CHAPTER 9INFERRING POPULATION MEANS

SYMBOLS... SAMPLE STATISTICS & POPULATION PARAMETERS

SAMPLING DISTRIBUTIONS... UNBIASED ESTIMATORS...

AS N INCREASES, SD (OR SE) DECREASES...

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?

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CENTRAL LIMIT THEOREM... http://blog.minitab.com/blog/michelle-p

aret/explaining-the-central-limit-theorem-with-bunnies-and-dragons-v2

http://onlinestatbook.com/stat_sim/sampling_dist/

≈ NORMAL DISTRIBUTIONS & CENTRAL LIMIT THEOREM

POPULATION DISTRIBUTION & SAMPLING DISTRIBUTIONS...

GUINNESS BEER...

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.

GUINNESS BEER... 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.

ONE-SAMPLE T-DISTRIBUTIONt-distribution, with SRS, size n, ≈ Normal, then

t =

with degrees of freedom = n – 1

remember, unknown σ is the reason for t-distribution

t-distribution has fatter tails (due to unknown σ); more “wiggle room”

‘T’ IS LIKE ‘Z’... WELL, KIND OF... FOR MEANS

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

INFERENCE FOR MEANS (T PROCEDURES)... DIFFERENT FLAVORS

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)

T DISTRIBUTIONS... STILL NEED CONDITIONS MET

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.

LET’S TRY A ONE-SAMPLE T-TEST ... The mean weight of all 20-year-old women is 128

pounds; and the weights of all 20-year-old women are Normally distributed. A random sample of 40 vegetarian women who are 20 years old showed a sample mean of 122 pounds with a standard deviation of 15 pounds. The women’s measurements were independent of each other.

Determine if the weight for all 20-year-old vegetarian women is different than 128 pounds, using a significance level of 5%.

ONE-SAMPLE T-TEST ... μ = 128; n = 40; = 122; s = 15

Ho: μ = 128 Ha: μ ≠ 128where μ is the mean weight of all vegetarian women

Conditions: SRS, Normal, Independent

Input into Minitab; your results should be (≈):t = - 2.529 P-value = 0.0156

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.

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HOW ABOUT A T-INTERVAL ... μ = 128; n = 40; = 122; s = 15

Ho: μ = 128 Ha: 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?

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HOW ABOUT A T-INTERVAL ...Ho: μ = 128 Ha: 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.

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?

SWEETNESS OF DIET COLAS ...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? Or is this variation just due to sampling variability/just by coincidence?

Are these data good evidence that the cola loses sweetness over time?

Hypotheses:Ho: μ before - after = 0Ha: 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

Results (≈): t = 2.6967 P-value = 0.0123

Reject Ho. 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.

NOTES... 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)

LET’S GO BACK TO ALPHA = 0.05 FOR COLA SWEETNESS TEST... P-VALUE = 0.0123

At α = 0.05, reject Ho; 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?

ABOUT CONDITIONS...

HEAT THROUGH GLASS...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. 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.

Ho: μ = 1 Ha: μ > 1where 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

Ho: μ = 1 Ha: μ > 1

t = 8.95 and df = 10 p-value = 0.00000216≈ 0

Interpretation:Reject Ho. 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.

POSSIBLE ERRORS... AND HOW ABOUT IF ALPHA LEVEL CHANGED?

Ho: μ = 1 Ha: μ > 1

p-value = 0.00000216 ≈ 0 Reject Ho

we conclude that the mean heat conductivity for this type of glass is > 1

How about if α was changed to 10%? 1%?

FLORAL SCENTS & LEARNING...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.

TASK: DETERMINE IF FLORAL SCENT IMPROVES TEST PERFORMANCE

H0: μ( unscented – scented)= 0 Ha: 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 Ha says that unscented times are longer than scented times on average.

Assume a Normal distribution.

TASK: DETERMINE IF FLORAL SCENT IMPROVES TEST PERFORMANCE

H0: μ( unscented – scented)= 0 Ha: μ(unscented – scented) > 0

We could perform a paired t test or a t test (why?)

SRS, Normal, Independence

Input values from the ‘difference’ column into Minitab & perform the t-test

TASK: DETERMINE IF FLORAL SCENT IMPROVES TEST PERFORMANCE

H0: μ( unscented – scented)= 0 Ha: μ(unscented – scented) > 0

Interpretation:Fail to reject Ho (don’t believe Ha). Since

P-value of 0.3652 is larger than any reasonable α (say α = 0.05), data doesn’t support that floral scent improves test performance.(outside rejection zone)

TASK: DETERMINE IF FLORAL SCENT IMPROVES TEST PERFORMANCE

H0: μ( unscented – scented)= 0 Ha: μ(unscented – scented) > 0

Possible errors in context:Reject Ho when Ho is really true. We would

determine that floral scent improves test performance when it really doesn’t. Consequences?

Fail to reject Ho, when Ha is really true. We would determine that floral scent does not improve test performance when it really does. Consequences?

TASK: DETERMINE IF FLORAL SCENT IMPROVES TEST PERFORMANCE

H0: μ( unscented – scented)= 0 Ha: μ(unscented – scented) > 0

What if alpha level was something other than 5%? Would our decision change? Why or why not?

