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One-Way Analysis ofVariance: ComparingSeveral Means

In this chapter we cover. . .Comparing several means

The analysis of variance Ftest

Using technology

The idea of analysis ofvariance

The ANOVA model

Some details of ANOVA*

The two-sample t procedures of Chapter 17 compare the means of two popu-lations or the mean responses to two treatments in an experiment. Of course,studies don’t always compare just two groups. We need a method for comparingany number of means.

EXAMPLE 22.1 Cars, SUVs, and pickup trucks

Sport-utility vehicles (SUVs) and pickup trucks have replaced cars in many Amer-ican driveways. SUVs and trucks are often larger and heavier than cars and so areunder attack for reasons of safety and fuel economy. Do SUVs and trucks have lowergas mileage than cars? Table 22.1 contains data on the highway gas mileage (in milesper gallon) for 31 midsize cars, 31 SUVs, and 14 standard-size pickup trucks. The gasmileages and vehicle classifications are compiled by the Environmental ProtectionAgency.1 Because the cars are two-wheel-drive, the table describes only two-wheel-drive SUVs and trucks. The popular four-wheel-drive versions get poorer mileage.

Figure 22.1 shows side-by-side stemplots of the gas mileages for the three types ofvehicle. We used the same stems in all three for easier comparison. It does appearthat gas mileage decreases as we move from cars to SUVs to pickups.

597



598 CHAPTER 22 � One-Way Analysis of Variance: Comparing Several Means

Here are the means, standard deviations, and five-number summaries for the threetypes of vehicle, from software:

N Mean Median StDev Minimum Maximum Q1 Q3Midsize 31 27.903 27.000 2.561 23.000 33.000 26.000 30.000SUV 31 22.677 22.000 3.673 17.000 29.000 19.000 26.000Pickup 14 21.286 20.000 2.758 17.000 26.000 19.750 23.500

We will use the mean to describe the center of the gas mileage distributions. Meangas mileage goes down as we move from midsize cars to SUVs to pickup trucks. Thedifferences among the means are not large. Are they statistically significant?

TABLE 22.1 Highway gas mileage for 2003 model vehicles

Midsize Cars Sport-Utility Vehicles Standard Pickup Trucks

Model MPG Model MPG Model MPG

Acura 3.5TL 29 Cadillac Escalade 18 Chevrolet Silverado 20Audi A6 27 Chevrolet Avalanche 18 Dodge Dakota 19Audi A8 25 Chevrolet Blazer 23 Dodge Ram 20Buick Century 29 Chevrolet Suburban 18 Ford Explorer 20Buick Regal 27 Chevrolet Tahoe 18 Ford F150 20Cadillac CTS 26 Chevrolet Tracker 26 Ford Ranger 26Cadillac Seville 27 Chevrolet Trailblazer 22 GMC Sierra 20Chevrolet Monte Carlo 32 Chrysler PT Cruiser 25 Lincoln Blackwood 17Chrysler Sebring 30 Dodge Durango 19 Mazda B2300 26Honda Accord 33 Ford Escape 28 Mazda B3000 22Hyundai Sonata 30 Ford Expedition 19 Mazda B4000 21Hyundai XG350 26 Ford Explorer 19 Nissan Frontier 23Infiniti I35 26 Honda CR-V 28 Toyota Tacoma 25Jaguar S-Type 26 Hyundai Santa Fe 27 Toyota Tundra 19Jaguar Vanden Plas 24 Infiniti QX4 21Kia Optima 30 Isuzu Axiom 21Lexus ES300 29 Isuzu Rodeo 23Lexus GS300 25 Jeep Grand Cherokee 21Lincoln LS 26 Jeep Liberty 24Mercedes-Benz E320 27 Kia Sorento 20Mercedes-Benz E500 23 Lincoln Navigator 17Mitsubishi Diamante 25 Mazda Tribute 28Mitsubishi Galant 27 Mitsubishi Montero 22Nissan Altima 29 Nissan Pathfinder 21Nissan Maxima 26 Nissan Xterra 24Saab 9-5 29 Saturn Vue 28Saturn L200 32 Suzuki Vitara 25Saturn L300 29 Toyota 4Runner 20Toyota Camry 32 Toyota Highlander 27Volkswagen Passat 31 Toyota Rav4 29Volvo S80 28 Volvo XC 90 24



599Comparing several means

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0

0000000000

000

Midsize SUV Pickup

Figure 22.1 Side-by-side stemplots comparing the highway gas mileages of midsizecars, sport-utility vehicles, and standard-size pickup trucks, from Table 22.1.

Comparing several meansCall the mean highway gas mileages for the three populations of vehicles µ1 formidsize cars, µ2 for SUVs, and µ3 for pickups. The subscript reminds us whichgroup a parameter or statistic describes. To compare these three populationmeans, we might use the two-sample t test several times:� Test H0: µ1 = µ2 to see if the mean miles per gallon for midsize cars

differs from the mean for SUVs.� Test H0: µ1 = µ3 to see if midsize cars differ from pickup trucks.� Test H0: µ2 = µ3 to see if SUVs differ from pickups.

The weakness of doing three tests is that we get three P -values, one foreach test alone. That doesn’t tell us how likely it is that three sample means arespread apart as far as these are. It may be that x1 = 27.903 and x3 = 21.286 aresignificantly different if we look at just two groups but not significantly differentif we know that they are the smallest and the largest means in three groups. Aswe look at more groups, we expect the gap between the smallest and the largestsample mean to get larger. (Think of comparing the tallest and shortest personin larger and larger groups of people.) We can’t safely compare many parametersby doing tests or confidence intervals for two parameters at a time.

The problem of how to do many comparisons at once with some overallmeasure of confidence in all our conclusions is common in statistics. This is theproblem of multiple comparisons. Statistical methods for dealing with multiple multiple comparisonscomparisons usually have two steps:1. An overall test to see if there is good evidence of any differences among

the parameters that we want to compare.2. A detailed follow-up analysis to decide which of the parameters differ and

to estimate how large the differences are.




The overall test, though more complex than the tests we met earlier, is oftenreasonably straightforward. The follow-up analysis can be quite elaborate. Inour basic introduction to statistical practice, we will concentrate on the overalltest, along with data analysis that points to the nature of the differences.

The analysis of variance F testWe want to test the null hypothesis that there are no differences among themean highway gas mileages for the three vehicle types:

H0: µ1 = µ2 = µ3

The alternative hypothesis is that there is some difference, that not all threepopulation means are equal:

Ha: not all of µ1, µ2, and µ3 are equal

The alternative hypothesis is no longer one-sided or two-sided. It is “many-sided,” because it allows any relationship other than “all three equal.” For ex-ample, Ha includes the case in which µ2 = µ3 but µ1 has a different value.The test of H0 against Ha is called the analysis of variance F test. Analysisanalysis of variance F testof variance is usually abbreviated as ANOVA. The ANOVA F test is almostalways carried out by software that reports the test statistic and its P-value.

EXAMPLE 22.2 Cars, SUVs, and pickup trucks: ANOVA

As we will soon see, software tells us that for the gas mileage data in Table 22.1,F = 31.61 with P-value P < 0.001. There is very strong evidence that the threetypes of vehicle do not all have the same mean gas mileage.

The F test does not say which of the three means are significantly different. Itappears from our preliminary data analysis that SUVs and pickups have similar fueleconomy and that both get distinctly poorer mileage than midsize cars. The gap ofmore than 5 miles per gallon between midsize cars and the other two types of vehicleis large enough to be of practical interest.

