This module will proceed the discussion of hypothesis testing, wright here a specific statement or hypothesis is produced around a population parameter, and sample statistics are provided to assess the likelihood that the hypothesis is true. The hypothesis is based on accessible indevelopment and also the investigator"s idea about the populace parameters. The certain test taken into consideration below is called evaluation of variance (ANOVA) and also is a test of hypothesis that is proper to compare means of a constant variable in two or even more independent comparikid teams. For example, in some clinical trials tbelow are even more than two comparichild groups. In a clinical trial to evaluate a new medication for asthma, investigators might compare an experimental medication to a placebo and also to a standard treatment (i.e., a medication currently being used). In an observational study such as the Framingham Heart Study, it might be of interest to compare expect blood push or intend cholesterol levels in persons who are underweight, normal weight, overweight and obese.

The method to test for a difference in more than two independent suggests is an expansion of the two independent samples procedure disputed previously which uses as soon as tbelow are specifically two independent comparikid teams. The ANOVA method applies once there are 2 or more than two independent teams. The ANOVA procedure is provided to compare the indicates of the comparichild groups and also is carried out making use of the exact same 5 action method supplied in the scenarios disputed in previous sections. Because there are even more than 2 groups, but, the computation of the test statistic is more connected. The test statistic should take into account the sample sizes, sample suggests and sample standard deviations in each of the comparison groups.

You are watching: If an anova test is conducted and the null hypothesis is rejected, what does this indicate?

If one is researching the suggests oboffered among, say 3 groups, it could be tempting to perdevelop 3 separate team to group comparisons, but this approach is incorrect bereason each of these comparisons falls short to take right into account the complete information, and it increases the likelihood of incorrectly concluding that there are statistically significate distinctions, considering that each compariboy adds to the probcapacity of a kind I error. Analysis of variance avoids these problemss by asking an extra global question, i.e., whether tbelow are considerable differences among the teams, without addressing distinctions between any kind of two groups in specific (although tright here are extra tests that can do this if the analysis of variance suggests that tbelow are differences among the groups).

The standard strategy of ANOVA is to systematically study varicapacity within groups being compared and also also research varicapacity among the groups being compared.

Learning Objectives

After completing this module, the student will be able to:

Perform analysis of variance by handAppropriately interpret outcomes of analysis of variance testsDistinguish in between one and also two factor evaluation of variance testsIdentify the proper hypothesis experimentation procedure based on kind of outcome variable and variety of samples

The ANOVA Approach

Consider an example through four independent teams and a consistent outcome measure.The independent groups can be defined by a details characteristic of the participants such as BMI (e.g., underweight, normal weight, overweight, obese) or by the investigator (e.g., randomizing participants to among 4 completing therapies, call them A, B, C and D). Suppose that the outcome is systolic blood push, and also we wish to test whether tbelow is a statistically substantial difference in expect systolic blood pressures among the four teams. The sample data are organized as follows:

Group 1

Group 2

Group 3

Group 4

Sample Size

n1

n2

n3

n4

Sample Mean

Sample Standard Deviation

s1

s2

s3

s4

The hypotheses of interemainder in an ANOVA are as follows:

H0: μ1 = μ2 = μ3 ... = μkH1: Means are not all equal.

wright here k = the number of independent comparison teams.

In this instance, the hypotheses are:

H0: μ1 = μ2 = μ3 = μ4H1: The suggests are not all equal.

The null hypothesis in ANOVA is always that tright here is no difference in means. The research or alternate hypothesis is constantly that the means are not all equal and is commonly composed in words quite than in mathematical symbols. The research study hypothesis captures any difference in indicates and consists of, for example, the instance where all 4 suggests are unequal, wright here one is different from the other 3, wright here two are different, and also so on. The alternate hypothesis, as presented above, capture all feasible situations various other than etop quality of all implies mentioned in the null hypothesis.

## Test Statistic for ANOVA

The test statistic for trial and error H0: μ1 = μ2 = ... = μk is:

and also the critical worth is uncovered in a table of probability values for the F distribution with (degrees of freedom) df1 = k-1, df2=N-k. The table have the right to be discovered in "Other Resources" on the left side of the peras.

