Sullivan ANOVA

8/10/2019 Sullivan ANOVA

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are normally distributed with minutes.In a study, thechain reconfigured five restaurants to have a single line andmeasured the wait times for 50 randomly selected customers.The sample standard deviation was determined to be

minute. Does the evidence indicate that the vari-ability in wait time is less for a single line than for multiplelines at the level of significance? Reject

19. Heights of Baseball Players Data obtained from the Nation-al Center for Health Statistics show that men between theages of 20 and 29 have a mean height of 69.3 inches, with astandard deviation of 2.9 inches.A baseball analyst believesthat the standard deviation of heights of major-league base-ball players is less than 2.9 inches.The heights (in inches) of20 randomly selected players are shown in the table.

H0a = 0.05

s = 0.84

s = 1.2

72 74 71 72 76

70 77 75 72 72

77 72 75 70 73

73 75 73 74 74

Source: espn.com

(a) Verify that the data are normally distributed bydrawing a normal probability plot.

(b) Compute the sample standard deviation. 2.059 inches(c) Test the belief at the level of significance.

20. NCAA Softball NCAA rules require the circumferenceof a softball to be inches. A softball manufac-turer bidding on an NCAA contract is shown to meet

the requirements for mean circumference. Suppose theNCAA also requires that the standard deviation of thesoftball circumferences not exceed 0.05 inch. A repre-sentative from the NCAA believes the manufacturerdoes not meet this requirement. She collects a randomsample of 20 softballs from the production line and findsthat Do you believe the balls conform?Use the level of significance?Yes; reject

21. P-Values Determine the exact P-value of the hypothe-sis test in Problem 9.

22. P-Values Determine the exact P-value of the hypothe-sis test in Problem 10. P-value 0.3199

P-value 0.0446

H0a = 0.05s = 0.09 inch.

12 ; 0.125

a = 0.01

Preparing for This Section Before getting started, review the following:

Completely randomized design (Section 1.5,pp. 4243)

Nature of hypothesis testing (Section 10.1,pp. 454460)

Comparing two population means (Section 11.2,pp. 521528)

Normal probability plots (Section 7.4, pp. 354358)

Boxplots (Section 3.5, pp. 161164)

Objectives Verify the requirements to perform a one-way ANOVA

Test hypotheses regarding three or more means using

one-way ANOVA

In Section 11.2, we compared two population means. Just as we extended theconcept of comparing two population proportions (Section 11.3) to comparingthree or more population proportions (Tests for Homogeneity of Proportions,Section 12.2), we now extend the concept of comparing two population meansto comparing three or more population means. The procedure for doing this is

calledAnalysis of Variance, orANOVA for short.

Definition Analysis of Variance (ANOVA) is an inferential method that is used totest the equality of three or more population means.

For example, a family doctor might want to show the mean HDL (so-called good) levels of cholesterol of males in the age groups 20 to 29 years old,40 to 49 years old, and 60 to 69 years old are different. To conduct a hypothe-sis test, we assume the mean HDL cholesterol of each age group is the same.If we call the 20- to 29-year-olds population 1, 40- to 49-year-olds population2, and 60 to 69-year-olds population 3, our null hypothesis would be

H0: m1 = m2 = m3

Note to InstructorIt is recommended that hand computa-tions be de-emphasized in this section.Concentrate on the concepts, and let

technology do the number crunching.

In Other WordsIn ANOVA, the null hypothesis is

always that the means of the differentpopulations are equal. The alternativehypothesis is always that the mean ofat least one population is different fromthe others.

Comparing Three or More Means (One-Way Anaylsis of Variance) C19

4. Comparing Three or More Means(One-Way Analysis of Variance)

(19c) Do not reject H0


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CAUTIONIt is vital that individuals be

randomly assigned to treatments.

CAUTION!Do not test by

conducting three separate hypothesis

tests,because the probability of

making a Type I error will be much

higher than a.

H0: m1 = m2 = m3

versus the alternative hypothesis,

As another example, a medical researcher might want to compare the effectdifferent levels of an experimental drug have on hair growth. The researchermight randomly divide a group of subjects into three different treatment groups.Group 1 might receive a placebo once a day, group 2 might receive 50 mg of theexperimental drug once a day, and group 3 might receive 100 mg of the experi-

mental drug once a day.The researcher then compares the mean numbers of newhair follicles for each of the three treatment groups. The three different treat-ment groups correspond to three different populations.

It is tempting to test the null hypothesis by comparing thepopulation means two at a time using the techniques introduced in Section 11.2.If we proceeded this way,we would need to test three different hypotheses:

Each hypothesis would have a probability of Type I error (rejecting the nullhypothesis when it is true) of If we used an level of significance,each hypothesis would have a 95% probability of rejecting the null hypothesiswhen the alternative hypothesis is true (i.e. a 95% probability of making a cor-rect decision). The probability that all three hypotheses correctly reject the null

hypothesis is (assuming the tests are independent). There is aor 14%, probability that at least one hypothesis

will lead to an incorrect rejection of A 14% probability of a Type I error ismuch higher than the desired 5% probability. As the number of populationsthat are to be compared increases, the probability of making a Type I error usingmultiple t-tests for a given value of also increases.

To address this problem, Sir Ronald A. Fisher (18901962) introduced themethod of analysis of variance.Although it seems strange to name a procedurethat is used to compare means analysis of variance, the justification for the namewill become clear as we see the procedure in action.

The procedure used in this section is called one-way analysis of variance be-cause there is only one factor that distinguishes the various populations in thestudy. For example, in comparing the mean HDL cholesterol levels of males, the

only factor that distinguishes the three groups is age. In comparing the effective-ness of the hair growth drug, the only factor that distinguishes the three groupsis the amount of the experimental drug received (0 mg, 50 mg, or 100 mg).

In performing one-way analysis of variance, care must be taken so that thesubjects are similar in all characteristics except for the level of the treatment.Fisher stated that this is easiest to accomplish through randomization, whichneutralizes the effect of the uncontrolled variables. In the hair growth example,the subjects should be similar in terms of eating habits, age, and so on, by ran-domly assigning the subjects to the three groups.

Verify the Requirements to Perform

a One-Way ANOVATo perform a one-way ANOVA test, certain requirements must be satisfied.

Requirements of a One-Way ANOVA Test

1. There are k simple random samples from k populations.

2. The k samples are independent of each other; that is, the subjects in onegroup cannot be related in any way to subjects in a second group.

3. The populations are normally distributed.

4. The populations have the same variance; that is, each treatment grouphas population variance s2.

a

H0 .1 - 0.953 = 1 - 0.86 = 0.14,

0.953

= 0.86

a = 0.05a.

H0: m1 = m2 and H0: m1 = m3 and H0: m2 = m3

H0: m1 = m2 = m3

H1: At least one of the population means is different from the others

C20 Topics to Discuss


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Suppose we are testing a hypothesis regarding population means sothat the null hypothesis is

and the alternative hypothesis is

Figure 1(a) shows the distribution of each population if the null hypothesis istrue, and Figure 1(b) shows what the distribution of each population might looklike if the alternative hypothesis is true.

H1: at least one of the population means is different from the others

H0: m1 = m2 = m3

k = 3

m1m2m3

(a)m1 m3 m2

(b)

Figure 1

The methods of one-way ANOVA are robust, so small departures from therequirement of normality will not significantly affect the results of the proce-dure. In addition, the requirement of equal population variances does not needto be strictly adhered to, especially if the sample size for each treatment group isthe same.Therefore, it is worthwhile to design an experiment in which the sam-ples from the populations are roughly equal in size.

