This module will proceed the discussion of hypothesis testing, where a details statement or theory is generated about a populace parameter, and also sample statistics are provided to evaluate the likelihood the the hypothesis is true. The hypothesis is based on accessible information and also the investigator"s belief about the populace parameters. The particular test considered here is called evaluation of variance (ANOVA) and also is a test of theory that is suitable to compare means of a consistent variable in 2 or much more independent comparison groups. For example, in part clinical trials there are more than two comparison groups. In a clinical attempt to evaluate a new medication for asthma, investigators might compare an experimental medication to a placebo and also to a conventional treatment (i.e., a medication right now being used). In an observational research such together the Framingham love Study, it could be of interest to compare median blood press or median cholesterol levels in people who are underweight, common weight, overweight and obese.

The method to test because that a difference in more than 2 independent way is an expansion of the two independent samples procedure discussed previously which uses when over there are precisely two independent comparison groups. The ANOVA an approach applies as soon as there are two or more than 2 independent groups. The ANOVA procedure is used to to compare the means of the compare groups and is conducted using the same 5 step method used in the scenarios discussed in previous sections. Due to the fact that there are much more than 2 groups, however, the computation the the test statistic is an ext involved. The check statistic should take into account the sample sizes, sample method and sample typical deviations in every of the comparison groups.

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If one is assessing the method observed among, say three groups, it might be tempting to carry out three separate team to team comparisons, yet this technique is incorrect due to the fact that each of these comparisons fails to take right into account the full data, and also it rises the likelihood of incorrectly concluding that there room statistically significate differences, due to the fact that each comparison adds come the probability that a kind I error. Evaluation of variance prevents these problemss by questioning a more worldwide question, i.e., whether there are far-ranging differences among the groups, there is no addressing distinctions between any two teams in particular (although over there are additional tests that have the right to do this if the analysis of variance suggests that there are differences among the groups).

The an essential strategy that ANOVA is come systematically examine variability within groups being contrasted and also examine variability amongst the groups being compared.

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

Consider an instance with 4 independent groups and a consistent outcome measure.The independent groups could be identified by a specific characteristic of the attendees such as BMI (e.g., underweight, typical weight, overweight, obese) or by the investigator (e.g., randomizing participants to one of four contending treatments, call them A, B, C and D). Mean that the outcome is systolic blood pressure, and also we wish to test whether there is a statistically far-reaching difference in median systolic blood pressures among the 4 groups. The sample data are organized as follows:

**Group 1**

**Group 2**

**Group 3**

**Group 4**

| n1 | n2 | n3 | n4 |

| ||||

| s1 | s2 | s3 | s4 |

The hypotheses of attention in an ANOVA room as follows:

H0: μ1 = μ2 = μ3 ... = μkH1: method are no all equal.where k = the number of independent compare groups.

In this example, the hypotheses are:

H0: μ1 = μ2 = μ3 = μ4H1: The way are not all equal.The null hypothesis in ANOVA is constantly that over there is no distinction in means. The study or different hypothesis is always that the method are not all equal and also is usually written in words rather than in math symbols. The research hypothesis captures any kind of difference in means and includes, because that example, the case where all four means are unequal, wherein one is different from the other three, wherein two are different, and so on. The alternate hypothesis, as shown above, catch all possible situations other than equality of all method specified in the null hypothesis.

## Test Statistic for ANOVA

The test statistic for testing H0: μ1 = μ2 = ... = μk is:

and the an important value is uncovered in a table the probability worths for the F circulation with (degrees the freedom) df1 = k-1, df2=N-k. The table have the right to be uncovered in "Other Resources" top top the left next of the pages.