In the test statistic, nj = the sample dimension in the jth group (e.g., j =1, 2, 3, and also 4 once tbelow are 4 comparikid groups),

is the sample suppose in the jth team, and
is the all at once suppose. k represents the variety of independent groups (in this instance, k=4), and also N represents the complete variety of observations in the evaluation. Note that N does not describe a populace dimension, however rather to the full sample size in the analysis (the sum of the sample sizes in the compariboy teams, e.g., N=n1+n2+n3+n4). The test statistic is complicated bereason it incorpoprices all of the sample data. While it is not basic to watch the extension, the F statistic presented over is a generalization of the test statistic used for testing the equality of precisely 2 implies.

NOTE: The test statistic F assumes equal variability in the k populations (i.e., the populace variances are equal, or s12 = s22 = ... = sk2 ). This indicates that the outcome is equally variable in each of the compariboy populaces. This assumption is the very same as that assumed for correct use of the test statistic to test equality of two independent suggests. It is possible to assess the likelihood that the assumption of equal variances is true and also the test have the right to be performed in a lot of statistical computing packages. If the varicapability in the k comparikid groups is not comparable, then alternative techniques should be used.

The F statistic is computed by taking the proportion of what is called the "in between treatment" varicapability to the "residual or error" varicapacity. This is wright here the name of the procedure originates. In evaluation of variance we are testing for a distinction in means (H0: means are all equal versus H1: implies are not all equal) by evaluating variability in the information. The numerator captures between treatment variability (i.e., differences among the sample means) and also the denominator contains an estimate of the varicapability in the outcome. The test statistic is a meacertain that enables us to assess whether the differences among the sample implies (numerator) are more than would certainly be supposed by chance if the null hypothesis is true. Respeak to in the 2 independent sample test, the test statistic was computed by taking the ratio of the difference in sample indicates (numerator) to the varicapacity in the outcome (approximated by Sp).

The decision dominion for the F test in ANOVA is set up in a similar way to decision rules we established for t tests. The decision rule again depends on the level of meaning and the degrees of freedom. The F statistic has actually two levels of flexibility. These are deprovided df1 and also df2, and also dubbed the numerator and denominator levels of liberty, respectively. The levels of flexibility are characterized as follows:

df1 = k-1 and also df2=N-k,

wright here k is the variety of compariboy groups and N is the full number of monitorings in the analysis. If the null hypothesis is true, the in between therapy variation (numerator) will certainly not exceed the residual or error variation (denominator) and also the F statistic will certainly small. If the null hypothesis is false, then the F statistic will certainly be large. The rejection region for the F test is constantly in the top (right-hand) tail of the distribution as displayed listed below.

Rejection Region for F Test through a =0.05, df1=3 and also df2=36 (k=4, N=40)

For the scenario illustrated right here, the decision preeminence is: Reject H0 if F > 2.87.

The ANOVA Procedure

We will following highlight the ANOVA procedure making use of the 5 action method. Because the computation of the test statistic is associated, the computations are frequently arranged in an ANOVA table. The ANOVA table breaks down the components of variation in the information right into variation in between treatments and error or residual variation. Statistical computer packperiods additionally produce ANOVA tables as component of their standard output for ANOVA, and the ANOVA table is put up as follows:

Source of Variation

Sums of Squares (SS)

Degrees of Freedom (df)

Median Squares (MS)

F

Between Treatments

Error (or Residual)

Total

k-1

N-k

N-1

wright here

X = individual monitoring,
= sample expect of the jth therapy (or group),
= as a whole sample intend,k = the variety of therapies or independent comparison groups, andN = complete variety of monitorings or complete sample dimension.

The ANOVA table over is organized as complies with.