We can verify the requirement of normality by constructing normal proba-bility plots. The requirement of equal population variances is more difficult toverify. However, a general rule of thumb is as follows:

Verifying the Requirement of Equal Population VariancesThe one-way ANOVA procedures may be used provided that the largestsample standard deviation is no more than two times larger than the small-est sample standard deviation.

Testing the Requirements of One-Way ANOVA

Problem: Researcher Jelodar Gholamali wanted to determine the effective-ness of various treatments on glucose levels of diabetic rats. He randomly as-signed diabetic albino rats into four treatment groups. Group 1 rats served as a

control group and were fed a regular diet. Group 2 rats were served a reg-

ular diet supplemented with a herb, fenugreek. Group 3 rats were served aregular diet supplemented with garlic. Group 4 rats were served a regulardiet supplemented with onion.The basis for the study is that Persian folk-lore states that diets supplemented with fenugreek, garlic, or onion help totreat diabetes.After 15 days of treatment,the blood glucose was measuredin milligrams per deciliter (mg/dL). The results presented in Table 1 arebased on the results published in the article. Verify that the requirementsto perform one-way ANOVA are satisfied. Source: Jelodar, G.A.,et.al.,Effectof Fenugreek, Onion and Garlic on Blood Glucose and Histopathology of Pan-

creas of Alloxan-Induced Diabetic Rats, Indian Journal of Medical Science,

2005;59:6469.

EXAMPLE 1

In Other WordsTry to design experiments that use

ANOVA so that each treatment group isthe same size.

Note to InstructorIt may seem contradictory that we re-quire equal population variances in one-way ANOVA, but are unwilling to make this

assumption when comparing two means.The reason lies in the fact that there isno parametric alternative to one-wayANOVA. The parametric alternative topooling is Welchs t.

Table 1Control Fenugreek Garlic Onion

288.1 229.1 177.4 299.7

296.8 240.7 202.2 258.3

267.8 239.4 163.1 286.8

256.7 207.7 184.7 244.0

292.1 225.7 197.9 267.1

282.9 230.8 164.6 297.1

260.3 206.6 193.9 249.9

283.8 213.3 158.1 265.1



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Variable

Control

Fenugreek

Garlic

Onion

Descriptive Statistics: Control, Fenugreek, Garlic, OnionN*

0

0

0

0

N

8

8

8

8

SE Mean

5.31

4.77

6.03

7.49

Maximum

296.80

240.70

202.20

299.70

Mean

278.56

224.16

180.24

271.00

StDev

15.03

13.49

17.06

21.18

Median

283.35

227.40

181.05

266.10

Q3

291.10

237.25

196.90

294.53

Minimum

256.70

206.60

158.10

244.00

Q1

262.18

209.10

163.48

252.06

Figure 3

(a)

Figure 2

(b) (c) (d)

Approach: We must verify the previously listed requirements.

Solution

1. The rats were randomly assigned to each treatment group.

2. None of the subjects selected is related in any way, so the samples are inde-pendent.

3. Figure 2 shows the normal probability plots for all four treatment

groups. All the normal probability plots are roughly linear, so we con-clude that the sample data come from populations that are approxi-mately normally distributed.

4. The sample standard deviations for each sample are computed usingMINITAB and presented in Figure 3. The largest standard deviation is21.18 mg/dL, and the smallest standard deviation is 13.49 mg/dL. Be-cause the largest standard deviation is not more than two times largerthan the smallest standard deviation the re-quirement of equal population variances is satisfied.

12 # 13.49 = 26.98 7 21.182,

Because all four requirements are satisfied, we can perform a one-wayANOVA.

Test Hypotheses Regarding Three or MoreMeans Using One-Way ANOVA

The computations in performing any analysis of variance test are tedious,so virtual-ly all researchers use statistical software to conduct the test.When using software, it

is easiest to use the P-value approach.The nice thing about P-values is that the de-cision rule is always the same, regardless of the type of hypothesis being tested.

Decision Rule in the One-Way ANOVA Test

If the P-value is less than the level of significance, reject the null hypothesis.

Performing One-Way ANOVA Using Technology

Problem: The researcher in Example 1 wishes to determine if there is a dif-ference in the mean glucose among the four treatment groups. Conduct the testat the level of significance.a = 0.05

EXAMPLE 2

a,



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Approach: We will use MINITAB, Excel, and a TI-84 Plus graphing calcula-tor to test the hypothesis. If the P-value is less than the level of significance, wereject the null hypothesis.

Result: The researcher wants to show there is a difference in the mean glucoseamong the four treatment groups. The null hypothesis is always a statement of nodifference.In this case it is a statement that the mean glucose among the four treat-ment groups is the same. So the null hypothesis is

versus the alternative hypothesis

Figure 4(a) shows the output from MINITAB, Figure 4(b) shows theoutput from Excel, and Figure 4(c) shows the output from a TI-84 Plusgraphing calculator.

H1: at least one of the population means is different from the others

H0: mcontrol = mfenugreek = mgarlic = monion

One-Way ANOVA: Control, Fenugreek, Garlic, Onion

S16.94 R Sq86.18% RSq(adj) 84.70%

Level

Control

Fenugreek

Garlic

Onion

Pooled St Dev 16.94

DF

3

28

31

Source

Factor

Error

Total

N

8

8

8

8

SS

50091

8032

58122

Mean

278.56

224.16

180.24

271.00

175

(a)MINITAB Output

245 280210

MS

16697

287

StDev

15.03

13.49

17.06

21.18

F

58.21

Individual 95% CIs For Mean

Based on Pooled StDev

--+---------+---------+---------+-------

(---*--)

(--*---)

(---*--)

(--*---)

--+---------+---------+---------+-------

P

0.000

Figure 4

(b)Excel Output

(c)TI84 Plus Output

P-value

Comparing Three or More Means (One-Way Anaylsis ofVariance) C23


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A Conceptual Understanding of One-Way ANOVALook again at Figure 4. You may have noticed that each of the three outputsincluded an F-value We now illustrate the idea behind the F-teststatistic. Remember, in testing any hypothesis, the null hypothesis is assumedto be true until the evidence indicates otherwise. In testing the hypothesisregarding k population means, we assume that Thatis, we assume that all k samples come from the same normal population whose

mean is and variance is Table 2 shows the statistics that result by sam-pling from each of the k populations.s

2

.m

m1 = m2 = = mk = m.

1F = 58.212.

Fenugreek

Garlic

Onion

Glucose (mg/dl)

Blood Glucose Levels

Control

225200175150 250 275 300

Figure 5

Now Work Problems 11(a)(d).

Table 2

Sample Standard

Population Sample Size Sample Mean Deviation

1

2

3

k skxknk

oooo

s3x3n3

s2x2n2

s1x1n1

Notice that MINITAB indicates that the P-value is 0.000. This does notmean that the P-value is 0, but instead means that the P-value is less than0.0001. The output of Excel and the TI-84 Plus confirm this by indicating theP-value is Because the P-value is less than the level of signifi-cance, we reject the null hypothesis. There is sufficient evidence to concludethat at least one of the population means for glucose levels is different fromthe others.

Whenever you perform analysis of variance, it is always a good idea to pres-ent visual evidence that supports the conclusions of the test. Side-by-side box-plots are a great way to help see the results of the ANOVA procedure. Figure 5shows the side-by-side boxplots of the data presented in Table 1. The boxplotssupport the ANOVA results from Example 2.

3.7 * 10-12.



