

REVIEW ARTICLE 

Year : 2020  Volume
: 6
 Issue : 3  Page : 207215 

Conceptual Framework on Conducting TwoWay Analysis of Variance
Onchiri Sureiman^{1}, Callen Moraa Mangera^{2}
^{1} Department of Educational Planning and Management, Masinde Muliro University of Science and Technology, Kakamega Webuye, Kenya ^{2} Department of Physiotherapy, School of Medicine, Jomo Kenyatta University of Science and Technology, Juja, Kenya
Date of Submission  31Jul2020 
Date of Decision  24Aug2020 
Date of Acceptance  26Sep2020 
Date of Web Publication  23Dec2020 
Correspondence Address: Mr. Onchiri Sureiman P.O. Box 1300217, Limuru Kenya
Source of Support: None, Conflict of Interest: None  Check 
DOI: 10.4103/jpcs.jpcs_75_20
The twoway analysis of variance (ANOVA) is perceived to many researchers as being complex and tedious. The perception is partly caused by not understanding the application procedures. To give solution to the problem, this paper made an overview of the independent ttest and oneway ANOVA before highlighting the procedure of twoway ANOVA by the use of examples and simplified methods. The authors also developed a simplified conceptual framework that can be applied. To ensure ease in the interpretation of results, the researchers must understand the concept of twoway ANOVA manually before using the online calculator.
Keywords: Analysis of variance, hypothesis, with replication, without replication
How to cite this article: Sureiman O, Mangera CM. Conceptual Framework on Conducting TwoWay Analysis of Variance. J Pract Cardiovasc Sci 2020;6:20715 
Introduction   
The comparison of means of independent samples can be done manually using a hand calculator or an online calculator. The comparison can be done using various tests. The tests used depend on the independent of the observations, type of variables, distribution of data, equality of variance, number of groups, number of factors, and the sample size. These factors form the basis of the development of the conceptual framework on conducting twoway analysis of variance (ANOVA). The independent ttests, oneway ANOVA, and twoway ANOVA are considered.
Independent TTest   
The twosample independent ttest is commonly used to compare two independent groups and tests the null hypotheses that the means are equal.^{[1]} Valid inferences within a twosample independent ttest is based on following assumptions:
 The dependent variable is continuous
 Observations are independent of each other
 The data are approximately normally distributed in each group
 The variances are approximately equal in both groups.^{[2]}
Hypothesis tests such Shapiro–Wilk and Levene's test are available to test these assumptions but have low power to detect the violations at small sample sizes.^{[3]} These tests can be used in combination with graphical methods and in a comparison with group standard deviations to establish whether the assumptions are met.^{[4]}
The ttest is robust against the assumptions of nonnormality with larger sample size (≥30). Smaller samples rely on a normal distribution to avoid erroneous conclusion.^{[3]} Nonparametric methods such Mann–Whitney Utest can be used if the normality assumptions are violated.
The sample size is also robust against unequal variance if the sample sizes per group are equal and if the sample size is large enough (>15 per group). Alternatives such as the Welch ttest are available if the variances are not equal.^{[4]}
If our data meet these assumptions, then t follows a tdistribution with (n_{1} + n_{2} − 2) degrees of freedom (d.f.). Besides inappropriately using ttests when assumptions are grossly violated, a common mistake is to use a multiple pairwise ttests to compare more than two groups because it may give falsepositive result.^{[5]} Other techniques are instead needed, like ANOVA with appropriate post hoc tests.^{[2]}
Example 1
A health psychologist wants to compare the calorie estimates of people who regularly eat junk food with the estimates of people who rarely eat junk food. She believes that the difference could come out in either direction, so she decided to conduct a twotailed test. She collected data from a sample of eight participants who eat junk food regularly and seven participants who rarely eat junk food. The data are as follows:
 Junk food eaters: 180, 220, 150, 85, 200, 170, 150, 190
 Nonjunk food eaters: 200, 240, 190, 175, 200, 300, 240.
The mean for the junk food eaters is 220.71 with a standard deviation of 41.23. The mean for the nonjunk food eaters is 168.12 with a standard deviation of 42.66. Test the null hypothesis that the mean of the food eaters are equal at 5% level of significance.