The first column is entitled "Source of Variation" and also delineates the in between therapy and also error or residual variation. The total variation is the sum of the in between treatment and error variation.The second column is entitled "Sums of Squares (SS)". The between treatment sums of squares is

and is computed by summing the squared differences in between each therapy (or group) suppose and also the overall suppose. The squared differences are weighted by the sample sizes per team (nj). The error sums of squares is:

and is computed by summing the squared differences between each observation and its group expect (i.e., the squared differences in between each observation in group 1 and also the team 1 expect, the squared distinctions between each observation in group 2 and also the team 2 suppose, and also so on). The double summation ( SS ) indicates summation of the squared differences within each treatment and then summation of these totals throughout treatments to produce a single value. (This will certainly be portrayed in the complying with examples). The full sums of squares is:

and also is computed by summing the squared distinctions in between each observation and the in its entirety sample suppose. In an ANOVA, information are organized by comparison or therapy groups. If every one of the data were pooled into a single sample, SST would certainly reflect the numerator of the sample variance computed on the pooled or total sample. SST does not figure right into the F statistic straight. However before, SST = SSB + SSE, therefore if 2 sums of squares are known, the third have the right to be computed from the various other 2.

The third column has degrees of freedom. The in between treatment levels of flexibility is df1 = k-1. The error levels of freedom is df2 = N - k. The complete levels of flexibility is N-1 (and it is also true that (k-1) + (N-k) = N-1).The fourth column consists of "Typical Squares (MS)" which are computed by splitting sums of squares (SS) by levels of liberty (df), row by row. Specifically, MSB=SSB/(k-1) and MSE=SSE/(N-k). Dividing SST/(N-1) produces the variance of the total sample. The F statistic is in the rightmost column of the ANOVA table and is computed by taking the proportion of MSB/MSE.

Example:

A clinical trial is run to compare weight loss programs and participants are randomly assigned to among the comparison programs and are counseled on the details of the assigned regimen. Participants follow the assigned routine for 8 weeks. The outcome of interest is weight loss, identified as the difference in weight measured at the start of the examine (baseline) and also weight measured at the finish of the examine (8 weeks), measured in pounds.

Three well-known weight loss programs are thought about. The first is a low calorie diet. The second is a low fat diet and the third is a low carbohydrate diet. For comparichild objectives, a fourth group is considered as a manage group. Participants in the fourth team are told that they are participating in a study of healthy and balanced habits through weight loss only one component of interemainder. The control team is included below to assess the placebo effect (i.e., weight loss due to ssuggest participating in the study). A total of twenty patients agree to participate in the study and also are randomly assigned to among the 4 diet teams. Weights are measured at baseline and patients are counseled on the appropriate implementation of the assigned diet (via the exception of the regulate group). After 8 weeks, each patient"s weight is aacquire measured and also the distinction in weights is computed by subtracting the 8 week weight from the baseline weight. Confident differences show weight losses and negative differences indicate weight gains. For interpretation purposes, we describe the distinctions in weights as weight losses and the observed weight losses are presented below.

Low Calorie

Low Fat

Low Carbohydrate

Control

8

2

3

2

9

4

5

2

6

3

4

-1

7

5

2

0

3

1

3

3

Is there a statistically considerable distinction in the expect weight loss among the 4 diets? We will run the ANOVA using the five-step strategy.

Step 1. Set up hypotheses and also identify level of significance

H0: μ1 = μ2 = μ3 = μ4 H1: Means are not all equal α=0.05

Tip 2. Select the correct test statistic.

The test statistic is the F statistic for ANOVA, F=MSB/MSE.

Step 3. Set up decision ascendancy.

The correct important worth have the right to be discovered in a table of probabilities for the F distribution(watch "Other Resources"). In order to determine the crucial worth of F we require degrees of flexibility, df1=k-1 and also df2=N-k. In this instance, df1=k-1=4-1=3 and df2=N-k=20-4=16. The important worth is 3.24 and the decision preeminence is as follows: Reject H0 if F > 3.24.

Step 4. Compute the test statistic.

To organize our computations we finish the ANOVA table. In order to compute the sums of squares we must first compute the sample suggests for each group and the as a whole intend based on the complete sample.

Low Calorie

Low Fat

Low Carbohydrate

Control

n

Group mean

5

5

5

5

6.6

3.0

3.4

1.2

If we pool all N=20 observations, the all at once intend is

= 3.6.