Our conclusion: There is strong evidence (P < 0.001) that the means are not allequal. The most important difference among the means is that midsize cars havebetter gas mileage than SUVs and pickups.

Example 22.2 illustrates our approach to comparing means. The ANOVAF test (done by software) assesses the evidence for some difference among thepopulation means. In most cases, we expect the F test to be significant. Wewould not undertake a study if we did not expect to find some effect. The formaltest is nonetheless important to guard against being misled by chance variation.We will not do the formal follow-up analysis that is often the most useful partof an ANOVA study. Follow-up analysis would allow us to say which meansdiffer and by how much, with (say) 95% confidence that all our conclusions are



601The analysis of variance F test

correct. We rely instead on examination of the data to show what differencesare present and whether they are large enough to be interesting.

APPLY YOUR KNOWLEDGE

22.1 Do fruit flies sleep? Mammals and birds sleep. Insects such as fruit fliesrest, but is this rest sleep? Biologists now think that insects do sleep.One experiment gave caffeine to fruit flies to see if it affected their rest.We know that caffeine reduces sleep in mammals, so if it reduces rest infruit flies that’s another hint that the rest is really sleep. The paperreporting the study contains a graph similar to Figure 22.2 and states,“Flies given caffeine obtained less rest during the dark period in adose-dependent fashion (n = 36 per group, P < 0.0001).”2

(a) The explanatory variable is amount of caffeine, in milligrams permilliliter of blood. The response variable is minutes of rest(measured by an infrared motion sensor) during a 12-hour darkperiod. Outline the design of this experiment.

(b) The P-value in the report comes from the ANOVA F test. Whatmeans does this test compare? State in words the null andalternative hypotheses for the test in this setting. What do thegraph and the statistical test together lead you to conclude?

Control

Caffeine

1 mg/ml 2.5 mg/ml 5 mg/ml

020

0

Rest

(min

)

400

600

Figure 22.2 Bar graph comparing the mean rest of fruit flies given different amountsof caffeine, for Exercise 22.1.




22.2 Road rage. A random-digit-dialing survey asked a random sample of1382 drivers questions about their driving. Responses to a number ofquestions were then combined to yield a measure of “angry/threateningdriving,” commonly called “road rage.” What driver characteristics gowith road rage? There were no significant differences among races orlevels of education. What about the effect of the driver’s age? Here arethe mean responses for three age-groups:3

<30 yr 30–55 yr >55 yr

2.22 1.33 0.66

The report says that F = 34.96, with P < 0.01.(a) What are the null and alternative hypotheses for the ANOVA F

test? Be sure to explain what means the test compares.(b) Based on the sample means and the F test, what do you conclude?

Using technologyAny technology used for statistics should perform analysis of variance.Figure 22.3 displays ANOVA output for the data of Table 22.1 from the Minitabstatistical software, the Excel spreadsheet program, and the TI-83 graphing cal-culator.

Minitab and Excel give the sizes of the three samples and their means. Theseagree with those in Example 22.1. The most important part of all three outputsreports the F test statistic, F = 31.61, and its P-value. Minitab sensibly reportsP as 0.000, that is, 0 to three decimal places. Excel and the TI-83 show that Pis about 0.0000000001. (The E-10 in these displays means move the decimalpoint 10 digits to the left.) There is very strong evidence that the three typesof vehicle do not all have the same mean gas mileage.

All three outputs also report sums of squares (SS) and mean squares (MS).We don’t need this information now. The optional section “Some Details ofANOVA” at the end of this chapter explains these terms.

Minitab also gives confidence intervals for all three means that help us seewhich means differ and by how much. The SUV and pickup intervals overlap,and the midsize interval lies well above them on the mileage scale. These are95% confidence intervals for each mean separately. We are not 95% confidentthat all three intervals cover the three means. This is another example of theperil of multiple comparisons.


22.3 Logging in the rainforest. How does logging in a tropical rainforestaffect the forest several years later? Researchers compared forest plots inBorneo that had never been logged (Group 1) with similar plots nearby


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603Using technology

Excel

TI-83

Minitab

Figure 22.3 Analysis of variance output for the gas mileage data from Minitab,Excel, and the TI-83.




that had been logged 1 year earlier (Group 2) and 8 years earlier(Group 3). Although the study was not an experiment, the authorsexplain why we can consider the plots to be randomly selected. Thedata appear in Table 22.2. The variable Trees is the count of trees in aplot; Species is the count of tree species in a plot. The variable Richnessis the ratio of number of species to number of individual trees,calculated as Species/Trees.4

TABLE 22.2 Data from a study of logging in Borneo

Observation Group Trees Species Richness

1 1 27 22 0.814812 1 22 18 0.818183 1 29 22 0.758624 1 21 20 0.952385 1 19 15 0.789476 1 33 21 0.636367 1 16 13 0.812508 1 20 13 0.650009 1 24 19 0.79167

10 1 27 13 0.4814811 1 28 19 0.6785712 1 19 15 0.7894713 2 12 11 0.9166714 2 12 11 0.9166715 2 15 14 0.9333316 2 9 7 0.7777817 2 20 18 0.9000018 2 18 15 0.8333319 2 17 15 0.8823520 2 14 12 0.8571421 2 14 13 0.9285722 2 2 2 1.0000023 2 17 15 0.8823524 2 19 8 0.4210525 3 18 17 0.9444426 3 4 4 1.0000027 3 22 18 0.8181828 3 15 14 0.9333329 3 18 18 1.0000030 3 19 15 0.7894731 3 22 15 0.6818232 3 12 10 0.8333333 3 12 12 1.00000


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605Using technology

Figure 22.4 Excel output for analysis of variance on the number of trees in forestplots, for Exercise 22.3.

(a) Make side-by-side stemplots of Trees for the three groups. Use stems0, 1, 2, and 3 and split the stems (see page 17). What effects oflogging are visible?

(b) Figure 22.4 shows Excel ANOVA output for Trees. What do thegroup means show about the effects of logging?

(c) What are the values of the ANOVA F statistic and its P-value?What hypotheses does F test? What conclusions about the effectsof logging on number of trees do the data lead to?

EESEE

22.4 Dogs, friends, and stress. If you are a dog lover, perhaps having yourdog along reduces the effect of stress. To examine the effect of pets instressful situations, researchers recruited 45 women who said they weredog lovers. The EESEE story “Stress among Pets and Friends” describesthe results. Fifteen of the subjects were randomly assigned to each ofthree groups to do a stressful task alone (the control group), with a goodfriend present, or with their dog present. The subject’s mean heart rateduring the task is one measure of the effect of stress. Table 22.3 containsthe data.(a) Make stemplots of the heart rates for the three groups (round to the

nearest whole number of beats). Do any of the groups show outliersor extreme skewness?

(b) Figure 22.5 gives the Minitab ANOVA output for these data. Dothe mean heart rates for the groups appear to show that the presenceof a pet or a friend reduces heart rate during a stressful task?

(c) What are the values of the ANOVA F statistic and its P-value?What hypotheses does F test? Briefly describe the conclusions youdraw from these data. Did you find anything surprising?