Solution
The ttest can be done using manual or online calculator methods. For the sake of convenience, the ttest can be done through online calculators which are easily available in the internet. For userfriendly online calculator, you may visit this uniform locator https://www.statskingdom.com/150MeanT2uneq.html. In the software, it is really easy to conduct a ttest and most of the assumptions are preloaded. The basic step for using an online calculator is to select the appropriate ttest and correctly fill in your data into it. For instance, in the above example, we have to select two sample ttest (Welch) since the means are assumed not equal and then fill in the raw data in the columns of two sample ttest calculator directly or from excel [Figure 1].  Figure 1: Setting up the raw data in the two sample ttest calculator (Welch's ttest).
Click here to view 
You can also enter summarized data (x¯, n, σ, S) as shown in [Figure 2].  Figure 2: Setting up the summarized data in the two sample ttest calculator (Welch's ttest).
Click here to view 
After entering the data, you click the calculate button to generate the results.
The output of the ttest is summarized below.
Validity checking
Normality assumptions
When entering the raw data, the tool will run the Shapiro–Wilk normality test as part of the test calculation. The assumption was checked based on the Shapiro–Wilk test (α = 0.05). It is assumed that Group 1 is normally distributed (P = 0.551), or more accurately, you cannot reject the normality assumption. It is assumed that Group 2 is normally distributed (P = 0.314), or more accurately, you cannot reject the normality assumption.
Outliers
When entering the raw data, the tool will run outliers' detection method (Tukey Fence) to test for outliers as part of the test calculation. For this case, k = 1.5. Therefore, the data do not have outliers.
Equality of variance assumption
Based on a twotailed F test, σ_{1} is considered as equal to σ_{2} (P = 0.917). F test assumes equal standard deviations, which is not your test assumption. Note that the F test cannot reject the unequal variance assumption, since it is based on the equal variance assumption, which is not the preliminary assumption. You should continue with the Welch's ttest, which is also robust to variances equality.
Calculated tstatistic
The test statistic t equals − 2.419393 and the P = 0.03148.
Decision
The test statistic t equals − 2.419393 and is not in the 95% critical value accepted range: [−2.1674: 2.1674]. P = 0.0314767 < 0.05. Since P value < α, null hypothesis is rejected. The average of Group 1 population is considered to be not equal to the average of the Group 2 population. In other words, the difference between the average of the Group 1 and Group 2 populations is big enough to be statistically significant.
OneWay Analysis of Variance   
ANOVA is a statistical procedure concerned with comparing more than two sample means.^{[3],[6]} It is an extension of the ttest of two independent samples to more than two groups.^{[7]} To apply the ANOVA, the data must be independent across the groups, normally distributed and homogenous.^{[8],[9]}
The simplest of the ANOVA is the one way. Oneway ANOVA is used when data are divided into groups according to one factor.^{[3],[10],[11]} Valid inferences from oneway ANOVA that rely on several assumptions met: the observations are independent of each other; the dependent variable is almost normally distributed in each group; and the variances in each group are normally distributed.^{[3],[8],[9],[12]}
While ANOVA tells us whether the group mean values are significantly different, it does not tell us which specific groups differ from each other. A variety of post hoc tests such Tukey, Dunnett, and Bonferroni are available to address this question.^{[6]} These post hoc tests are conceptually similar to performing multiple pairwise ttests, but they adjust for inflation of Type I error risk due to multiple testing. The exception is the Fisher least significance difference test. The choice depends on which groups are being compared and how conservative one wishes to be on the multiplicity adjustment.^{[3]}
We now illustrate oneway ANOVA using an Example 2.
Example 2
[Table 1] illustrates the sample psychological health ratings of hospital staff in the departments of oncology, physiotherapy, and orthopedics. Can we consider the psychological health rating of the staff in the given departments to be equal at 5% level of significance?