We have the right to currently compute

So, in this case:

Next we compute,

SSE needs computing the squared distinctions in between each observation and also its team expect. We will certainly compute SSE in parts. For the participants in the low calorie diet:

Low Calorie

(X - 6.6)

(X - 6.6)2

8

1.4

2.0

9

2.4

5.8

6

-0.6

0.4

7

0.4

0.2

3

-3.6

13.0

Totals

0

21.4

Therefore,

For the participants in the low fat diet:

Low Fat

(X - 3.0)

(X - 3.0)2

2

-1.0

1.0

4

1.0

1.0

3

0.0

0.0

5

2.0

4.0

1

-2.0

4.0

Totals

0

10.0

Thus,

For the participants in the low carbohydrate diet:

Low Carbohydrate

(X - 3.4)

(X - 3.4)2

3

-0.4

0.2

5

1.6

2.6

4

0.6

0.4

2

-1.4

2.0

3

-0.4

0.2

Totals

0

5.4

Hence,

For the participants in the control group:

Control

(X - 1.2)

(X - 1.2)2

2

0.8

0.6

2

0.8

0.6

-1

-2.2

4.8

0

-1.2

1.4

3

1.8

3.2

Totals

0

10.6

Thus,

Therefore,

We deserve to currently construct the ANOVA table.

Source of Variation

Sums of Squares

(SS)

Degrees of Freedom

(df)

Means Squares

(MS)

F

Between Treatmenst

Error (or Residual)

Total

75.8

4-1=3

75.8/3=25.3

25.3/3.0=8.43

47.4

20-4=16

47.4/16=3.0

123.2

20-1=19

Step 5. Conclusion.

We disapprove H0 because 8.43 > 3.24. We have actually statistically substantial proof at α=0.05 to display that there is a difference in expect weight loss among the 4 diets.

ANOVA is a test that gives a worldwide assessment of a statistical distinction in more than two independent implies. In this instance, we discover that there is a statistically substantial difference in suppose weight loss among the 4 diets thought about. In addition to reporting the results of the statistical test of hypothesis (i.e., that there is a statistically significant distinction in expect weight losses at α=0.05), investigators have to additionally report the oboffered sample indicates to facilitate interpretation of the results. In this instance, participants in the low calorie diet shed an average of 6.6 pounds over 8 weeks, as compared to 3.0 and also 3.4 pounds in the low fat and low carbohydprice groups, respectively. Participants in the regulate group lost an average of 1.2 pounds which might be referred to as the placebo result bereason these participants were not participating in an active arm of the trial particularly targeted for weight loss. Are the observed weight losses clinically meaningful?

Anvarious other ANOVA Example

Calcium is an important mineral that regulates the heart, is vital for blood clotting and for building healthy and balanced bones. The National Osteoporosis Foundation recommends a day-to-day calcium intake of 1000-1200 mg/day for adult guys and also women. While calcium is had in some foodstuffs, the majority of adults perform not get sufficient calcium in their diets and also take supplements. Unfortunately some of the supplements have side effects such as gastric distress and anxiety, making them challenging for some patients to take on a regular basis.

A research is designed to test whether there is a distinction in intend daily calcium intake in adults with normal bone density, adults via osteopenia (a low bone density which might lead to osteoporosis) and adults with osteoporosis. Adults 60 years of age through normal bone thickness, osteopenia and also osteoporosis are schosen at random from hospital records and also invited to get involved in the examine. Each participant"s daily calcium intake is measured based upon reported food intake and also supplements. The data are presented below.

Normal Bone Density

Osteopenia

Osteoporosis

1200

1000

890

1000

1100

650

980

700

1100

900

800

900

750

500

400

800

700

350

Is tbelow a statistically considerable distinction in suppose calcium intake in patients via normal bone density as compared to patients with osteopenia and osteoporosis? We will certainly run the ANOVA making use of the five-step approach.

Step 1. Set up hypotheses and determine level of significance

H0: μ1 = μ2 = μ3 H1: Means are not all equal α=0.05

Step 2. Select the appropriate test statistic.

The test statistic is the F statistic for ANOVA, F=MSB/MSE.

Tip 3. Set up decision dominion.

In order to recognize the crucial value of F we require degrees of liberty, df1=k-1 and also df2=N-k. In this example, df1=k-1=3-1=2 and also df2=N-k=18-3=15. The crucial worth is 3.68 and the decision ascendancy is as follows: Reject H0 if F > 3.68.