TABLE 22.3 Mean heart rates during stress with a pet (P),with a friend (F), and for the control group (C)

Group Rate Group Rate Group Rate

P 69.169 P 68.862 C 84.738F 99.692 C 87.231 C 84.877P 70.169 P 64.169 P 58.692C 80.369 C 91.754 P 79.662C 87.446 C 87.785 P 69.231P 75.985 F 91.354 C 73.277F 83.400 F 100.877 C 84.523F 102.154 C 77.800 C 70.877P 86.446 P 97.538 F 89.815F 80.277 P 85.000 F 98.200C 90.015 F 101.062 F 76.908C 99.046 F 97.046 P 69.538C 75.477 C 62.646 P 70.077F 88.015 F 81.600 F 86.985F 92.492 P 72.262 P 65.446

Analysis of Variance for BeatsSourceGroupErrorTotal

DF24244

SS2387.73561.35949.0

MS1193.884.8

LevelControlFriendPet

Pooled StDev =

N151515

Mean82.52491.32573.483

9.208 72.0 80.0 88.0 96.0

StDev9.2428.3419.970

F14.08

Individual 95% CIs For MeanBased on Pooled StDev

P0.000

Figure 22.5 Minitab output for the data in Table 22.3 on heart rates (beats perminute) during stress, for Exercise 22.4. The “Control” group worked alone, the“Friend” group had a friend present, and the “Pet” group had a pet dog present.



607The idea of analysis of variance

(a) (b)

Figure 22.6 Boxplots for two sets of three samples each. The sample means are thesame in (a) and (b). Analysis of variance will find a more significant difference amongthe means in (b) because there is less variation among the individuals within thosesamples.

The idea of analysis of varianceHere is the main idea for comparing means: what matters is not how far apartthe sample means are but how far apart they are relative to the variability ofindividual observations. Look at the two sets of boxplots in Figure 22.6. Forsimplicity, these distributions are all symmetric, so that the mean and medianare the same. The centerline in each boxplot is therefore the sample mean.Both sets of boxplots compare three samples with the same three means. Likethe three vehicle types in Example 22.1, the means are different but not verydifferent. Could differences this large easily arise just due to chance, or are theystatistically significant?� The boxplots in Figure 22.6(a) have tall boxes, which show lots of

variation among the individuals in each group. With this much variationamong individuals, we would not be surprised if another set of samplesgave quite different sample means. The observed differences among thesample means could easily happen just by chance.� The boxplots in Figure 22.6(b) have the same centers as those

in Figure 22.6(a), but the boxes are much shorter. That is, there is muchless variation among the individuals in each group. It is unlikely that anysample from the first group would have a mean as small as the mean of thesecond group. Because means as far apart as those observed would rarelyarise just by chance in repeated sampling, they are good evidence of real dif-ferences among the means of the three populations we are sampling from.

APPLET

You can use the One-Way ANOVA applet to demonstrate the analysis ofvariance idea for yourself. The applet allows you to change both the groupmeans and the spread within groups. You can watch the ANOVA F statisticand its P-value change as you work.

This comparison of the two parts of Figure 22.6 is too simple in one way.It ignores the effect of the sample sizes, an effect that boxplots do not show.Small differences among sample means can be significant if the samples arevery large. Large differences among sample means may fail to be significant if




the samples are very small. All we can be sure of is that for the same sample size,Figure 22.6(b) will give a much smaller P-value than Figure 22.6(a). Despitethis qualification, the big idea remains: if sample means are far apart relative tothe variation among individuals in the same group, that’s evidence that some-thing other than chance is at work.

THE ANALYSIS OF VARIANCE IDEA

Analysis of variance compares the variation due to specific sources withthe variation among individuals who should be similar. In particular,ANOVA tests whether several populations have the same mean bycomparing how far apart the sample means are with how much variationthere is within the samples.

It is one of the oddities of statistical language that methods for compar-ing means are named after the variance. The reason is that the test works bycomparing two kinds of variation. Analysis of variance is a general methodfor studying sources of variation in responses. Comparing several means is thesimplest form of ANOVA, called one-way ANOVA. One-way ANOVA is theone-way ANOVAonly form of ANOVA that we will study.

THE ANOVA F STATISTIC

The analysis of variance F statistic for testing the equality of severalmeans has this form:

F = variation among the sample meansvariation among individuals in the same sample

We give more detail later. Because ANOVA is in practice done by software,the idea is more important than the detail. The F statistic can only take val-ues that are zero or positive. It is zero only when all the sample means areidentical and gets larger as they move farther apart. Large values of F are evi-dence against the null hypothesis H0 that all population means are the same.Although the alternative hypothesis Ha is many-sided, the ANOVA F test isone-sided because any violation of H0 tends to produce a large value of F .

How large must F be to provide significant evidence against H0? To an-swer questions of statistical significance, compare the F statistic with criticalvalues from an F distribution. The F distributions are described on page 456F distributionin Chapter 17. A specific F distribution is specified by two parameters: a nu-merator degrees of freedom and a denominator degrees of freedom. Table D inthe back of the book contains critical values for F distributions with variousdegrees of freedom.



609The idea of analysis of variance

EXAMPLE 22.3 Using the F table

Look again at the software output for the gas mileage data in Figure 22.3. All threeoutputs give the degrees of freedom for the F test, labeled “df ” or “DF.” There are2 degrees of freedom in the numerator and 73 in the denominator.

In Table D, find the numerator degrees of freedom 2 at the top of the table. Thenlook for the denominator degrees of freedom 73 at the left of the table. There isno entry for 73, so we use the next smaller entry, 50 degrees of freedom. The uppercritical values for 2 and 50 degrees of freedom are

Criticalp value

0.100 2.410.050 3.180.025 3.970.010 5.060.001 7.96

Figure 22.7 shows the F density curve with 2 and 50 degrees of freedom and theupper 5% critical value 3.18. The observed F = 31.61 lies far to the right on thiscurve. We see from the table that F = 31.61 is larger than the 0.001 critical value,so P < 0.001.

Software, of course, uses the F distribution with 2 and 73 degrees of freedom tocalculate P-values. Excel also tells us how large F must be to be significant at theα = 0.05 level. This is the 5% critical value for F with 2 and 73 degrees of freedom.It is 3.122. Approximating the P-value from Table D is usually adequate in practice,as it is in this example.

10 3.18 10.00

Area = 0.05

F density curve, df = 2 and 50

Figure 22.7 The density curve of the F distribution with 2 degrees of freedom in thenumerator and 50 degrees of freedom in the denominator, for Example 22.3. Thevalue F ∗ = 3.18 is the 0.05 critical value.




The degrees of freedom of the F statistic depend on the number of meanswe are comparing and the number of observations in each sample. That is, theF test takes into account the number of observations. Here are the details.

DEGREES OF FREEDOM FOR THE F TEST

We want to compare the means of I populations. We have an SRS of sizeni from the ith population, so that the total number of observations in allsamples combined is

N = n1 + n2 + · · · + nI

If the null hypothesis that all population means are equal is true, theANOVA F statistic has the F distribution with I − 1 degrees of freedomin the numerator and N − I degrees of freedom in the denominator.

EXAMPLE 22.4 Degrees of freedom for F

In Examples 22.1 and 22.2, we compared the mean highway gas mileage for threetypes of vehicle, so I = 3. The three sample sizes are

n1 = 31 n2 = 31 n3 = 14

The total number of observations is therefore

N = 31 + 31 + 14 = 76

The ANOVA F test has numerator degrees of freedom

I − 1 = 3 − 1 = 2

and denominator degrees of freedom

N − I = 76 − 3 = 73

These degrees of freedom are given in the outputs in Figure 22.3.