RIGHT’IES VS. LEFT’IESThe 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.

CARRY OUT PAIRED T-TEST AT ALPHA = 0.05

μ(right – left)

PAIRED T-TEST; ALPHA 5%; SAMPLE STATISTICS IN SECONDS...

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

ALPHA = 0.05; TIME IN SECONDSHo: μ(right – left) = 0Ha: μ(right – left) < 0

We want to assess if right-handed people find right-hand threads easier (faster) to use.

CONDITIONS... 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.

CALCULATIONS... Use Minitab 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

INTERPRETATION...Ho: μ(right – left) = 0 Ha: μ(right – left) < 0

Reject Ho. With a p-value of 0.0039 (well below any reasonable α of, say, 0.05), data does supports Ha, right-ies find right-handed threads easier & faster

Possible errors? What if we changed our alpha level?

NOW WE ARE ON TO TWO-SAMPLE MEAN PROCEDURES...

CALCIUM & BLOOD PRESSURE

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; 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.

DOES INCREASING CALCIUM IN THE DIET REDUCE 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.

Ho: μ1 = μ2 OR μ1 – μ2 = 0Ha: μ1 > μ2 OR μ1 – μ2 > 0

CONDITIONS... 2-sample t-procedure

SRS – Subjects were randomly assigned (good)

Normality/Big Sample – Generally we must look at n1 & n2 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

CALCULATIONS... Use Minitab; 2-sample t-procedure

Don’t ever pool (don’t ever assume equal variances)

t = 1.60 P-Value = 0.0644 df = 15

INTERPRETATION... Fail to reject Ho. 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?

A SIDE NOTE...ROBUSTNESS OF T-DISTRIBUTIONS... 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 (n1 = n2)

RED WINE VS. WHITE WINE

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:Ho: μR = μW or μR – μW = 0Ha: μR ≠ μW or μR – μW ≠ 0

where μR & μW are the mean percent change in polyphenols for men who drink red and white wine, respectfully.

Ho: μR = μW or μR – μW = 0Ha: μR ≠ μW or μR – μW ≠ 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.

Ho: μR = μW or μR – μW = 0Ha: μR ≠ μW or μR – μW ≠ 0

Calculations:2-sample t-test (do not assume

variances =)t = 3.81 p = 0.0019 df = 14

HO: ΜR = ΜW OR ΜR – ΜW = 0HA: ΜR ≠ ΜW OR ΜR – ΜW ≠ 0

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.

HO:ΜR = ΜW OR ΜR – ΜW = 0HA: ΜR ≠ ΜW OR ΜR – ΜW ≠ 0

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.

HO:ΜR = ΜW OR ΜR – ΜW = 0HA: ΜR ≠ ΜW OR ΜR – ΜW ≠ 0

... 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?

THE EFFECT OF LOGGING...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:

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 Ho: μU = μL OR μU – μL = 0 Ha: μU ≠ μL OR μU – μL ≠ 0

where μU & μL are the mean number of species in unlogged and logged plots, respectfully

Ho: μU = μL OR μU – μL = 0Ha: μU ≠ μL OR μU – μL ≠ 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.

Ho: μU = μL OR μU – μL = 0Ha: μU ≠ μL OR μU – μL ≠ 0

Use Minitab to run your 2-sample t procedure

t = 2.114 P-value = 0.0519

HO: ΜU = ΜL OR ΜU – ΜL = 0HA: ΜU ≠ ΜL OR ΜU – ΜL ≠ 0

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?

HO: ΜU = ΜL OR ΜU – ΜL = 0HA: ΜU ≠ ΜL OR ΜU – ΜL ≠ 0

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.

EACH DAY I AM GETTING BETTER IN MATH....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.

EACH DAY I’M GETTING BETTER IN MATH...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.

EACH DAY I’M GETTING BETTER IN MATH... ENTER STATISTICS INTO MINITAB

sample mean improvement/change for treatment group = 11.4

sample standard deviation treatment group = 3.17sample size n = 10 for treatment group

sample mean improvement/change for control group =8.25

sample standard deviation for control group = 3.69sample size n = 8 for control group

HO: ΜPOST - PRE FOR TREATMENT GROUP = Μ POST – PRE FOR CONTROL GROUPHA:ΜPOST - PRE FOR TREATMENT GROUP > Μ POST – PRE FOR CONTROL GROUP

t = 1.91 p-value = 0.038 df = 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.

Errors? Different alpha? Confidence interval?

IS OUR MEAN RESTING HEART RATE DIFFERENT FROM THE NATIONAL MEAN RESTING HEART RATE?

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.

IS OUR BODY TEMPERATURE DIFFERENT FROM THE MEAN HEALTHY BODY TEMPERATURE FOR ADULTS IN US (WHICH IS 98.6 DEGREES F)?

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.

NEXT UP? EXAM #3... Homework quiz ....

Will cover chapters 7, 8, & 9 List of review topics, randomly

assigned to teams of 2 students

Not sure if we will have a 4th exam... Need to see how are timing is...