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The computation of the F test statistic requires that we understand meansquares. A mean square is an average (mean) of squared values. For example, anyvariance is a mean square.The F-test statistic is the ratio of two mean squares.

If the null hypothesis is true, then each treatment group comes from the samepopulation whose mean is and whose variance is The sample mean of theentire set of data (all treatment groups combined) is a good estimate of We willcall this sample mean Similarly, the sample mean of Sample 1 (or treatment 1)will be the sample mean of Sample 2 (or treatment 2) will be and so on.

Finding an estimate of is somewhat more complicated. One approach isto estimate by computing a measure of variation in sample means from onetreatment group to the next, weighted by the sample size of the correspondingtreatment group.We call this value the mean square due to treatment, denotedMST. It is computed as follows:

(1)

If the null hypothesis is true, then MST is an unbiased estimator of the vari-ance of the population.

A second approach to estimating is to compute the sample variance foreach sample (or treatment), and then to find a weighted average of the sample

variances. We call this the mean square due to error, denoted MSE. The meansquare due to error is an unbiased estimator of whether or not the null hy-pothesis is true. It is computed as follows:

(2)

where n is the total size of the sample. In other words,The F-test statistic is the ratio of the two estimates of

If the null hypothesis is true, both MST and MSE provide unbiased estimates

for the population variance. So, if the null hypothesis is true, we would expectthe F-test statistic to be close to 1.However,if the null hypothesis is not true,at leastone of the sample means from a treatment group will be far away from thesample mean of the entire data set.This will cause MST to be large relative to MSE,which ultimately leads to an F-test statistic substantially larger than 1.

We now present the steps to be used in the computation of the F-test statistic.

Computing theF-Test Statistic

Step 1: Compute the sample mean of the combined data set by adding upall the observations and dividing by the number of observations. Call thisvalue

Step 2: Find the sample mean for each sample (or treatment). Let repre-sent the sample mean of sample 1, represent the sample mean of sample2, and so on.

Step 3: Find the sample variance for each sample (or treatment). Letrepresent the sample variance for sample 1, represent the sample vari-ance for sample 2, and so on.

Step 4: Compute the mean square due to treatment, MST.

Step 5: Compute the mean square due to error, MSE.

Step 6: Compute the F-test statistic:

F =mean square due to treatment

mean square due to error =

MST

MSE

s22

s12

x2

x1

x.

x,

s2

,



MST

MSE

s2.

n = n1 + n2 + + nk .

MSE =1n1 - 12s12 +1n2 - 12s22 + +1nk - 12sk2

n - k

s2

s2

s2,

MST = n11x1 - x22 + n21x2 - x22 + + nk1xk - x22

k - 1

s2

s2

x2 ,x1 ,x.

m.s

2.m



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Computing the F-Test Statistic

Problem: Compute the F-test statistic for the data presented in Example 1.

Approach: We follow Steps 16 just presented.

Solution

Step 1: Compute the mean of the entire data set.

Step 2: Find the sample mean of each treatment. Call the control group popu-lation 1, the fenugreek group population 2, the garlic group population 3, andthe onion group population 4. Then

Step 3: Find the sample variance for each treatment group.

Step 4: Compute MST.

Step 5: Compute MSE.

Step 6: Compute the F-test statistic.

This is the same result provided by the statistical software and graphing calcula-tor in Example 2.



MST

MSE =

16,695.80

286.84 = 58.21

= 286.82

=8030.89

28

MSE =18 - 12225.77 +18 - 12181.99 +18 - 12291.03 +18 - 12448.48

32 - 4

= 16,695.80

=50,087.4112

3

MST =81278.56 - 238.5422 + 81224.16 - 238.5422 + 81180.24 - 238.5422 + 81271 - 238.5422

4 - 1

s42=1299.7 - 27122 +1258.3 - 27122 + +1265.1 - 27122

8 - 1 = 448.58

s32=1177.4 - 180.2422 +1202.2 - 180.2422 + +1158.1 - 180.2422

8 - 1 = 291.03

s22 =

1229.1 - 224.1622 +1240.7 - 224.1622 + +1213.3 - 224.16228 - 1

= 181.99

s12=1288.1 - 278.5622 +1296.8 - 278.5622 + +1283.8 - 278.5622

8 - 1 = 225.77

x3 =177.4 + 202.2 + + 158.1

8 = 180.24 x4 =

299.7 + 258.3 + + 265.1

8 = 271.00

x1 =288.1 + 296.8 + + 283.8

8 = 278.56 x2 =

229.1 + 240.7 + + 213.3

8 = 224.16

x =288.1 + 296.8 + + 249.9 + 265.1

32 =

7633.30

32 = 238.54

EXAMPLE 3



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Looking back at the formula for the mean square due to treatment, we noticethat if the null hypothesis is true the overall mean, should be close to the meanscomputed from each sample (treatment), and so on. If one or more of themeans computed from each sample is substantially different from the overallmean, MST will be large, which in turn makes the F-statistic large. In Example 3,we can see that the control group has a sample mean much larger than the overallmean and the garlic group has a sample meanmuch smaller than the overall mean

The results of the computations that lead to the F-test statistic are present-ed in an ANOVA table, the form of which is shown in Table 3.1

x3 = 180.24 versus x = 238.54

2.

1x1 = 278.56 versus x = 238.542,

x1 , x2 ,x,

Table 3

Source of Sum of Degrees of Mean F-Test

Variation Squares Freedom Squares Statistic

Treatment 50,087.41 16,695.80 58.21

Error 8030.89 286.82

Total 58,118.3 n - 1 = 32 - 1 = 31

n - k = 32 - 4 = 28

k - 1 = 4 - 1 = 3

Notice that the sum of squares treatment is the numerator of the compu-

tation for the mean square due to treatment.The sum of squares error is thenumerator of the computation for the mean square due to error. Each entry inthe mean square column is the sum of squares divided by the corresponding de-grees of freedom. In addition,

This result is a consequence of the requirement of independence of the obser-vations within the groups.

If we were conducting this ANOVA test by hand, we would compare theF-test statistic with a critical F-value. The critical F-value is the F-value whosearea in the right tail is with degrees of freedom in the numerator and

degrees of freedom in the denominator.The critical F-value for the claimmade in Example 1 at the level of significance is Be-

cause the F-test statistic, 58.21, is greater than the critical F, reject the null hy-pothesis. See Figure 6.Suppose the null hypothesis of equal population means is rejected. This

conclusion tells us that at least one of the population means is different from theothers, but we dont know which one. We can determine which populationmeans differ using Tukeys tests, which is not discussed in this text.

F0.05, 3.28 L 2.99.a = 0.05n - k

k - 1a

Sum of squares total = sum of squares treatment + sum of squares error

Area 0.05

F0.05,3,28 2.99

Figure 6

In Other WordsIf we reject the null hypothesis when

doing ANOVA, we are rejecting theassumption that the population means

are all equal. However, the test doesnttell us which means differ.

ASSESS YOUR UNDERSTANDING

Concepts and Vocabulary

1. The acronym ANOVA stands for _____ _____ _____.2. What are the requirements to perform a one-way

ANOVA? Is the test robust?

3. What is the mean square due to treatment estimate ofWhat is the mean square due to error estimate of

4. Why does a large value of the Fstatistic provide evidence

against the null hypothesisH

0: m1 =

m2 = =

mk?

s2?

s2?

Skill Building

In Problems 5 and 6, fill in the ANOVA table.

5. 6.