22.5 Logging in the rainforest, continued. Exercise 22.3 compares thenumber of tree species in rainforest plots that had never been logged(Group 1) with similar plots nearby that had been logged 1 year earlier(Group 2) and 8 years earlier (Group 3).(a) What are I , the ni , and N for these data? Identify these quantities

in words and give their numerical values.(b) Find the degrees of freedom for the ANOVA F statistic. Check

your work against the Excel output in Figure 22.4.(c) For these data, F = 11.43. What does Table D tell you about the

P-value of this statistic?22.6 Dogs, friends, and stress, continued. Exercise 22.4 compares the

mean heart rates for women performing a stressful task under threeconditions.



611The ANOVA model

(a) What are I , the ni , and N for these data? State in words what eachof these numbers means.

(b) Find both degrees of freedom for the ANOVA F statistic. Checkyour results against the Minitab output in Figure 22.5.

(c) The output shows that F = 14.08. What does Table D tell youabout the significance of this result?

22.7 Smoking among French men. A study of smoking categorized a sampleof French men aged 20 to 60 years as nonsmokers (146 men), formersmokers (125 men), moderate smokers (104 men), or heavy smokers(84 men). One set of response variables was the amount of several typesof food consumed, in grams.5

(a) What are the degrees of freedom for an ANOVA F statisticthat compares the mean weights of a food consumed in the4 populations? Use Table D with 3 and 200 degrees of freedom togive an approximate P-value for each of the following foods:

(b) For milk, F = 3.58.(c) For eggs, F = 1.08.(d) For vegetables, F = 6.02.

Where are the details?

Papers reporting scientificresearch must be short. Brevityallows researchers to hidedetails about their data. Didthey choose their subjects in abiased way? Did they reportdata on only some of theirsubjects? Did they try severalstatistical analyses and reportthe one that looked best? Thestatistician John Bailarscreened more than 4000papers for the New EnglandJournal of Medicine. He says,“When it came to thestatistical review, it was oftenclear that critical informationwas lacking, and the gapsnearly always had the practicaleffect of making the authors’conclusions look stronger thanthey should have.”

The ANOVA modelLike all inference procedures, ANOVA is valid only in some circumstances.Here is the model that describes when we can use ANOVA to compare popu-lation means.

THE ANOVA MODEL� We have I independent SRSs, one from each of I populations.� The i th population has a Normal distribution with unknown mean

µi . The means may be different in the different populations. TheANOVA F statistic tests the null hypothesis that all of thepopulations have the same mean:

H0: µ1 = µ2 = · · · = µI

Ha: not all of the µi are equal� All of the populations have the same standard deviation σ , whose

value is unknown.The ANOVA model has I + 1 parameters, the I population means andthe standard deviation σ .

The first two requirements are familiar from our study of the two-samplet procedures for comparing two means. As usual, the design of the data pro-duction is the most important foundation for inference. Biased sampling orconfounding can make any inference meaningless. If we do not actually draw




separate SRSs from each population or carry out a randomized comparative ex-periment, it is often unclear to what population the conclusions of inferenceapply. This is the case in Example 22.1, for example. ANOVA, like other in-ference procedures, is often used when random samples are not available. Youmust judge each use on its merits, a judgment that usually requires some knowl-edge of the subject of the study in addition to some knowledge of statistics. Wemight consider the 2003 model year vehicles in Example 22.1 to be samplesfrom vehicles of their types produced in recent years.

Because no real population has exactly a Normal distribution, the usefulnessof inference procedures that assume Normality depends on how sensitive theyare to departures from Normality. Fortunately, procedures for comparing meansare not very sensitive to lack of Normality. The ANOVA F test, like the t pro-cedures, is robust. What matters is Normality of the sample means, so ANOVArobustnessbecomes safer as the sample sizes get larger, because of the central limit theoremeffect. Remember to check for outliers that change the value of sample meansand for extreme skewness. When there are no outliers and the distributions areroughly symmetric, you can safely use ANOVA for sample sizes as small as 4or 5. (Don’t confuse the ANOVA F , which compares several means, with theF statistic discussed in Chapter 17, which compares two standard deviationsand is not robust against non-Normality.)

The third assumption is annoying: ANOVA assumes that the variability ofobservations, measured by the standard deviation, is the same in all popula-tions. You may recall from Chapter 17 (page 454) that there is a special versionof the two-sample t test that assumes equal standard deviations in both popu-lations. The ANOVA F for comparing two means is exactly the square of thisspecial t statistic. We prefer the t test that does not assume equal standard devi-ations, but for comparing more than two means there is no general alternativeto the ANOVA F . It is not easy to check the assumption that the populationshave equal standard deviations. Statistical tests for equality of standard devia-tions are very sensitive to lack of Normality, so much so that they are of littlepractical value. You must either seek expert advice or rely on the robustness ofANOVA.

How serious are unequal standard deviations? ANOVA is not too sensitiveto violations of the assumption, especially when all samples have the same orsimilar sizes and no sample is very small. When designing a study, try to takesamples of the same size from all the groups you want to compare. The sam-ple standard deviations estimate the population standard deviations, so checkbefore doing ANOVA that the sample standard deviations are similar to eachother. We expect some variation among them due to chance. Here is a rule ofthumb that is safe in almost all situations.

CHECKING STANDARD DEVIATIONS IN ANOVA

The results of the ANOVA F test are approximately correct when thelargest sample standard deviation is no more than twice as large as thesmallest sample standard deviation.



613The ANOVA model

EXAMPLE 22.5 Do the standard deviations allow ANOVA?

In the gas mileage study, the sample standard deviations for midsize cars, SUVs, andpickups are

s1 = 2.561 s2 = 3.673 s3 = 2.758

These standard deviations easily satisfy our rule of thumb. We can safely useANOVA to compare the mean gas mileage for the three vehicle types.

The report from which Table 22.1 was taken also contained data on 22 subcom-pact cars. Can we use ANOVA to compare the means for all four types of vehicle?The standard deviation for subcompact cars is s4 = 5.262 miles per gallon. This isslightly more than twice the smallest standard deviation:

largest ssmallest s

= 5.2622.561

= 2.05

Our rule of thumb is conservative, so many statisticians would go ahead withANOVA in this circumstance.

A large standard deviation is often due to skewness or outliers. It turns out thatthe government’s “subcompact” category includes three rather exotic cars, a Bentley,a Ferrari, and a Maserati. All three get poor gas mileage. The standard deviationfor the remaining 19 subcompact cars is s = 3.304, well within the guidelines forANOVA.

EXAMPLE 22.6 Which color attracts beetles best?

To detect the presence of harmful insects in farm fields, we can put up boards coveredwith a sticky material and examine the insects trapped on the boards. Which colorsattract insects best? Experimenters placed six boards of each of four colors at randomlocations in a field of oats and measured the number of cereal leaf beetles trapped.Here are the data:6

Board color Insects trapped

Blue 16 11 20 21 14 7Green 37 32 20 29 37 32White 21 12 14 17 13 20Yellow 45 59 48 46 38 47

We would like to use ANOVA to compare the mean numbers of beetles that wouldbe trapped by all boards of each color. Because the samples are small, we plot thedata in side-by-side stemplots in Figure 22.8. Minitab output for ANOVA appears

0

1

2

3

4

5

7

146

01

0

1

2

3

4

5

09

2277

0

1

2

3

4

5

2347

01

Blue Green White

0

1

2

3

4

5

8

5678

9

Yellow

Figure 22.8 Side-by-sidestemplots comparing thecounts of insects attracted bysix boards for each of fourboard colors, for Example22.6.




in Figure 22.9. The yellow boards attract by far the most insects (x4 = 47.167), withgreen next (x2 = 31.167) and blue and white far behind.