Tip 4. Compute the test statistic.

To organize our computations we will certainly complete the ANOVA table. In order to compute the sums of squares we should initially compute the sample means for each team and the as a whole expect.

Typical Bone Density

Osteopenia

Osteoporosis

n1=6

n2=6

n3=6

If we pool all N=18 monitorings, the in its entirety intend is 817.8.

We can currently compute:

Substituting:

Finally,

Next off,

SSE needs computing the squared distinctions in between each monitoring and also its team mean. We will certainly compute SSE in components. For the participants through normal bone density:

Common Bone Density

(X - 938.3)

(X - 938.3333)2

1200

261.6667

68,486.9

1000

61.6667

3,806.9

980

41.6667

1,738.9

900

-38.3333

1,466.9

750

-188.333

35,456.9

800

-138.333

19,126.9

Total

0

130,083.3

Hence,

For participants via osteopenia:

Osteopenia

(X - 800.0)

(X - 800.0)2

1000

200

40,000

1100

300

90,000

700

-100

10,000

800

0

0

500

-300

90,000

700

-100

10,000

Total

0

240,000

Hence,

For participants via osteoporosis:

Osteoporosis

(X - 715.0)

(X - 715.0)2

890

175

30,625

650

-65

4,225

1100

385

148,225

900

185

34,225

400

-315

99,225

350

-365

133,225

Total

0

449,750

Therefore,

We have the right to currently construct the ANOVA table.

Source of Variation

Sums of Squares (SS)

Degrees of freedom (df)

Mean Squares (MS)

F

Between Treatments

152,477.7

2

76,238.6

1.395

Error or Residual

819,833.3

15

54,655.5

Total

972,311.0

17

Step 5. Conclusion.

We execute not refuse H0 because 1.395 One-Way ANOVA in R

The video below by Mike Marin demonstrates just how to perform analysis of variance in R. It likewise covers some other statistical problems, however the initial component of the video will be advantageous to you.

Two-Factor ANOVA

The ANOVA tests explained over are dubbed one-element ANOVAs. Tright here is one treatment or grouping variable via k>2 levels and we wish to compare the indicates throughout the different categories of this factor. The element might reexisting various diets, different classifications of risk for illness (e.g., osteoporosis), different medical therapies, various age teams, or various racial/ethnic groups. Tright here are cases wbelow it may be of interest to compare indicates of a continuous outcome throughout 2 or even more factors. For example, intend a clinical trial is designed to compare 5 different therapies for joint pain in patients with osteoarthritis. Investigators might additionally hypothedimension that tright here are distinctions in the outcome by sex. This is an instance of a two-variable ANOVA wright here the components are treatment (with 5 levels) and also sex (via 2 levels). In the two-element ANOVA, investigators deserve to assess whether tbelow are distinctions in suggests because of the treatment, by sex or whether tbelow is a distinction in outcomes by the combination or interaction of therapy and sex. Higher order ANOVAs are performed in the exact same way as one-aspect ANOVAs presented right here and also the computations are again organized in ANOVA tables with even more rows to distinguish the various sources of variation (e.g., between therapies, between guys and also women). The adhering to instance illustprices the method.

Example:

Consider the clinical trial outlined over in which three contending treatments for joint pain are compared in terms of their intend time to pain relief in patients through osteoarthritis. Since investigators hypothesize that tright here might be a distinction in time to pain relief in guys versus women, they randomly assign 15 participating men to among the three competing therapies and randomly assign 15 participating womales to one of the 3 competing therapies (i.e., stratified randomization). Participating men and also woguys do not know to which treatment they are assigned. They are instructed to take the assigned medication as soon as they endure joint pain and to record the time, in minutes, until the pain subsides. The information (times to pain relief) are displayed below and also are arranged by the assigned treatment and sex of the participant.

Table of Time to Pain Relief by Treatment and also Sex

Treatment

Male

Female

A

12

21

15

19

16

18

17

24

14

25

B

14

21

17

20

19

23

20

27

17

25

C

25

37

27

34

29

36

24

26

22

29

The analysis in two-element ANOVA is similar to that depicted above for one-factor ANOVA. The computations are aacquire organized in an ANOVA table, however the complete variation is partitioned right into that due to the major impact of therapy, the main impact of sex and the interaction impact. The outcomes of the evaluation are presented listed below (and were created with a statistical computer package - here we focus on interpretation).