Source of Sum of Degrees of Mean F-TestVariation Squares Freedom Squares Statistic

Treatment 387 2

Error 8042 27

Total

Source of Sum of Degrees of Mean F-TestVariation Squares Freedom Squares Statistic

Treatment 2814 3

Error 4915 36

Total

Comparing Three or More Means (One-WayAnaylsis of Variance) C27


10/19

11. Corn Production The data in the table represent the num-ber of corn plants in randomly sampled rows (a 17-foot by5-inch strip) for various types of plot. An agricultural re-searcher wants to know whether the mean numbers ofplants for each plot type are different.

NW

In Problems 7 and 8, determine the F-test statistic based on the given summary statistics.

7.

cHint: x = nixini

. d8.

9. The following data represent a simple random sampleof from three populations that are known tobe normally distributed. Verify that the F-test sta-tistic is 2.04.

n = 4

Applying the Concepts

(a) Write the null and alternative hypotheses.(b) State the requirements that must be satisfied to use

the one-way ANOVA procedure.(c) Use the following partial MINITAB output to test

the hypothesis at the level of significance.Reject H0

a = 0.05

(d) Shown are side-by-side boxplots of each type of plot.Do these boxplots support the results obtained inpart (c)?

(e) Verify that the F-test statistic is 7.10.

12. Soybean Yield The data in the table represent the num-ber of pods on a random sample of soybean plants for var-ious plot types. An agricultural researcher wants to know

whether the mean numbers of pods for each plot type aredifferent.

Sample Sample SamplePopulation Size Mean Variance

1 10 40 48

2 10 42 31

3 10 44 25

Sample Sample SamplePopulation Size Mean Variance

1 15 105 34

2 15 110 40

3 15 108 30

4 15 90 38

Sample 1 Sample 2 Sample 3

28 22 25

23 25 24

30 17 19

27 23 30

Sample 1 Sample 2 Sample 3

73 67 72

82 77 80

82 66 87

81 67 77

97 83 96

Plot Type Number of Plants

Sludge plot 25 27 33 30 28 27

Spring disk 32 30 33 35 34 34

No till 30 26 29 32 25 29

Source: Andrew Dieter and Brad Schmidgall. Joliet Junior College

One-way ANOVA: Sludge Plot, Spring Disk, No TillDF

2

15

17

Source

Factor

Error

Total

MS

42.06

5.92

SS

84.11

88.83

172.94

F

7.10

P

0.007

25

Sludge Plot

Number of Plants

PlotType

30 35

Spring Disk

No Till

Plot Type Pods

Liberty 32 31 36 35 41 34 39 37 38

No till 34 30 31 27 40 33 37 42 39

Chisel plowed 34 37 24 23 32 33 27 34 30

Source: Andrew Dieter and Brad Schmidgall, Joliet Junior College

10. The following data represent a simple random sample offrom three populations that are known to be nor-

mally distributed.Verify that the F-test statistic is 2.599.n = 5



11/19


the one-way ANOVA procedure.(c) Use the following MINITAB output to test the hy-

pothesis at the level of significance.Reject H0

a = 0.05


(e) Verify that the F-test statistic is 3.77.(f) Based on the boxplots, which type of plot appears to

have a significantly different mean number of plants?Liberty or No Till

13. Births by Day of Week An obstetrician knew that therewere more live births during the week than on weekends.She wanted to determine whether the mean number ofbirths was different for each of the five days of the week.She randomly selected eight dates for each of the five daysof the week and obtained the following data:


the one-way ANOVA procedure.(c) Use the following MINITAB output to determine if

the number of births differs by day of the weekusing the level of significance:a = 0.01


(e) Verify that the F-test statistic is 9.80.(f) Based on the boxplots, which day of the week ap-

pears to have a significantly different number ofbirths? Monday

14. Punkin Chunkin The World Championship PunkinChunkin contest is held every fall in Millsboro, Delaware.Contestants build devices meant to hurl 8 to 10-poundpumpkins across a field. One class of entry is the air can-non, which must use compressed air to fire a pumpkin.The following data represent a simple random sample ofdistances that pumpkins have traveled (in feet) for theyears 2001 to 2004. Is there evidence to conclude that the

One-way ANOVA: Liberty, No Till Chisel PlowedDF

2

24

26

Source

Factor

Error

Total

MS

74.5

19.8

SS

149.0

474.7

623.6

F

3.77

P

0.038

20

Liberty

Number of Plants

PlotType

30 40

No Till

Chisel Plowed

Monday Tuesday Wednesday Thursday Friday

10,456 11,621 11,084 11,171 11,54510,023 11,944 11,570 11,745 12,321

10,691 11,045 11,346 12,023 11,749

10,283 12,927 11,875 12,433 12,192

10,265 12,577 12,193 12,132 12,422

11,189 11,753 11,593 11,903 11,627

11,198 12,509 11,216 11,233 11,624

11,465 13,521 11,818 12,543 12,543

Source: National Center for Health Statistics

One-way ANOVA: Mon, Tues, Wed, Thurs, FriDF

4

35

39

Source

Factor

Error

Total

MS

2876908

293451

SS

11507633

10270781

21778414

F

9.80

P

0.000

Wednesday

Tuesday

Monday

Number of Births

Day

Thursday

Friday

10,000 11,000 12,000 13,000



12/19

mean distance that a pumpkin is fired is different for thevarious years?


the one-way ANOVA procedure.(c) Use the following MINITAB output to test the claim

at the level of significance.a = 0.05


(e) Verify that the F-test statistic is 2.82.

15. Rates of Return A stock analyst wondered whether themean rate of return of financial, energy, and utility stocksdiffered over the past 5 years. He obtained a simple ran-dom sample of eight companies from each of the three

sectors and obtained the 5-year rates of return shown inthe following table (in percent):

(a) State the null and alternative hypothesis.(b) Verify that the requirements to use the one-way

ANOVA procedure are satisfied. Normal probabilityplots indicate that the sample data come from normalpopulations.

(c) Test whether the mean rates of return are different at

the level of significance. Do not reject(d) Draw boxplots of the three sectors to support the re-

sults obtained in part (c).

16. Reaction Time In an online psychology experiment spon-sored by the University of Mississippi, researchers askedstudy participants to respond to various stimuli. Partici-pants were randomly assigned to one of three groups.Sub-

jects in group 1 were in the simple group. They wererequired to respond as quickly as possible after a stimuluswas presented. Subjects in group 2 were in the go/no-gogroup. These subjects were required to respond to a par-ticular stimulus while disregarding other stimuli. Finally,subjects in group 3 were in the choice group.They neededto respond differently, depending on the stimuli present-ed. Depending on the type of whistle sound, the subjectmust press a certain button.The reaction time (in seconds)for each stimulus is presented in the table.

H0a = 0.05

The researcher wants to determine if the mean reactiontimes for each stimulus are different.

(a) State the null and alternative hypotheses.(b) Verify that the requirements to use the one-way

ANOVA procedure are satisfied. Normal probabilityplots indicate that the sample data come from a nor-mal population.

(c) Test whether the mean reaction times to the three stim-uli differ at the level of significance.