Check that we can safely use ANOVA to test equality of the four means. Thelargest of the four sample standard deviations is 6.795 and the smallest is 3.764. Theratio

largest ssmallest s

= 6.7953.764

= 1.8

is less than 2, so these data satisfy our rule of thumb. The shapes of the four distri-butions are irregular, as we expect with only 6 observations in each group, but thereare no outliers. The ANOVA results will be approximately correct.

There are I = 4 groups and N = 24 observations overall, so the degrees of free-dom for F are

numerator: I − 1 = 4 − 1 = 3denominator: N − I = 24 − 4 = 20

This agrees with Minitab’s results. The F statistic is F = 42.84, a large Fwith P < 0.001. Despite the small samples, the experiment gives very strong evi-dence of differences among the colors. Yellow boards appear best at attracting leafbeetles.

Analysis of Variance for BeetlesSourceColorErrorTotal

DF32023

SS4134.0643.34777.3

MS1378.032.2

LevelBlueGreenWhiteYellow

N6666

Mean14.83331.16716.16747.167

StDev5.3456.3063.7646.795

F42.84


P0.000

Pooled StDev = 5.672 12 24 36 48

Figure 22.9 Minitab ANOVA output for comparing the four board colors inExample 22.6.



615The ANOVA model


22.8 Checking standard deviations. Verify that the sample standarddeviations for these sets of data do allow use of ANOVA to comparethe population means.(a) The counts of trees in Exercise 22.3 and Figure 22.4.(b) The heart rates of Exercise 22.4 and Figure 22.5.

22.9 Species richness after logging. Table 22.2 gives data on the speciesrichness in rainforest plots, defined as the number of tree species in aplot divided by the number of trees in the plot. ANOVA may not betrustworthy for the richness data. Do data analysis: make side-by-sidestemplots to examine the distributions of the response variable in thethree groups and also compare the standard deviations. What aboutthe data makes ANOVA risky?

22.10 Marital status and salary. Married men tend to earn more than singlemen. An investigation of the relationship between marital status andincome collected data on all 8235 men employed as managers orprofessionals by a large manufacturing firm in 1976. Suppose (this isrisky) we regard these men as a random sample from the population ofall men employed in managerial or professional positions in largecompanies. Here are summary statistics for the salaries of thesemen:7

Single Married Divorced Widowed

ni 337 7,730 126 42xi $21,384 $26,873 $25,594 $26,936si $5,731 $7,159 $6,347 $8,119

(a) Briefly describe the relationship between marital status and salary.(b) Do the sample standard deviations allow use of the ANOVA

F test? (The distributions are skewed to the right. We expectright-skewness in income distributions. The investigators actuallyapplied ANOVA to the logarithms of the salaries, which are moresymmetric.)

(c) What are the degrees of freedom of the ANOVA F test?(d) The F test is a formality for these data, because we are sure that

the P-value will be very small. Why are we sure?(e) Single men earn less on the average than men who are or have

been married. Do the highly significant differences in mean salaryshow that getting married raises men’s mean income? Explain youranswer.




Some details of ANOVA*Now we will give the actual formula for the ANOVA F statistic. We haveSRSs from each of I populations. Subscripts from 1 to I tell us which sample astatistic refers to:

Sample Sample SamplePopulation size mean std. dev.

1 n1 x1 s12 n2 x2 s2· · · ·· · · ·· · · ·I nI x I s I

You can find the F statistic from just the sample sizes ni , the sample meansxi , and the sample standard deviations si . You don’t need to go back to theindividual observations.

The ANOVA F statistic has the form

F = variation among the sample meansvariation among individuals within samples

The measures of variation in the numerator and denominator of F are calledmean squares. A mean square is a more general form of a sample variance. Anmean squaresordinary sample variance s 2 is an average (or mean) of the squared deviationsof observations from their mean, so it qualifies as a “mean square.”

The numerator of F is a mean square that measures variation among the Isample means x1, x2, . . . , x I . Call the overall mean response (the mean of allN observations together) x . You can find x from the I sample means by

x = n1x1 + n2x2 + · · · + nI x I

N

The sum of each mean multiplied by the number of observations it representsis the sum of all the individual observations. Dividing this sum by N, the totalnumber of observations, gives the overall mean x . The numerator mean squarein F is an average of the I squared deviations of the means of the samplesfrom x . We call it the mean square for groups, abbreviated as MSG.MSG

MSG = n1(x1 − x)2 + n2(x2 − x)2 + · · · + nI (x I − x)2

I − 1

Each squared deviation is weighted by ni , the number of observations it repre-sents.

*This more advanced section is optional if you are using software to find the F statistic.



617Some details of ANOVA

The mean square in the denominator of F measures variation among in-dividual observations in the same sample. For any one sample, the samplevariance s 2

i does this job. For all I samples together, we use an average of theindividual sample variances. It is again a weighted average in which each s 2

i isweighted by one fewer than the number of observations it represents, ni − 1.Another way to put this is that each s 2

i is weighted by its degrees of freedomni − 1. The resulting mean square is called the mean square for error, MSE. MSE

MSE =(n1 − 1)s 2

1 + (n2 − 1)s 22 + · · · + (nI − 1)s 2

I

N − I

“Error” doesn’t mean a mistake has been made. It’s a traditional term for chancevariation. Here is a summary of the ANOVA test.

THE ANOVA F TEST

Draw an independent SRS from each of I populations. The i thpopulation has the N(µi , σ ) distribution, where σ is the commonstandard deviation in all the populations. The i th sample has size ni ,sample mean xi , and sample standard deviation si .The ANOVA F statistic tests the null hypothesis that all I populationshave the same mean:

H0: µ1 = µ2 = · · · = µI

Ha: not all of the µi are equal

The statistic is

F = MSGMSE

The numerator of F is the mean square for groups

MSG = n1(x1 − x)2 + n2(x2 − x)2 + · · · + nI (x I − x)2

I − 1

The denominator of F is the mean square for error

MSE =(n1 − 1)s 2

1 + (n2 − 1)s 22 + · · · + (nI − 1)s 2

I

N − I

When H0 is true, F has the F distribution with I − 1 and N − Idegrees of freedom.

The denominators in the recipes for MSG and MSE are the two degrees offreedom I − 1 and N − I of the F test. The numerators are called sums of sums of squaressquares, from their algebraic form. It is usual to present the results of ANOVAin an ANOVA table. Output from software usually includes an ANOVA table. ANOVA table




EXAMPLE 22.7 ANOVA calculations: software

Look again at the three outputs in Figure 22.3 (page 603). Minitab and Excel give theANOVA table. The TI-83, with its small screen, gives the degrees of freedom, sumsof squares, and mean squares separately. Each output uses slightly different language.The basic ANOVA table is:

Source of variation df SS MS F statistic

Variation among samples 2 606.45 MSG = 303.22 31.61Variation within samples 73 700.34 MSE = 9.59

You can check that each mean square MS is the corresponding sum of squares SSdivided by its degrees of freedom df. The F statistic is MSG divided by MSE.