In the test statistic, nj = the sample size in the jth team (e.g., j =1, 2, 3, and also 4 as soon as there room 4 compare groups),

is the sample median in the jth group, and is the overall mean. K to represent the number of independent teams (in this example, k=4), and also N to represent the total variety of observations in the analysis. Keep in mind that N go not refer to a populace size, however instead to the total sample size in the evaluation (the sum of the sample sizes in the compare groups, e.g., N=n1+n2+n3+n4). The check statistic is complex because that incorporates all of the sample data. While the is not simple to see the extension, the F statistic shown over is a generalization of the check statistic provided for testing the equality of precisely two means.**NOTE:** The test statistic F assumes equal variability in the k populaces (i.e., the population variances are equal, or s12 = s22 = ... = sk2 ). This way that the outcome is same variable in each of the to compare populations. This assumption is the very same as that assumed for ideal use of the check statistic to test equality of two independent means. That is feasible to assess the likelihood the the assumption of same variances is true and the test have the right to be conducted in many statistical computer packages. If the variability in the k comparison teams is not similar, then different techniques need to be used.

The F statistic is computed by taking the ratio of what is referred to as the "between treatment" variability come the "residual or error" variability. This is wherein the name of the procedure originates. In evaluation of variance us are testing for a difference in way (H0: means are all equal versus H1: method are not all equal) by examining variability in the data. The molecule captures in between treatment variability (i.e., differences among the sample means) and the denominator includes an calculation of the variability in the outcome. The check statistic is a measure up that enables us to evaluate whether the differences amongst the sample method (numerator) are an ext than would be intended by chance if the null theory is true. Recall in the 2 independent sample test, the check statistic to be computed by taking the ratio of the distinction in sample means (numerator) come the variability in the result (estimated by Sp).

The decision ascendancy for the F test in ANOVA is collection up in a similar way to decision rule we developed for t tests. The decision ascendancy again depends on the level of significance and the degrees of freedom. The F statistic has actually two levels of freedom. These space denoted df1 and df2, and also called the numerator and denominator degrees of freedom, respectively. The degrees of flexibility are characterized as follows:

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

where k is the number of comparison groups and N is the total variety of observations in the analysis. If the null hypothesis is true, the between treatment sport (numerator) will not exceed the residual or error variation (denominator) and the F statistic will certainly small. If the null theory is false, then the F statistic will be large. The rejection an ar for the F check is constantly in the upper (right-hand) tail the the distribution as displayed below.

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

For the scenario illustrated here, the decision rule is: reject H0 if F > 2.87.

The ANOVA ProcedureWe will next show the ANOVA procedure making use of the 5 step approach. Since the computation the the check statistic is involved, the computations are often organized in an ANOVA table. The ANOVA table breaks under the components of sports in the data into variation between treatments and error or residual variation. Statistical computer packages also produce ANOVA tables as part of their typical output for ANOVA, and also the ANOVA table is collection up together follows:

Source the Variation

Sums of Squares (SS)

Degrees of freedom (df)

Mean Squares (MS)

F

Between Treatments

Error (or Residual)

Total

k-1 | |||

N-k | |||

N-1 |

where

X = separation, personal, instance observation, = sample average of the jth therapy (or group), = as whole sample mean,k = the number of treatments or independent compare groups, andN = total number of observations or total sample size.**The ANOVA table above is arranged as follows.**

**"Source of Variation"**and also delineates the between treatment and also error or residual variation. The full variation is the amount of the in between treatment and also error variation.The 2nd column is licensed has been granted

**"Sums that Squares (SS)"**. The in between treatment sums the squares is

and is computed by summing the squared differences in between each treatment (or group) mean and the all at once mean. The squared differences are weight by the sample size per team (nj). The error sums that squares is:

and is computed by summing the squared differences between each observation and its team mean (i.e., the squared differences between each observation in team 1 and the group 1 mean, the squared differences between each monitoring in team 2 and the group 2 mean, and also so on). The double summation ( SS ) shows summation that the squared distinctions within each treatment and then summation of this totals across treatments to produce a single value. (This will be depicted in the following examples). The complete sums the squares is:

and is computed through summing the squared differences between each observation and also the all at once sample mean. In one ANOVA, data are arranged by to compare or therapy groups. If every one of the data to be pooled into a solitary sample, SST would reflect the molecule of the sample variance computed top top the pooled or total sample. SST go not figure into the F statistic directly. However, SST = SSB + SSE, for this reason if 2 sums the squares are known, the third can be computed indigenous the various other two.