(d) Draw boxplots of the three stimuli to support the ana-lytic results obtained in part (c).

a = 0.05

2001 2002 2003 2004

3494.70 3881.54 3895.85 4224.00

3360.20 3232.74 4434.28 4065.60

3911.02 3696.19 3448.56 3967.95

3124.40 3414.54 3665.89 3596.09

3718.77 3816.64 3539.66 3970.00

3453.12 3631.78 3591.16 3877.68

Source: www.punkinchunkin.com/main.htm

One-way ANOVA: 2001, 2002, 2003, 2004DF

3

20

23

Source

Factor

Error

Total

MS

219747

77973

SS

659242

1559456

2218698

F

2.82

P

0.065

2002

2003

2004

Distance (feet)

Boxplot of 2001, 2002, 2003, 2004

2001

3600340032003000 3800 4000 4200 4400 4600

Financial Energy Utilities

10.76 12.72 11.88

15.05 13.91 5.86

17.01 6.43 13.46

5.07 11.19 9.9019.50 18.79 3.95

8.16 20.73 3.44

10.38 9.60 7.11

6.75 17.40 15.70

Source: Morningstar.com

Simple Go/No Go Choice

0.430 0.588 0.561

0.498 0.375 0.498

0.480 0.409 0.519

0.376 0.613 0.538

0.402 0.481 0.464

0.329 0.355 0.725

Source: PsychExperiments; The University of Mississippi;

www.olemiss.edu/psychexps/

(16a) vs. at least one mean is not equal (16b) Do not rejectH

0H

1:H

0:m

S m

G m

C



13/19

(a) The researcher wants to know if the means for chest compression for each classof vehicle differ. State the null and alternative hypotheses.

(b) Verify that the requirements to use the one-way ANOVA procedure are satis-fied. Normal probability plots indicate that the sample data come from normalpopulations.

(c) Test whether the mean chest compression for each vehicle type is different atthe level of significance. Do not reject

(d) Draw boxplots of the three types of vehicle to support the analytic results ob-tained in part (c).

18. Crash Data The Insurance Institute for Highway Safety conducts experiments in

which cars are crashed into a fixed barrier at 40 mph. In the Institutes 40-mph off-set test, 40% of the total width of each vehicle strikes a barrier on the drivers side.The barriers deformable face is made of aluminum honeycomb, which makes theforces in the test similar to those involved in a frontal offset crash between two ve-hicles of the same weight, each going just less than 40 mph. Suppose you are in themarket to buy a new family car. You want to know if the mean head injury resultingfrom this offset crash is the same for large family cars, passenger vans, and midsizeutility vehicles. The following data were collected from the institutes study.

H0a = 0.01

17. Crash Data The Insurance Institute for Highway Safety conducts experiments inwhich cars are crashed into a fixed barrier at 40 mph. In the Institutes 40-mph offsettest, 40% of the total width of each vehicle strikes a barrier on the drivers side.Thebarriers deformable face is made of aluminum honeycomb, which makes the forces inthe test similar to those involved in a frontal offset crash between two vehicles of thesame weight, each going just less than 40 mph.Suppose you are in the market to buy anew family car.You want to know whether the mean chest compression resulting fromthis offset crash is the same for large family cars,passenger vans,and midsize utility ve-hicles.The following data were collected from the institutes study.

(a) The researcher wants to know whether the means for head injury for each classof vehicle differ. State the null and alternative hypotheses.

Chest Chest Chest Large Compression Compression Midsize CompressionFamily Cars (mm) Passenger Vans (mm) Utility Vehicles (mm)

Hyundai XG350 33 Toyota Sienna 29 Honda Pilot 31

Ford Taurus 28 Honda Odyssey 28 Toyota 4Runner 36

Buick LeSabre 28 Ford Freestar 27 Mitsubishi Endeavor 35

Chevrolet Impala 26 Mazda MPV 30 Nissan Murano 29

Chrysler 300 34 Chevrolet Uplander 26 Ford Explorer 29

Pontiac Grand Prix 34 Nissan Quest 33 Jeep Liberty 36

Toyota Avalon 31 Kia Sedona 21 Buick Randezvous 29

Source: Insurance Institute for Highway Safety

Head Head HeadLarge Injury Injury Midsize Injury Family Cars (hic) Passenger Vans (hic) Utility Vehicles (hic)

Hyundai XG350 264 Toyota Sienna 148 Honda Pilot 202

Ford Taurus 170 Honda Odyssey 238 Toyota 4Runner 216

Buick LeSabre 409 Ford Freestar 340 Mitsubishi Endeavor 186

Chevrolet Impala 204 Mazda MPV 693 Nissan Murano 517

Chrysler 300 149 Chevrolet Uplander 550 Ford Explorer 202

Pontiac Grand Prix 627 Nissan Quest 470 Kia Sorento 552

Toyota Avalon 166 Kia Sedona 332 Chevy Trailblazer 386

Source: Insurance Institute for Highway Safety

(17a) vs. at least one mean is not equalH1:H0: mL mP mM

(18a) vs. at least one of the means is not equalH1:H0: mL mP mM



14/19

(b) Verify that the requirements to use the one-wayANOVA procedure are satisfied. Normal probabilityplots indicate that the sample data come from normalpopulations.

(c) Test whether the mean head injury for each vehicletype differs at the level of significance. Donot reject

(d) Draw boxplots of the three vehicle types to supportthe analytic results obtained in part (c).

19. pH in Rain An environmentalist wanted to determineif the mean acidity of rain differed among Alaska,Florida, and Texas. He randomly selected six rain datesat each of the three locations and obtained the data inthe following table.

H0

a = 0.01

(a) State the null and alternative hypothesis.(b) Verify that the requirements to use the one-way

ANOVA procedure are satisfied. Normal probability

plots indicate that the sample data come from a nor-mal population.

(c) Test if the mean pHs in the rainwater are different atthe level of significance. Reject

(d) Draw boxplots of the pH in rain for the three states tosupport the results obtained in part (c).

20. Lower Your Cholesterol Researchers FranciscoFuentes and his colleagues wanted to determine themost effective diet for reducing LDL cholesterol, the so-called bad cholesterol, among three diets: (1) asaturated-fat diet, (2) the Mediterranean diet, and(3) the U.S. National Cholesterol Education Program or

NCEP-1 Diet. The participants in the study were shownto have the same levels of LDL cholesterol before thestudy. Participants were randomly assigned to one of thethree treatment groups. Individuals in group 1 receivedthe saturated fat diet, which is 15% protein, 47% carbo-hydrates, 38% fat (20% saturated fat, 12% monounsatu-rated fat, and 6% polyunsaturated fat). Individuals ingroup 2 received the Mediterranean diet, which is 47%carbohydrates, 38% fat ( saturated fat, 22% mo-nounsaturated fat, and 6% polyunsaturated fat). Indi-

610%

H0a = 0.05

(a) Does the evidence suggest the cholesterol levels dif-fer? State the null and alternative hypothesis.

(b) Verify that the requirements to use the one-wayANOVA procedure are satisfied. Normal probabilityplots indicate that the sample data come from normalpopulations.

(c) Test if the mean LDL cholesterol levels are different at

the level of significance. Reject(d) Draw boxplots of the three LDL cholesterol levels for

the three groups to support the analytic results ob-tained in part (c).

21. Concrete Strength An engineer wants to know if the meanstrengths of three different concrete mix designs differ sig-nificantly. He randomly selects 9 cylinders that measure 6inches in diameter and 12 inches in height in which mixture67-0-301 is poured, 9 cylinders of mixture 67-0-400, and 9cylinders of mixture 67-0-353. After 28 days, he measuresthe strength (in pounds per square inch) of the cylinders.The results are presented in the following table:

H0a = 0.05

Alaska Florida Texas

5.41 4.87 5.46

5.39 5.18 6.29

4.90 4.40 5.57

5.14 5.12 5.15

4.80 4.89 5.45

5.24 5.06 5.30

Source: National Atmospheric Deposition Program

Saturated Fat Mediterranean NCEP-1

245 56 125123 78 100

166 101 140

104 158 151

196 145 138

300 118 268

140 145 75

240 211 71

218 131 184

173 125 116

223 160 144

177 130 101

193 83 135

224 263 144

149 150 130

Mixture Mixture Mixture67-0-301 67-0-400 67-0-353

3960 4090 4070 4120 4150 3820

4040 3830 4330 4640 3820 3750

3780 3940 4620 4190 4010 3990

3890 4080 3730 3850 4150 4320

3990 4890 4190

(19a) vs. at least one of themeans is not equal

H1:H0: mA mF mT

viduals in group 3 received the NCEP-1 Diet (saturated fat, 12% monounsaturated fat, and 6%polyunsaturated fat). After 28 days, their LDL choles-terol levels were recorded. The data in the followingtable are based on this study.