Because MSE is an average of the individual sample variances, it is also calledthe pooled sample variance, written as s 2

p . When all I populations have the samepopulation variance σ 2 (ANOVA assumes that they do), s 2

p estimates the com-mon variance σ 2. The square root of MSE is the pooled standard deviation s p .pooled standard deviationIt estimates the common standard deviation σ of observations in each group.The Minitab and TI-83 outputs in Figure 22.3 give the value s p = 3.097.

The pooled standard deviation s p is a better estimator of the common σ

than any individual sample standard deviation si because it combines (pools)the information in all I samples. We can get a confidence interval for any ofthe means µi from the usual form

estimate ± t∗SEestimate

using s p to estimate σ . The confidence interval for µi is

xi ± t∗ s p√ni

Use the critical value t∗ from the t distribution with N − I degrees of freedom,because s p has N − I degrees of freedom. These are the confidence intervalsthat appear in Minitab ANOVA output.

EXAMPLE 22.8 ANOVA calculations: without software

We can do the ANOVA test comparing the gas mileages of midsize cars, SUVs, andpickup trucks using only the sample sizes, sample means, and sample standard devi-ations. These appear in Example 22.1, but it is easy to find them with a calculator.There are I = 3 groups with a total of N = 76 vehicles.

The overall mean of the 76 gas mileages in Table 22.1 (page 598) is

x = n1x1 + n2x2 + n3x3

N

= (31)(27.903) + (31)(22.677) + (14)(21.286)76

= 1865.98476

= 24.552



619Some details of ANOVA

The mean square for groups is

MSG = n1(x1 − x)2 + n2(x2 − x)2 + n3(x3 − x)2

I − 1

= 13 − 1

[(31)(27.903 − 24.552)2 + (31)(22.677 − 24.552)2

+ (14)(21.286 − 24.552)2]

= 606.4242

= 303.212

The mean square for error is

MSE =(n1 − 1)s 2

1 + (n2 − 1)s 22 + (n3 − 1)s 2

3

N − I

= (30)(2.5612) + (30)(3.6732) + (13)(2.7582)73

= 700.37573

= 9.594

Finally, the ANOVA test statistic is

F = MSGMSE

= 303.2129.594

= 31.60

Our work differs slightly from the output in Figure 22.3 because of roundoff error.We don’t recommend doing these calculations, because tedium and roundoff errorscause frequent mistakes.


The calculations of ANOVA use only the sample sizes ni , the samplemeans xi , and the sample standard deviations si . You can thereforere-create the ANOVA calculations when a report gives these summariesbut does not give the actual data. These optional exercises ask you to dothe ANOVA calculations starting with the summary statistics.

22.11 Road rage. Exercise 22.2 describes a study of road rage. Here are themeans and standard deviations for a measure of “angry/threateningdriving” for random samples of drivers in three age-groups:

Age-group n x s

Less than 30 years 244 2.22 3.1130 to 55 years 734 1.33 2.21Over 55 years 364 0.66 1.60

(a) The distributions of responses are somewhat right-skewed.ANOVA is nonetheless safe for these data. Why?

(b) Check that the standard deviations satisfy the guideline forANOVA inference.




(c) Calculate the overall mean response x , the mean squares MSGand MSE, and the ANOVA F statistic.

(d) What are the degrees of freedom for F ? How significant are thedifferences among the three mean responses?

22.12 Exercise and weight loss. What conditions help overweight peopleexercise regularly? Subjects were randomly assigned to threetreatments: a single long exercise period 5 days per week; several10-minute exercise periods 5 days per week; and several 10-minuteperiods 5 days per week on a home treadmill that was provided to thesubjects. The study report contains the following information aboutweight loss (in kilograms) after six months of treatment:8

Treatment n x s

Long exercise periods 37 10.2 4.2Short exercise periods 36 9.3 4.5Short periods with equipment 42 10.2 5.2

(a) Do the standard deviations satisfy the rule of thumb for safe use ofANOVA?

(b) Calculate the overall mean response x and the mean square forgroups MSG.

(c) Calculate the mean square for error MSE.(d) Find the ANOVA F statistic and its approximate P-value. Is

there evidence that the mean weight losses of people who followthe three exercise programs differ?

22.13 Weights of newly hatched pythons. A study of the effect of nesttemperature on the development of water pythons separated pythoneggs at random into nests at three temperatures: cold, neutral, andhot. Exercise 20.22 (page 555) shows that the proportions of eggs thathatched at each temperature did not differ significantly. Now we willexamine the little pythons. In all, 16 eggs hatched at the coldtemperature, 38 at the neutral temperature, and 75 at the hottemperature. The report of the study summarizes the data in thecommon form “mean ± standard error” as follows:9

Temperature n Weight (grams) at hatching Propensity to strike

Cold 16 28.89 ± 8.08 6.40 ± 5.67Neutral 38 32.93 ± 5.61 5.82 ± 4.24Hot 75 32.27 ± 4.10 4.30 ± 2.70

(a) We will compare the mean weights at hatching. Recall that thestandard error of the mean is s/

√n. Find the standard deviations



621Statistics in Summary

of the weights in the three groups and verify that they satisfy ourrule of thumb for using ANOVA.

(b) Starting from the sample sizes ni , the means xi , and the standarddeviations si , carry out an ANOVA. That is, find MSG, MSE, andthe F statistic, and use Table D to approximate the P-value. Isthere evidence that nest temperature affects the mean weight ofnewly hatched pythons?

22.14 Python strikes. The data in the previous exercise also describe the“propensity to strike” of the hatched pythons at 30 days of age. This isthe number of taps on the head with a small brush until the pythonlaunches a strike. (Don’t try this with adult pythons.) The data areagain summarized in the form “sample mean ± standard error of themean.” Follow the outline in (a) and (b) of the previous exercise forpropensity to strike. Does nest temperature appear to influencepropensity to strike?

Chapter 22 SUMMARY

One-way analysis of variance (ANOVA) compares the means of severalpopulations. The ANOVA F test tests the overall H0 that all thepopulations have the same mean. If the F test shows significant differences,examine the data to see where the differences lie and whether they are largeenough to be important.The ANOVA model states that we have an independent SRS from eachpopulation; that each population has a Normal distribution; and that allpopulations have the same standard deviation.

In practice, ANOVA inference is relatively robust when the populations arenon-Normal, especially when the samples are large. Before doing the F test,check the observations in each sample for outliers or strong skewness. Alsoverify that the largest sample standard deviation is no more than twice aslarge as the smallest standard deviation.When the null hypothesis is true, the ANOVA F statistic for comparingI means from a total of N observations in all samples combined has theF distribution with I − 1 and N − I degrees of freedom.ANOVA calculations are reported in an ANOVA table that gives sums ofsquares, mean squares, and degrees of freedom for variation among groupsand for variation within groups. In practice, we use software to do thecalculations.

STATISTICS IN SUMMARY

After studying this chapter, you should be able to do the following.A. RECOGNITION

1. Recognize when testing the equality of several means is helpful inunderstanding data.




2. Recognize that the statistical significance of differences amongsample means depends on the sizes of the samples and on how muchvariation there is within the samples.

3. Recognize when you can safely use ANOVA to compare means.Check the data production, the presence of outliers, and the samplestandard deviations for the groups you want to compare.

B. INTERPRETING ANOVA1. Explain what null hypothesis F tests in a specific setting.2. Locate the F statistic and its P-value on the output of analysis of

variance software.3. Find the degrees of freedom for the F statistic from the number and

sizes of the samples. Use Table D of the F distributions toapproximate the P-value when software does not give it.