The third column contains**degrees that freedom**. The between treatment levels of liberty is df1 = k-1. The error levels of freedom is df2 = N - k. The complete degrees of freedom is N-1 (and the is likewise true the (k-1) + (N-k) = N-1).The 4th column consists of

**"Mean Squares (MS)"**which room computed by splitting sums of squares (SS) by degrees of flexibility (df), heat by row. Specifically, MSB=SSB/(k-1) and also MSE=SSE/(N-k). Dividing SST/(N-1) to produce the variance that the full sample. The F statistic is in the rightmost obelisk of the ANOVA table and also is computed by acquisition the proportion of MSB/MSE.

Example:

A clinical attempt is operation to compare load loss programs and also participants room randomly assigned to one of the compare programs and also are counseled on the details of the assigned program. Participants follow the assigned program for 8 weeks. The outcome of interest is weight loss, defined as the difference in weight measured in ~ the start of the study (baseline) and weight measured at the end of the study (8 weeks), measured in pounds.

Three popular weight ns programs are considered. The very first is a low calorie diet. The second is a short fat diet and the third is a short carbohydrate diet. For comparison purposes, a 4th group is taken into consideration as a regulate group. Entrants in the 4th group space told the they are participating in a examine of healthy behaviors with weight loss only one component of interest. The regulate group is contained here to assess the placebo effect (i.e., weight loss because of simply participating in the study). A full of twenty patients agree to take part in the study and also are randomly assigned to one of the four diet groups. Weights room measured in ~ baseline and patients room counseled on the proper implementation of the assigned diet (with the exemption of the control group). After ~ 8 weeks, each patient"s load is again measured and the difference in weights is computed by subtracting the 8 week load from the baseline weight. Positive distinctions indicate load losses and an adverse differences show weight gains. For translate purposes, we describe the distinctions in weights as weight losses and also the observed load losses are shown 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 significant difference in the median weight loss among the 4 diets? We will run the ANOVA utilizing the five-step approach.

**Step 1.**set up hypotheses and also determine level that significance

H0: μ1 = μ2 = μ3 = μ4 H1: way are not all same α=0.05

**Step 2.**pick the proper test statistic.

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

**Step 3.**collection up decision rule.

The appropriate an important value deserve to be discovered in a table the probabilities for the F distribution(see "Other Resources"). In stimulate to determine the an important value the F we need levels of freedom, df1=k-1 and df2=N-k. In this example, df1=k-1=4-1=3 and df2=N-k=20-4=16. The an important value is 3.24 and the decision dominance is as follows: reject H0 if F > 3.24.

**Step 4.**Compute the check statistic.

To organize our computations we complete the ANOVA table. In order come compute the sums that squares us must an initial compute the sample means for every group and the in its entirety mean 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 overall mean is

= 3.6.We have the right to now compute

So, in this case:

Next we compute,

SSE requires computer the squared differences between each observation and its team mean. We will 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 |

Thus,

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 short 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 |

Thus,

For the entrants 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 now construct the **ANOVA table**.

Source of Variation

Sums the Squares

(SS)

Degrees that 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 due to the fact that 8.43 > 3.24. We have statistically far-reaching evidence in ~ α=0.05 to present that there is a distinction in median weight loss among the 4 diets.

ANOVA is a check that offers a worldwide assessment the a statistical difference in more than two independent means. In this example, we find that over there is a statistically far-reaching difference in average weight loss among the 4 diets considered. In addition to report the results of the statistical test of hypothesis (i.e., that there is a statistically far-reaching difference in typical weight losses in ~ α=0.05), investigators should also report the observed sample means to facilitate translate of the results. In this example, entrants in the short calorie diet shed an average of 6.6 pounds end 8 weeks, as compared to 3.0 and 3.4 pounds in the short fat and low carbohydrate groups, respectively. Entrants in the manage group lost an median of 1.2 pounds which could be referred to as the placebo effect due to the fact that these participants were no participating in an energetic arm of the trial especially targeted for weight loss. Are the observed load losses clinically meaningful?

