- temporal reasoning of a SRS of 34 preschool children before and after six months of piano lessons. (The study also included children who took computer lessons and a control group; but we are not concerned with those here.) The changes in the reasoning scores are on the following slide.

 Construct and interpret a 95% confidence interval for the mean difference in reasoning scores.  Assume all conditions have been met and checked  Positive means improvement; negative means worsen

 Construct and interpret a 95% confidence interval for the mean difference in reasoning scores.  The population of interest is all preschool children. We want to estimate the mean difference μ DIFF = μ before – after in reasoning scores that would be reported if all preschoolers in a population took piano lessons.

 Interpretation: We are 95% confident that the true, unknown, population parameter, μ, of the difference μ DIFF = μ before – after in reasoning scores that would be reported if all preschoolers in a population took piano lessons is between 2.55 and 4.68.

 Class activity: Let’s gather our data and calculate a 95% confidence interval based on our class data. You may assume that the population is Normally distributed; and that conditions have been met and checked.

 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.

 Will cover chapters 7, 8, & 9  List of review topics, randomly assigned to teams of 2 students  Not sure if we will have a 4 th exam... Need to see how are timing is...