610%

(20a) vs. at least one of the means isnot equal

H1:H0: mSF mM mNCE



15/19

(a) State the null and alternative hypotheses.(b) Explain why we cannot use one-way ANOVA to test

these hypotheses.

22. Analyzing Journal Article Results Researchers (Brian G.Feagan et al., Erythropoietin with Iron Supplementationto Prevent Allogeneic Blood Transfusion in Total HipJoint Arthroplasty,Annals of Internal Medicine, Vol. 133,

No. 11) wanted to determine whether epoetin alfa was ef-fective in increasing the hemoglobin concentration in pa-tients undergoing hip arthroplasty. The researchersscreened patients for eligibility by performing a completemedical history and physical of the patients. Once eligiblepatients were identified, the researchers used a computer-generated schedule to assign the patients to the high-doseepoetin group, low-dose epoetin group, or placebo group.

The study was double-blind. Based on an analysis of vari-ance, it was determined that there were significant differ-ences in the increase in hemoglobin concentration in thethree groups with a P-value less than 0.001.The mean in-crease in hemoglobin in the high-dose epoetin group was19.5 g/L, the mean increase in hemoglobin in the low-doseepoetin group was 17.2 g/L, and mean increase in hemo-globin in the placebo group was 1.2 g/L.

(a) Why do you think it was necessary to screen patientsfor eligibility?(b) Why was a computer-generated schedule used to as-

sign patients to the various treatment groups?(c) What does it mean for a study to be double-blind?

Why do you think the researchers desired a double-blind study?

(d) Interpret the reported P-value.

ANOVA

TI-83/84 PlusStep 1: Enter the raw data into L1, L2, L3, and so on, for each population ortreatment.

Step 2: Press STAT, highlight TESTS, and select F:ANOVA(.

Step 3: Enter the list names for each sample or treatment after ANOVA(.For example, if there are three treatments in L1, L2, and L3, enter

ANOVA(L1,L2,L3)

Press ENTER.MINITAB Step 1: Enter the raw data into C1, C2, C3, and so on, for each sample or

treatment.

Step 2: Select Stat, then highlight ANOVA, and select One-way (Unstacked).

Step 3: Enter the column names in the cell marked Responses. Click OK.

Excel Step 1: Enter the raw data in columns A, B, C, and so on, for each sample ortreatment.

Step 2: Be sure the Data Analysis Tool Pak is activated. This is done by se-lecting the Tools menu and highlighting Add-Ins Check the box for theAnalysis ToolPak and select OK. Select Tools, then highlight Data Analysis.Select ANOVA: Single Factor and click OK.

Step 3: With the cursor in the Input Range: cell, highlight the data. ClickOK.

.

Technology Step by Step

(21a) vs. atleast one of the means is not equal

H1:H0: m670301 m670400 m670353



16/19

0.100 39.86 49.59 53.59 55.83 57.24 58.20 58.91 59.44 0.050 161.45 199.50 215.71 224.58 230.16 233.99 236.77 238.881 0.025 647.79 799.50 864.16 899.58 921.85 937.11 948.22 956.66 0.010 4052.20 4999.50 5403.40 5624.60 5763.60 5859.00 5928.40 5981.10 0.001 405284.00 500000.00 540379.00 562500.00 576405.00 585937.00 592873.00 598144.00

0.100 8.53 9.00 9.16 9.24 9.29 9.33 9.35 9.370.050 18.51 19.00 19.16 19.25 19.30 19.33 19.35 19.37

2 0.025 38.51 39.00 39.17 39.25 39.30 39.33 39.36 39.37

0.010 98.50 99.00 99.17 99.25 99.30 99.33 99.36 99.37 0.001 998.50 999.00 999.17 999.25 999.30 999.33 999.36 999.37

0.100 5.54 5.46 5.39 5.34 5.31 5.28 5.27 5.25 0.050 10.13 9.55 9.28 9.12 9.01 8.94 8.89 8.853 0.025 17.44 16.04 15.44 15.10 14.88 14.73 14.62 14.54 0.010 34.12 30.82 29.46 28.71 28.24 27.91 27.67 27.49 0.001 167.03 148.50 141.11 137.10 134.58 132.85 131.58 130.62

0.100 4.54 4.32 4.19 4.11 4.05 4.01 3.98 3.95 0.050 7.71 6.94 6.59 6.39 6.26 6.16 6.09 6.044 0.025 12.22 10.65 9.98 9.60 9.36 9.20 9.07 8.98 0.010 21.20 18.00 16.69 15.98 15.52 15.21 14.98 14.80 0.001 74.14 61.25 56.18 53.44 51.71 50.53 49.66 49.00

0.100 4.06 3.78 3.62 3.52 3.45 3.40 3.37 3.34 0.050 6.61 5.79 5.41 5.19 5.05 4.95 4.88 4.82

5 0.025 10.01 8.43 7.76 7.39 7.15 6.98 6.85 6.76 0.010 16.26 13.27 12.06 11.39 10.97 10.67 10.46 10.29 0.001 47.18 37.12 33.20 31.09 29.75 28.83 28.16 27.65

0.100 3.78 3.46 3.29 3.18 3.11 3.05 3.01 2.98 0.050 5.99 5.14 4.76 4.53 4.39 4.28 4.21 4.156 0.025 8.81 7.26 6.60 6.23 5.99 5.82 5.70 5.60 0.010 13.75 10.92 9.78 9.15 8.75 8.47 8.26 8.10 0.001 35.51 27.00 23.70 21.92 20.80 20.03 19.46 19.03

0.100 3.59 3.26 3.07 2.96 2.88 2.83 2.78 2.75 0.050 5.59 4.74 4.35 4.12 3.97 3.87 3.79 3.737 0.025 8.07 6.54 5.89 5.52 5.29 5.12 4.99 4.90 0.010 12.25 9.55 8.45 7.85 7.46 7.19 6.99 6.84 0.001 29.25 21.69 18.77 17.20 16.21 15.52 15.02 14.63

0.100 3.46 3.11 2.92 2.81 2.73 2.67 2.62 2.59 0.050 5.32 4.46 4.07 3.84 3.69 3.58 3.50 3.448 0.025 7.57 6.06 5.42 5.05 4.82 4.65 4.53 4.43 0.010 11.26 8.65 7.59 7.01 6.63 6.37 6.18 6.03 0.001 25.41 18.49 15.83 14.39 13.48 12.86 12.40 12.05

F-Distribution Critical Values

Table VII

1 2 3 4 5 6 7 8

Degrees of Freedom in the Numerator

Area

F

DegreesofFreedomi

ntheDenominator

Area in

Right Tail



17/19

0.100 59.86 60.19 61.22 61.74 62.26 62.79 63.06 63.30 0.050 240.54 241.88 245.95 248.01 250.10 252.20 253.25 254.191 0.025 963.28 968.63 984.87 993.10 1001.4 1009.8 1014.0 1017.7 0.010 6022.5 6055.8 6157.3 6208.7 6260.6 6313.0 6339.4 6362.7 0.001 602284.0 605621.0 615764.0 620908.0 626099.0 631337.0 633972.0 636301.0