Another ANOVA ExampleCalcium is crucial mineral the regulates the heart, is important for blood clotting and also for structure healthy bones. The national Osteoporosis structure recommends a daily calcium intake of 1000-1200 mg/day for adult men and also women. When calcium is included in some foods, many adults execute not get sufficient calcium in your diets and also take supplements. Unfortunately some of the supplements have side impacts such together gastric distress, make them an overwhelming for part patients to take it on a regular basis.

A research is draft to check whether there is a distinction in mean everyday calcium intake in adults with normal bone density, adults with osteopenia (a low bone thickness which may lead to osteoporosis) and adults v osteoporosis. Adults 60 years of age with normal bone density, osteopenia and also osteoporosis are selected at random from hospital records and also invited to participate in the study. Every participant"s everyday calcium intake is measured based on reported food intake and 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 there a statistically far-reaching difference in average calcium entry in patients through normal bone density as compared to patients v osteopenia and osteoporosis? We will certainly run the ANOVA making use of the five-step approach.

**Step 1.**collection up hypotheses and also determine level that significance

H0: μ1 = μ2 = μ3 H1: means are no all equal α=0.05

**Step 2.**pick the suitable test statistic.

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

**Step 3.**set up decision rule.

In order to determine the an essential value of F us need levels of freedom, 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 critical value is 3.68 and the decision preeminence is as follows: refuse H0 if F > 3.68.

**Step 4.**Compute the check statistic.

To organize our computations us will finish the ANOVA table. In order to compute the sums that squares us must very first compute the sample way for each group and also the all at once mean.

Normal Bone Density

**Osteopenia**

**Osteoporosis**

n1=6 | n2=6 | n3=6 |

If we pool all N=18 observations, the overall mean is 817.8.

We have the right to now compute:

Substituting:

Finally,

Next,

SSE requires computing the squared differences in between each observation and its group mean. We will certainly compute SSE in parts. For the participants through normal bone density:

**Normal 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 |

Thus,

For participants with 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 |

Thus,

for participants through 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 |

Thus,

We deserve to now construct the **ANOVA table**.

**Source that Variation**

**Sums the 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 carry out not disapprove H0 since 1.395 One-Way ANOVA in R

The video below through Mike Marin demonstrates exactly how to perform evaluation of variance in R. It also covers some other statistical issues, yet the initial part of the video will be helpful to you.

Two-Factor ANOVAThe ANOVA tests described above are referred to as one-factor ANOVAs. There is one treatment or grouping element with k>2 levels and we wish to compare the way across the different categories the this factor. The factor could represent different diets, various classifications of danger for condition (e.g., osteoporosis), different medical treatments, different age groups, or various racial/ethnic groups. Over there are cases where it may be of attention to compare method of a consistent outcome throughout two or an ext factors. For example, mean a clinical trial is designed to compare five various treatments because that joint pain in patients v osteoarthritis. Investigators might likewise hypothesize that there are differences in the outcome by sex. This is an example of a two-factor ANOVA where the factors are treatment (with 5 levels) and sex (with 2 levels). In the two-factor ANOVA, investigators have the right to assess whether there are differences in method due to the treatment, by sex or whether over there is a difference in outcomes by the combination or communication of treatment and sex. Higher order ANOVAs are performed in the same method as one-factor ANOVAs gift here and also the computations are again organized in ANOVA tables with an ext rows to distinguish the different sources of sport (e.g., between treatments, in between men and also women). The following instance illustrates the approach.

Example:

Consider the clinical psychological outlined above in which three contending treatments because that joint ache are compared in terms of their mean time to pain relief in patients through osteoarthritis. Because investigators hypothesize that there may be a difference in time to pain relief in males versus women, they randomly assign 15 participating males to among the three completing treatments and randomly assign 15 participating women to among the three completing treatments (i.e., stratified randomization). Participating men and women do not recognize to which therapy they room assigned. They room instructed to take the assigned medication when they suffer joint pain and to record the time, in minutes, till the ache subsides. The data (times come pain relief) are shown below and are organized by the assigned treatment and also sex of the participant.