4. If the test is significant, use graphs and descriptive statistics to seewhat differences among the means are most important.

Chapter 22 EXERCISES

Exercises 22.15 to 22.19 describe situations in which we want to comparethe mean responses in several populations. For each setting, identify thepopulations and the response variable. Then give I , the ni , and N.Finally, give the degrees of freedom of the ANOVA F test.

22.15 Smoking and sleep. A study of the effects of smoking classifiessubjects as nonsmokers, moderate smokers, or heavy smokers. Theinvestigators interview a sample of 200 people in each group. Amongthe questions is “How many hours do you sleep on a typical night?”

22.16 Which package design is best? A maker of detergents wants tocompare the attractiveness to consumers of six package designs. Eachpackage is shown to 120 different consumers who rate theattractiveness of the design on a 1 to 10 scale.

(Shmuel Thaler/Index Stock Imager/PictureQuest)

22.17 Growing tomatoes. Do four tomato varieties differ in mean yield?Grow 10 plants of each variety and record the yield of each plant inpounds of tomatoes.

(Harry Sieplinga/HMS Images/The ImageBank/Getty Images)

22.18 Strong concrete. The strength of concrete depends on the mixture ofsand, gravel, and cement used to prepare it. A study compares fivedifferent mixtures. Workers prepare six batches of each mixture andmeasure the strength of the concrete made from each batch.

22.19 Teaching sign language. Which of four methods of teachingAmerican Sign Language is most effective? Assign 10 of the 42students in a class at random to each of three methods. Teach theremaining 12 students by the fourth method. Record the students’scores on a standard test of sign language after a semester’s study.



623Chapter 22 Exercises

22.20 Plant diversity. Does the presence of more species of plants increasethe productivity of a natural area, as measured by the total mass ofplant material? An experiment to investigate this question started asfollows:

In a 7-year experiment, we controlled one component of diversity, thenumber of plant species, in 168 plots, each 9 m by 9 m. We seeded theplots, in May 1994, to have 1, 2, 4, 8, or 16 species, with 39, 35, 29,30, and 35 replicates, respectively.10

Total plant mass in each plot was measured in 2000.(a) What null and alternative hypotheses does ANOVA test here?(b) What are the values of N, I , and the ni?(c) What are the degrees of freedom of the ANOVA F statistic?

22.21 Plants defend themselves. When some plants are attacked byleaf-eating insects, they release chemical compounds that attract otherinsects that prey on the leaf-eaters. A study carried out on plantsgrowing naturally in the Utah desert demonstrated both the release ofthe compounds and that they not only repel the leaf-eaters but attractpredators that act as the plants’ bodyguards.11 The investigators chose8 plants attacked by each of three leaf-eaters and 8 more that wereundamaged, 32 plants of the same species in all. They then measuredemissions of several compounds during seven hours. Here are data(mean ± standard error of the mean for eight plants) for onecompound. The emission rate is measured in nanograms (ng) per hour.

Emission rateGroup (ng/hr)

Control 9.22 ± 5.93Hornworm 31.03 ± 8.75Leaf bug 18.97 ± 6.64Flea beetle 27.12 ± 8.62

(a) Make a graph that compares the mean emission rates for the fourgroups. Does it appear that emissions increase when the plant isattacked?

(b) What hypotheses does ANOVA test in this setting?(c) We do not have the full data. What would you look for in deciding

whether you can safely use ANOVA?(d) What is the relationship between the standard error of the mean

(SEM) and the standard deviation for a sample? Do the standarddeviations satisfy our rule of thumb for safe use of ANOVA?

22.22 Does polyester decay? How quickly do synthetic fabrics such aspolyester decay in landfills? A researcher buried polyester strips in the




soil for different lengths of time, then dug up the strips and measuredthe force required to break them. Breaking strength is easy to measureand is a good indicator of decay; lower strength means the fabric hasdecayed.

Part of the study buried 20 polyester strips in well-drained soil inthe summer. Five of the strips, chosen at random, were dug up after2 weeks; another 5 were dug up after 4 weeks, 8 weeks, and 16 weeks.Here are the breaking strengths in pounds:12

2 weeks 118 126 126 120 1294 weeks 130 120 114 126 1288 weeks 122 136 128 146 140

16 weeks 124 98 110 140 110

(a) Find the mean strength for each group and plot the means againsttime. Does it appear that polyester loses strength consistently overtime after it is buried?

(b) Find the standard deviations for each group. Do they meet ourcriterion for applying ANOVA?

(c) In Examples 17.2 and 17.3 (pages 441 and 445), we used thetwo-sample t test to compare the mean breaking strengths forstrips buried for 2 weeks and for 16 weeks. The ANOVA F testextends the two-sample t to more than two groups. Explaincarefully why use of the two-sample t for two of the groups wasacceptable but using the F test on all four groups is not acceptable.

(Alix Phanie, Rex Interstock/StockConnection/PictureQuest)

22.23 Can you hear these words? To test whether a hearing aid is right fora patient, audiologists play a tape on which words are pronounced atlow volume. The patient tries to repeat the words. There are severaldifferent lists of words that are supposed to be equally difficult. Are thelists equally difficult when there is background noise? To find out, anexperimenter had subjects with normal hearing listen to four lists witha noisy background. The response variable was the percent of the50 words in a list that the subject repeated correctly. The data setcontains 96 responses.13 Here are two study designs that could producethese data:

Design A. The experimenter assigns 96 subjects to 4 groups atrandom. Each group of 24 subjects listens to one of the lists. Allindividuals listen and respond separately.Design B. The experimenter has 24 subjects. Each subject listensto all four lists in random order. All individuals listen and respondseparately.

Does Design A allow use of one-way ANOVA to compare the lists?Does Design B allow use of one-way ANOVA to compare the lists?Briefly explain your answers.




Analysis of Variance for PercentSourceListErrorTotal

DF39295

SS920.55738.26658.6

MS306.862.4

Level1234

N24242424

Mean32.75029.66725.25025.583

StDev7.4098.0588.3167.779

F4.92


P0.003

Pooled StDev = 7.898 24.0 28.0 32.0 36.0

Figure 22.10 Minitab ANOVA output for comparing the percents heard correctlyin four lists of words, for Exercise 22.24.

22.24 Can you hear these words? Figure 22.10 displays the Minitab outputfor one-way ANOVA applied to the hearing data described in theprevious exercise. The response variable is “Percent,” and “List”identifies the four lists of words. Based on this analysis, is there goodreason to think that the four lists are not all equally difficult? Write abrief summary of the study findings.

(Craig Aurness/CORBIS)

22.25 How much corn should I plant? How much corn per acre should afarmer plant to obtain the highest yield? Too few plants will give a lowyield. On the other hand, if there are too many plants, they willcompete with each other for moisture and nutrients, and yields willfall. To find out, plant at different rates on several plots of ground andmeasure the harvest. (Treat all the plots the same except for theplanting rate.) Use software to analyze these data from such anexperiment:14

Plants per acre Yield (bushels per acre)

12,000 150.1 113.0 118.4 142.616,000 166.9 120.7 135.2 149.820,000 165.3 130.1 139.6 149.924,000 134.7 138.4 156.128,000 119.0 150.5




(a) Do data analysis to see what the data appear to show about theinfluence of plants per acre on yield and also to check theconditions for ANOVA.