Solution
One of the intermediaries in the ANOVA is the calculation of F statistic. It can be obtained by (a) direct, (b) shortcut, (c) coding, and (d) online calculator methods. For the sake of convenience, the F test can be done through online calculators which are easily available in internet. For userfriendly online calculator, you may visit this uniform locator https://www.statskingdom.com/180Anova1way.html. In the software, most of the assumptions and post hoc tests are preloaded. The basic step for using and online calculator is to correctly fill in your data into it [Figure 3]. In the above example, we have to key in the raw data in the columns of oneway ANOVA calculator directly or from excel.  Figure 3: Setting up the data in the oneway analysis of variance calculator.
Click here to view 
On clicking the calculate button of the application, the output is summarized below.
Validity checking
Equality of variances
The tool used the Levene's test to assess the equality of variances. The population's variances consider to be equal (P = 0.680). Levene's test power considers to be weak (0.11). The groups' size considers similar. The ratio between the bigger group and the smaller group is 1.00.
Normality assumption
The assumption was checked based on the Shapiro–Wilk test (α = 0.05). It is assumed that all the groups distribute normally.
Calculation of the F statistic
The calculation of F statistic is shown on [Table 2].
Decision
The table value, using F distribution, d.f. (n_{1} = 2, n_{2} = 12) is 3.8853. The test statistic F equals 19.500 and is not in the 95% critical value accepted range: [−∞: 3.8853]; hence, the null hypothesis is rejected. Alternatively, since the P = 0.00017 < 0.05, the result is significant at P < 0.05. Therefore, we also reject the null hypothesis.
Post hoc test
From the above decision, at least two groups are on average different. However, oneway ANOVA gives no information on which groups are different. The Tukey honestly significant difference (HSD) test established that the means of the following pairs are statistically significantly different: X_{1}–X_{2}, X_{2}–X_{3} [Table 3]. This is because 0.0026 < 0.05 and 0.000155 < 0.05.
TwoWay Analysis of Variance   
Twoway ANOVA compares the means of populations that are classified into two ways or the mean responses into twofactor experiments.^{[13]} When we are interested in the effects of two factors, a twoway design offers great advantages over two singlefactor studies. It is more efficient to study two factors interconnectedly, rather than separately, we can reduce the residual variation in a model by including a secondfactor thought to influence the response, and we can investigate interactions between factors.^{[14]}
The four assumptions for a twofactor ANOVA when there is only one observed measurement at each combination of levels of the factors are as follows:
 The population mean at each factor level combination is (approximately) normally distributed
 The variances of the populations must be equal
 The samples must be independent
 The groups must have the same sample size.^{[15]}
Twoway ANOVA designs are of two types: (i) with replication if there is more than one observation in each group and (ii) without replication if there is only one.^{[13]} We shall now explain the twoway ANOVA technique in the context of both the said designs, with the help of relevant examples. The most convenient and simplified methods are also used.
Design without Repeated Measures   
You can do twoway ANOVA without replication (only one observation for each combination of the nominal variables), but this is less informative (you cannot test the interaction term) and requires you to assume that there is no interaction.^{[16]}
Test procedure for designs without repeated measures
Set up hypotheses and determine level of significance
The design has two null hypotheses: (i) for between columns and (ii) for between rows, while the level of significance (α) is normally 0.05.
Prepare twoway analysis of variance table for design without repeated measures
The table can be prepared by computerbased program or manual method (direct, shortcut, or coded). The basic step for using an online calculator is to correctly fill in your data into it and then click the calculate button. The following steps are involved in manual calculation.^{[13]}
 Take the total of the values of individual items (or their coded value) in all the samples and call it T
 Determine the correction factor. Use correction factor =
 Find out the square of all the item values (or their coded values as the case may be) one by one and then take its total. Subtract the correction factor from this total to obtain the sum of squares of deviations for total variance (SS_{t}). Symbolically, we can write it as: SS_{t} = ΣX_{ij}^{2} − correction factor
 Take the total of different columns and then obtain the square of each column total and divide such squared values of each column by the number of items in the concerning column and take the total of the result thus obtained. Finally, subtract the correction factor from this total to obtain the sum of squares of deviations for variance between columns or (SS_{b})
 Take the total of different rows and then obtain the square of each row total and divide such squared values of each row by the number of items in the corresponding row and take the total of the result thus obtained. Finally, subtract the correction factor from this total to obtain the sum of squares of deviations for variance between rows (or SS_{r}). Symbolically, we can write:
 Sum of squares of deviations for residual or error variance (SS_{e}) can be worked out by subtracting the result of the sum of iv^{th} and v^{th} steps from the result of iii^{th} step stated above. In other words, SS_{e} = SS_{t} – (SS_{b} + SS_{r}).