0.100 9.38 9.39 9.42 9.44 9.16 9.47 9.48 9.490.050 19.38 19.40 19.43 19.45 19.46 19.48 19.49 19.49

2 0.025 39.39 39.40 39.43 39.45 39.46 39.48 39.49 39.50 0.010 99.39 99.40 99.43 99.45 99.47 99.48 99.49 99.50 0.001 999.39 999.40 999.43 999.45 999.47 999.48 999.49 999.50

0.100 5.24 5.23 5.20 5.18 5.17 5.15 5.14 5.13 0.050 8.81 8.79 8.70 8.66 8.62 8.57 8.55 8.533 0.025 14.47 14.42 14.25 14.17 14.08 13.99 13.95 13.91 0.010 27.35 27.23 26.87 26.69 26.50 26.32 26.22 26.14 0.001 129.86 129.25 127.37 126.42 125.45 124.47 123.97 123.53

0.100 3.94 3.92 3.87 3.84 3.82 3.79 3.78 3.76 0.050 6.00 5.96 5.86 5.80 5.75 5.69 5.66 5.634 0.025 8.90 8.84 8.66 8.56 8.46 8.36 8.31 8.26 0.010 14.66 14.55 14.20 14.02 13.84 13.65 13.56 13.47 0.001 48.47 48.05 46.76 46.10 45.43 44.75 44.40 44.09

0.100 3.32 3.30 3.24 3.21 3.17 3.14 3.12 3.11 0.050 4.77 4.74 4.62 4.56 4.50 4.43 4.40 4.375 0.025 6.68 6.62 6.43 6.33 6.23 6.12 6.07 6.02 0.010 10.16 10.05 9.72 9.55 9.38 9.20 9.11 9.03 0.001 27.24 26.92 25.91 25.39 24.87 24.33 24.06 23.82

0.100 2.96 2.94 2.87 2.84 2.80 2.76 2.74 2.72 0.050 4.10 4.06 3.94 3.87 3.81 3.74 3.70 3.676 0.025 5.52 5.46 5.27 5.17 5.07 4.96 4.90 4.86 0.010 7.98 7.87 7.56 7.40 7.23 7.06 6.97 6.89 0.001 18.69 18.41 17.56 17.12 16.67 16.21 15.98 15.77

0.100 2.72 2.70 2.63 2.59 2.56 2.51 2.49 2.47 0.050 3.68 3.64 3.51 3.44 3.38 3.30 3.27 3.237 0.025 4.82 4.76 4.57 4.47 4.36 4.25 4.20 4.15 0.010 6.72 6.62 6.31 6.16 5.99 5.82 5.74 5.66 0.001 14.33 14.08 13.32 12.93 12.53 12.12 11.91 11.72

0.100 2.56 2.54 2.46 2.42 2.38 2.34 2.32 2.30 0.050 3.39 3.35 3.22 3.15 3.08 3.01 2.97 2.938 0.025 4.36 4.30 4.10 4.00 3.89 3.78 3.73 3.68 0.010 5.91 5.81 5.52 5.36 5.20 5.03 4.95 4.87 0.001 11.77 11.54 10.84 10.48 10.11 9.73 9.53 9.36


Table VII (continued)


Area

F

9 10 15 20 30 60 120 1000

Area in

Right Tail

DegreesofFreedomi

ntheDenominator

Comparing Three or More Means (One-WayAnaylsis ofVariance) C35


18/19

0.100 3.36 3.01 2.81 2.69 2.61 2.55 2.51 2.47 2.44 2.42

0.050 5.12 4.26 3.86 3.63 3.48 3.37 3.29 3.23 3.18 3.14 9 0.025 7.21 5.71 5.08 4.72 4.48 4.32 4.20 4.10 4.03 3.96 0.010 10.56 8.02 6.99 6.42 6.06 5.80 5.61 5.47 5.35 5.26 0.001 22.86 16.39 13.90 12.56 11.71 11.13 10.70 10.37 10.11 9.89

0.100 3.29 2.92 2.73 2.61 2.52 2.46 2.41 2.38 2.35 2.320.050 4.96 4.10 3.71 3.48 3.33 3.22 3.14 3.07 3.02 2.98

10 0.025 6.94 5.46 4.83 4.47 4.24 4.07 3.95 3.85 3.78 3.72 0.010 10.04 7.56 6.55 5.99 5.64 5.39 5.20 5.06 4.94 4.85 0.001 21.04 14.91 12.55 11.28 10.48 9.93 9.52 9.20 8.96 8.75

0.100 3.18 2.81 2.61 2.48 2.39 2.33 2.28 2.24 2.21 2.19 0.050 4.75 3.89 3.49 3.26 3.11 3.00 2.91 2.85 2.80 2.75 12 0.025 6.55 5.10 4.47 4.12 3.89 3.73 3.61 3.51 3.44 3.37 0.010 9.33 6.93 5.95 5.41 5.06 4.82 4.64 4.50 4.39 4.30 0.001 18.64 12.97 10.80 9.63 8.89 8.38 8.00 7.71 7.48 7.29

0.100 3.07 2.70 2.49 2.36 2.27 2.21 2.16 2.12 2.09 2.06 0.050 4.54 3.68 3.29 3.06 2.90 2.79 2.71 2.64 2.59 2.54 15 0.025 6.20 4.77 4.15 3.80 3.58 3.41 3.29 3.20 3.12 3.06 0.010 8.68 6.36 5.42 4.89 4.56 4.32 4.14 4.00 3.89 3.80 0.001 16.59 11.34 9.34 8.25 7.57 7.09 6.74 6.47 6.26 6.08

0.100 2.97 2.59 2.38 2.25 2.16 2.09 2.04 2.00 1.96 1.94 0.050 4.35 3.49 3.10 2.87 2.71 2.60 2.51 2.45 2.39 2.35 20 0.025 5.87 4.46 3.86 3.51 3.29 3.13 3.01 2.91 2.84 2.77 0.010 8.10 5.85 4.94 4.43 4.10 3.87 3.70 3.56 3.46 3.37 0.001 14.82 9.95 8.10 7.10 6.46 6.02 5.69 5.44 5.24 5.08

0.100 2.92 2.53 2.32 2.18 2.09 2.02 1.97 1.93 1.89 1.87 0.050 4.24 3.39 2.99 2.76 2.60 2.49 2.40 2.34 2.28 2.24 25 0.025 5.69 4.29 3.69 3.35 3.13 2.97 2.85 2.75 2.68 2.61 0.010 7.77 5.57 4.68 4.18 3.85 3.63 3.46 3.32 3.22 3.13

0.001 13.88 9.22 7.45 6.49 5.89 5.46 5.15 4.91 4.71 4.56 0.100 2.81 2.41 2.20 2.06 1.97 1.90 1.84 1.80 1.76 1.73 0.050 4.03 3.18 2.79 2.56 2.40 2.29 2.20 2.13 2.07 2.03 50 0.025 5.34 3.97 3.39 3.05 2.83 2.67 2.55 2.46 2.38 2.32 0.010 7.17 5.06 4.20 3.72 3.41 3.19 3.02 2.89 2.78 2.70 0.001 12.22 7.96 6.34 5.46 4.90 4.51 4.22 4.00 3.82 3.67