(b) Carry out the ANOVA F test. State hypotheses; give F and itsP-value. What do you conclude?

(c) The observed differences among the mean yields in the samplesare quite large. Why are they not statistically significant?

22.26 Logging in the rainforest: species counts. Table 22.2 (page 604)gives data on the number of trees per forest plot, the number of speciesper plot, and species richness. Exercise 22.3 analyzed the effect oflogging on number of trees. Exercise 22.9 concludes that it would berisky to use ANOVA to analyze richness. Use software to analyze theeffect of logging on the number of species.(a) Make a table of the group means and standard deviations. Do the

standard deviations satisfy our rule of thumb for safe use ofANOVA? What do the means suggest about the effect of loggingon the number of species?

(b) Carry out the ANOVA. Report the F statistic and its P-value andstate your conclusion.

(Eye of Science/Photo Researchers)

22.27 Nematodes and tomato plants. How do nematodes (microscopicworms) affect plant growth? A botanist prepares 16 identical plantingpots and then introduces different numbers of nematodes into thepots. He transplants a tomato seedling into each pot. Here are data onthe increase in height of the seedlings (in centimeters) 16 days afterplanting:15

Nematodes Seedling growth

0 10.8 9.1 13.5 9.21,000 11.1 11.1 8.2 11.35,000 5.4 4.6 7.4 5.0

10,000 5.8 5.3 3.2 7.5

(a) Make a table of means and standard deviations for the fourtreatments. Make side-by-side stemplots to compare thetreatments. What do the data appear to show about the effect ofnematodes on growth?

(b) State H0 and Ha for the ANOVA test for these data, and explainin words what ANOVA tests in this setting.

(c) Use software to carry out the ANOVA. Report your overallconclusions about the effect of nematodes on plant growth.




22.28 Earnings of athletic trainers (Optional). How much do newly hiredathletic trainers earn? Earnings depend on the educational and otherqualifications of the trainer and on the type of job. Here are summariesfrom a sample survey that contacted institutions advertising fortrainers and asked about the people they hired:16

Type of Mean salary Std. dev.institution n x s

High school 57 $26,470 $9507Clinic 108 $30,610 $4504College 94 $30,019 $7158

Calculate the ANOVA table and the F statistic. Are there significantdifferences among the mean salaries for the three types of job in thepopulation of all newly hired trainers? (Comments: The study authorsdid an ANOVA. As is often the case, the published work does notanswer questions about the suitability of the analysis. We wonder ifresponses from 60% of employers who advertised positions generate arandom sample of new hires. The actual data do not appear in thereport. The salary distributions are no doubt right-skewed; we hopethat strong skewness and outliers are absent because the subjects ineach group hold similar jobs. The large samples will then justifyANOVA. Also, the sample standard deviations do not quite satisfy ourrule of thumb for safe use of ANOVA.)

22.29 Plant defenses (Optional). The calculations of ANOVA use only thesample sizes ni , the sample means xi , and the sample standarddeviations si . You can therefore re-create the ANOVA calculationswhen a report gives these summaries but does not give the actual data.Use the information in Exercise 22.21 to calculate the ANOVA table(sums of squares, degrees of freedom, mean squares, and the Fstatistic). Note that the report gives the standard error of the mean(SEM) rather than the standard deviation. Are there significantdifferences among the mean emission rates for the four populations ofplants?

22.30 F versus t (Optional). We have two methods to compare the meansof two groups: the two-sample t test of Chapter 17 and the ANOVAF test with I = 2. We prefer the t test because it allows one-sidedalternatives and does not assume that both populations have the samestandard deviation. Let us apply both tests to the same data.

There are two types of life insurance companies. “Stock” companieshave shareholders, and “mutual” companies are owned by theirpolicyholders. Take an SRS of each type of company from those listedin a directory of the industry. Then ask the annual cost per $1000 of




insurance for a $50,000 policy insuring the life of a 35-year-old manwho does not smoke. Here are the data summaries:17

Stock Mutualcompanies companies

ni 13 17xi $2.31 $2.37si $0.38 $0.58

(a) Calculate the two-sample t statistic for testing H0: µ1 = µ2against the two-sided alternative. Use the conservative method tofind the P-value.

(b) Calculate MSG, MSE, and the ANOVA F statistic for the samehypotheses. What is the P-value of F ?

(c) How close are the two P-values? (The square root of the Fstatistic is a t statistic with N − I = n1 + n2 − 2 degrees offreedom. This is the “pooled two-sample t” mentioned on page454. So F for I = 2 is exactly equivalent to a t statistic, but it is aslightly different t from the one we use.)

22.31 Exercising to lose weight (Optional). Exercise 22.12 describes arandomized, comparative experiment that assigned subjects to threetypes of exercise program intended to help them lose weight. Some ofthe results of this study were analyzed using the chi-square test fortwo-way tables, and some others were analyzed using one-wayANOVA. For each of the following excerpts from the study report, saywhich analysis is appropriate and explain how you made your choice.(a) “Overall, 115 subjects (78% of 148 subjects randomized)

completed 18 months of treatment, with no significant differencein attrition rates between the groups (P = .12).”

(b) “In analyses using only the 115 subjects who completed 18 monthsof treatment, there were no significant differences in weight loss at6 months among the groups.”

(c) “The duration of exercise for weeks 1 through 4 was significantlygreater in the SB compared with both LB and SBEQ groups(P < .05). . . .However, exercise duration was greater in SBEQcompared with both LB and SB groups for months 13 through 18(P < .05).”

Chapter 22 MEDIA EXERCISES

Analysis of variance works by comparing the differences among the samplemeans in several groups to the spread of observations within the groups. A



629Chapter 22 Media Exercises

set of means is significantly different if they are farther apart than randomvariation would allow to happen just by chance. The extent of randomvariation is assessed by the random spread among observations in the samegroup. These exercises use an applet to demonstrate this ANOVA idea.

22.32 ANOVA compares several means. The One-Way ANOVA appletdisplays the observations in three groups, with the group meanshighlighted by black dots. When you open or reset the applet, thescale at the bottom of the display shows that for these groups theANOVA F statistic is F = 31.74, with P < 0.001.

APPLET

(a) The middle group has larger mean than the other two. Grab itsmean point with the mouse. How small can you make F ? Whatdid you do to the mean to make F -small? Roughly how significantis your small F ?

(b) Starting with the three means aligned from your configuration atthe end of (a), drag any one of the group means either up or down.What happens to F ? What happens to P ? Convince yourself thatthe same thing happens if you move any one of the means, or ifyou move one slightly and then another slightly in the oppositedirection.

APPLET

22.33 ANOVA uses within-group variation. Reset the One-Way ANOVAapplet to its original state. As in Figure 22.6(b) (page 607), thedifferences among the three means are highly significant (large F ,small P ) because the observations in each group cluster tightly aboutthe group mean.(a) Use the mouse to slide the Pooled Standard Error at the top of the

display to the right. You see that the group means do not change,but the spread of the observations in each group increases. Whathappens to F and P as the spread among the observations in eachgroup increases? What are the values of F and P when the slider isall the way to the right? This is similar to Figure 22.6(a): variationwithin groups hides the differences among the group means.

(b) Leave the Standard Error slider at the extreme right of its scale, sothat spread within groups stays fixed. Use the mouse to move thegroup means apart. What happens to F and P as you do this?

one-way analysis of variance: comparing several …virtual.yosemite.cc.ca.us/jcurl/math 134 3...

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