**Table the Time come Pain Relief by Treatment and Sex**

**Treatment**

**Male**

**Female**

| 12 | 21 |

15 | 19 | |

16 | 18 | |

17 | 24 | |

14 | 25 | |

| 14 | 21 |

17 | 20 | |

19 | 23 | |

20 | 27 | |

17 | 25 | |

| 25 | 37 |

27 | 34 | |

29 | 36 | |

24 | 26 | |

22 | 29 |

The evaluation in two-factor ANOVA is comparable to that illustrated over for one-factor ANOVA. The computations are again arranged in one ANOVA table, but the full variation is partitioned into that due to the main impact of treatment, the main effect of sex and also the communication effect. The outcomes of the analysis are shown below (and were produced with a statistical computer package - right here we emphasis on interpretation).

**ANOVA Table because that Two-Factor ANOVA**

**Source of Variation**

**Sums the Squares**

**(SS)**

**Degrees the freedom**

**(df)**

**Mean 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 |

There room 4 statistics tests in the ANOVA table above. The first test is an as whole test to evaluate whether there is a difference amongst the 6 cell way (cells are characterized by treatment and sex). The F statistic is 20.7 and also is highly statistically significant with p=0.0001. Once the in its entirety test is significant, focus then turns to the determinants that might be driving the significance (in this example, treatment, sex or the interaction between the two). The following three statistics tests assess the significance of the main result of treatment, the main effect of sex and the communication effect. In this example, over there is a highly far-reaching main result of treatment (p=0.0001) and a highly far-reaching main impact of sex (p=0.0001). The interaction in between the 2 does not reach statistical definition (p=0.91). The table below contains the average times come pain relief in every of the treatments for men and women (Note the each sample average is computed ~ above the 5 observations measured under that experimental condition).

**Mean Time to Pain Relief through 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 many efficacious therapy for both men and women. The median times to relief are reduced in therapy A for both men and also women and also highest in treatment C for both men and women. Throughout all treatments, females report much longer times come pain relief (See below).

Notice that there is the very same pattern of time to pain relief throughout treatments in both men and women (treatment effect). Over there is also a sex impact - specifically, time to pain relief is much longer in women in every treatment.

Suppose that the same clinical attempt is replicated in a second clinical site and the complying with data space observed.

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

**Treatment**

**Male**

**Female**

| 22 | 21 |

25 | 19 | |

26 | 18 | |

27 | 24 | |

24 | 25 | |

| 14 | 21 |

17 | 20 | |

19 | 23 | |

20 | 27 | |

17 | 25 | |

| 15 | 37 |

17 | 34 | |

19 | 36 | |

14 | 26 | |

12 | 29 |

The ANOVA table because that the data measure in clinical website 2 is presented below.

See more: What Is The Oxidation Number Of Elements In Group 17, Oxidation States Of Group 17 Elements

**Table - summary of Two-Factor ANOVA - Clinical website 2**

Source of Variation

Sums of Squares

(SS)

Degrees of freedom

(df)

Mean 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 the the overall test is significant (F=19.4, p=0.0001), over there is a significant treatment effect, sex effect and also a highly significant interaction effect. The table listed below contains the median times to relief in every of the treatments for men and also women.

**Table - median Time to Pain Relief through Treatment and also Gender - Clinical site 2**

**Treatment**

**Male**

**Female**

| 24.8 | 21.4 |

| 17.4 | 23.2 |

| 15.4 | 32.4 |

Notice that currently the distinctions in average time come pain relief amongst the treatments rely on sex. Among men, the median time come pain relief is highest possible in treatment A and also lowest in therapy C. Among women, the turning back is true.This is an interaction result (see below).

Notice above that the treatment effect varies depending upon sex. Thus, we cannot summarize an all at once treatment effect (in men, therapy C is best, in women, therapy A is best).

When interaction effects are present, some investigators carry out not examine main effects (i.e., perform not test for therapy effect since the result of treatment counts on sex). This concern is complex and is discussed in more detail in a later on module.