ANOVA Table for Two-Factor ANOVA

Source of Variation

Sums of Squares

(SS)

Degrees of freedom

(df)

Median Squares

(MS)

F

P-Value

Model

967.0

5

193.4

20.7

0.0001

Treatment

651.5

2

325.7

34.8

0.0001

Sex

313.6

1

313.6

33.5

0.0001

Treatment * Sex

1.9

2

0.9

0.1

0.9054

Error or Residual

224.4

24

9.4

Total

1191.4

29

Tbelow are 4 statistical tests in the ANOVA table over. The initially test is an all at once test to assess whether there is a distinction among the 6 cell suggests (cells are identified by therapy and sex). The F statistic is 20.7 and is extremely statistically substantial through p=0.0001. When the overall test is considerable, emphasis then transforms to the components that may be driving the definition (in this instance, therapy, sex or the interactivity in between the two). The following 3 statistical tests assess the definition of the major effect of treatment, the primary effect of sex and the interactivity impact. In this instance, tright here is a extremely substantial primary result of therapy (p=0.0001) and a highly substantial main result of sex (p=0.0001). The interaction between the two does not reach statistical definition (p=0.91). The table listed below has the mean times to pain relief in each of the treatments for guys and woguys (Note that each sample expect is computed on the 5 monitorings measured under that speculative condition).

Median Time to Pain Relief by Treatment and Gender

Treatment

Male

Female

A

14.8

21.4

B

17.4

23.2

C

25.4

32.4

Treatment A appears to be the a lot of efficacious therapy for both males and women. The intend times to relief are reduced in Treatment A for both men and also womales and highest possible in Treatment C for both men and women. Across all therapies, women report longer times to pain relief (See below).

Notice that tright here is the very same pattern of time to pain relief across treatments in both males and woguys (therapy effect). Tbelow is also a sex impact - particularly, time to pain relief is longer in womales in eincredibly therapy.

Suppose that the exact same clinical trial is replicated in a second clinical website and also the complying with information are observed.

Table - Time to Pain Relief by Treatment and Sex - Clinical Site 2

Treatment

Male

Female

A

22

21

25

19

26

18

27

24

24

25

B

14

21

17

20

19

23

20

27

17

25

C

15

37

17

34

19

36

14

26

12

29

The ANOVA table for the information measured in clinical site 2 is displayed below.

See more: The Key To Managing Conflict In A Multicultural Workforce Is:

Table - Rundown of Two-Factor ANOVA - Clinical Site 2

Source of Variation

Sums of Squares

(SS)

Degrees of freedom

(df)

Typical Squares

(MS)

F

P-Value

Model

907.0

5

181.4

19.4

0.0001

Treatment

71.5

2

35.7

3.8

0.0362

Sex

313.6

1

313.6

33.5

0.0001

Treatment * Sex

521.9

2

260.9

27.9

0.0001

Error or Residual

224.4

24

9.4

Total

1131.4

29

Notice that the as a whole test is substantial (F=19.4, p=0.0001), tright here is a significant therapy effect, sex result and also a highly considerable interactivity impact. The table listed below consists of the intend times to relief in each of the treatments for men and also womales.

Table - Average Time to Pain Relief by Treatment and Gender - Clinical Site 2

Treatment

Male

Female

A

24.8

21.4

B

17.4

23.2

C

15.4

32.4

Notice that currently the distinctions in suppose time to pain relief among the therapies depfinish on sex. Amongst guys, the expect time to pain relief is highest possible in Treatment A and lowest in Treatment C. Amongst woguys, the reverse is true.This is an interaction effect (check out below).

Notice over that the treatment impact varies relying on sex. Thus, we cannot summarize an all at once therapy result (in males, therapy C is ideal, in womales, treatment A is best).

When interaction results are present, some investigators perform not examine primary results (i.e., do not test for treatment effect because the result of treatment counts on sex). This worry is facility and is discussed in even more detail in a later on module.