0.100 2.76 2.36 2.14 2.00 1.91 1.83 1.78 1.73 1.69 1.66 0.050 3.94 3.09 2.70 2.46 2.31 2.19 2.10 2.03 1.97 1.93 100 0.025 5.18 3.83 3.25 2.92 2.70 2.54 2.42 2.32 2.24 2.18 0.010 6.90 4.82 3.98 3.51 3.21 2.99 2.82 2.69 2.59 2.50 0.001 11.50 7.41 5.86 5.02 4.48 4.11 3.83 3.61 3.44 3.30

0.100 2.73 2.33 2.11 1.97 1.88 1.80 1.75 1.70 1.66 1.63 0.050 3.89 3.04 2.65 2.42 2.26 2.14 2.06 1.98 1.93 1.88 200 0.025 5.10 3.76 3.18 2.85 2.63 2.47 2.35 2.26 2.18 2.11

0.010 6.76 4.71 3.88 3.41 3.11 2.89 2.73 2.60 2.50 2.41 0.001 11.15 7.15 5.63 4.81 4.29 3.92 3.65 3.43 3.26 3.12

0.100 2.71 2.31 2.09 1.95 1.85 1.78 1.72 1.68 1.64 1.61 0.050 3.85 3.00 2.61 2.38 2.22 2.11 2.02 1.95 1.89 1.841000 0.025 5.04 3.70 3.13 2.80 2.58 2.42 2.30 2.20 2.13 2.06 0.010 6.66 4.63 3.80 3.34 3.04 2.82 2.66 2.53 2.43 2.34 0.001 10.89 6.96 5.46 4.65 4.14 3.78 3.51 3.30 3.13 2.99




Degrees

ofFreedomi

ntheDenominator

1 2 3 4 5 6 7 8 9 10Area inRight Tail



19/19

0.100 2.38 2.34 2.30 2.27 2.25 2.23 2.22 2.21 2.18 2.16

0.050 3.07 3.01 2.94 2.89 2.86 2.83 2.80 2.79 2.75 2.71 9 0.025 3.87 3.77 3.67 3.60 3.56 3.51 3.47 3.45 3.39 3.34 0.010 5.11 4.96 4.81 4.71 4.65 4.57 4.52 4.48 4.40 4.32 0.001 9.57 9.24 8.90 8.69 8.55 8.37 8.26 8.19 8.00 7.84

0.100 2.28 2.24 2.20 2.17 2.16 2.13 2.12 2.11 2.08 2.060.050 2.91 2.85 2.77 2.73 2.70 2.66 2.64 2.62 2.58 2.54

10 0.025 3.62 3.52 3.42 3.35 3.31 3.26 3.22 3.20 3.14 3.09 0.010 4.71 4.56 4.41 4.31 4.25 4.17 4.12 4.08 4.00 3.92 0.001 8.45 8.13 7.80 7.60 7.47 7.30 7.19 7.12 6.94 6.78

0.100 2.15 2.10 2.06 2.03 2.01 1.99 1.97 1.96 1.93 1.91 0.050 2.69 2.62 2.54 2.50 2.47 2.43 2.40 2.38 2.34 2.30 12 0.025 3.28 3.18 3.07 3.01 2.96 2.91 2.87 2.85 2.79 2.73 0.010 4.16 4.01 3.86 3.76 3.70 3.62 3.57 3.54 3.45 3.37 0.001 7.00 6.71 6.40 6.22 6.09 5.93 5.83 5.76 5.59 5.44

0.100 2.02 1.97 1.92 1.89 1.87 1.85 1.83 1.82 1.79 1.76 0.050 2.48 2.40 2.33 2.28 2.25 2.20 2.18 2.16 2.11 2.07 15 0.025 2.96 2.86 2.76 2.69 2.64 2.59 2.55 2.52 2.46 2.40 0.010 3.67 3.52 3.37 3.28 3.21 3.13 3.08 3.05 2.96 2.88 0.001 5.81 5.54 5.25 5.07 4.95 4.80 4.70 4.64 4.47 4.33

0.100 1.89 1.84 1.79 1.76 1.74 1.71 1.69 1.68 1.64 1.61 0.050 2.28 2.20 2.12 2.07 2.04 1.99 1.97 1.95 1.90 1.85 20 0.025 2.68 2.57 2.46 2.40 2.35 2.29 2.25 2.22 2.16 2.09 0.010 3.23 3.09 2.94 2.84 2.78 2.69 2.64 2.61 2.52 2.43 0.001 4.82 4.56 4.29 4.12 4.00 3.86 3.77 3.70 3.54 3.40

0.100 1.82 1.77 1.72 1.68 1.66 1.63 1.61 1.59 1.56 1.52 0.050 2.16 2.09 2.01 1.96 1.92 1.87 1.84 1.82 1.77 1.72 25 0.025 2.51 2.41 2.30 2.23 2.18 2.12 2.08 2.05 1.98 1.91 0.010 2.99 2.85 2.70 2.60 2.54 2.45 2.40 2.36 2.27 2.18

0.001 4.31 4.06 3.79 3.63 3.52 3.37 3.28 3.22 3.06 2.91 0.100 1.68 1.63 1.57 1.53 1.50 1.46 1.44 1.42 1.38 1.33 0.050 1.95 1.87 1.78 1.73 1.69 1.63 1.60 1.58 1.51 1.45 50 0.025 2.22 2.11 1.99 1.92 1.87 1.80 1.75 1.72 1.64 1.56 0.010 2.56 2.42 2.27 2.17 2.10 2.01 1.95 1.91 1.80 1.70 0.001 3.44 3.20 2.95 2.79 2.68 2.53 2.44 2.38 2.21 2.05

0.100 1.61 1.56 1.49 1.45 1.42 1.38 1.35 1.34 1.28 1.22 0.050 1.85 1.77 1.68 1.62 1.57 1.52 1.48 1.45 1.38 1.30 100 0.025 2.08 1.97 1.85 1.77 1.71 1.64 1.59 1.56 1.46 1.36 0.010 2.37 2.22 2.07 1.97 1.89 1.80 1.74 1.69 1.57 1.45 0.001 3.07 2.84 2.59 2.43 2.32 2.17 2.08 2.01 1.83 1.64

0.100 1.58 1.52 1.46 1.41 1.38 1.34 1.31 1.29 1.23 1.16 0.050 1.80 1.72 1.62 1.56 1.52 1.46 1.41 1.39 1.30 1.21 200 0.025 2.01 1.90 1.78 1.70 1.64 1.56 1.51 1.47 1.37 1.25

0.010 2.27 2.13 1.97 1.87 1.79 1.69 1.63 1.58 1.45 1.30 0.001 2.90 2.67 2.42 2.26 2.15 2.00 1.90 1.83 1.64 1.43

0.100 1.55 1.49 1.43 1.38 1.35 1.30 1.27 1.25 1.18 1.08 0.050 1.76 1.68 1.58 1.52 1.47 1.41 1.36 1.33 1.24 1.111000 0.025 1.96 1.85 1.72 1.64 1.58 1.50 1.45 1.41 1.29 1.13 0.010 2.20 2.06 1.90 1.79 1.72 1.61 1.54 1.50 1.35 1.16 0.001 2.77 2.54 2.30 2.14 2.02 1.87 1.77 1.69 1.49 1.22



12 15 20 25 30 40 50 60 120 1000


Degrees

ofFreedomi

ntheDenominator

Area in

Right Tail

Comparing Three or More Means (One-Way Anaylsis ofVariance) C37

Sullivan ANOVA

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