 d.f. can be worked out as under:
 d.f for total variance = number of columns (c) × number of rows (r) − 1
 d.f. for variance between columns = c − 1
 d.f. for variance between rows = r − 1
 d.f. for residual variance = (c − 1) (r − 1)
The d.f. are used to obtain the critical values of the variation from a table of probability values for the F distribution.
The information is then represented in [Table 4].  Table 4: General format of twoway analysis of variance table for designs without repeated measures
Click here to view 
Decision
For columns
If the calculated F_{c} value is greater than the corresponding critical value, then we reject the null hypothesis and conclude that there is significant difference among the treatment means, in at least one pair.^{[9]}
For rows
If the calculated F_{r} value is greater than the corresponding critical value, then we reject the null hypothesis and conclude that there is significant difference among the block means, in at least one pair.^{[9]}
The above test procedure for designs without replication is illustrated by Example 3.
Example 3
A hospital doctor wished to compare four brands of painkillers A, B, C, and D. He arranged that when patients in a surgical ward requested for painkillers, they would be asked whether pain was mild, severe, or very severe. The first patient who said mild would be given A, the second patient who said mild would be give B, the third patient who said mild would be give C, and the fourth patient who said mild would be give D. The pain killer would also be allocated to the first four patients who said their pain was severe and to the first four patients who said their pain was very severe. The patients were then asked the time, in minutes, for which the pain killers were effective. The following data are recorded in [Table 5].
Solution
Hypotheses
 H_{01}: There is no significant difference between the means of the brands (columns)
 H_{01}: There is no significant difference between the means of the brands (columns).
Preparation of analysis of variance table by manual methods
To minimize the calculation work, we make use of the coding method. We subtract three from all the given figures and obtain a coded value as shown in [Table 6].
T = 24, N = 12, correction factor = = 48
SS_{t} =(2^{2} + 2^{2} + 3^{2}+ 1^{2} + 2^{2} + 4^{2} + 0^{2} + 0^{2} + 2^{2} + 0^{2} + 3^{2} + 5^{2}) − 48= (4 + 4 + 9 + 1 + 4++ 16 + 0 + 0 + 4 + 0 + 9 + 25) −48 = 76 − 48 = 28
SS_{c}, i.e., between brands, = 55.333 − 48 = 7.333
SS_{r}, i.e., between symptoms, = 63.5 − 48 = 15.5
SS_{e} = 28 − (7.333 + 15.5) = 5.167
The information is summarized in [Table 7].
Preparation of analysis of variance table using online calculator
For userfriendly online calculator, you may visit this uniform locator http://home.ubalt.edu/ntsbarsh/Businessstat/otherapplets/ANOVATwo.htm. Fill in your data into [Table 8].
Click the calculate button. The output from the online calculator is summarized in [Table 9].
Decision
The calculated F ratio (2.833) for between columns (brands) is less than the critical value (4.76). Hence, we conclude that there is no significant difference between the brands. The calculated F ratio (2.833) for between symptoms is greater than the critical value (5.14). Hence, we conclude that there is a significant difference between the symptoms.
The P value (0.13614) for between brands is greater than α (0.05), implying that we do not reject the null hypothesis. On the other hand, the P value (0.01271) for between rows (symptoms) is less than α, implying that we reject the null hypothesis.
Design with Repeated Measures   
One experimental design that people analyze with a twoway ANOVA is repeated measures, where an observation has been made on the same individual more than once.^{[16],[17]} This usually involves measurements taken at different time points. For example, you could be studying the cholesterol levels of the same group of patients at 1, 2, and 3 months after changing their diet. For this example, the independent variable is “time” and the dependent variable is “cholesterol.” The independent variable is usually called the withinsubject factor. Sometimes, the repeated measures are repeated at different places rather than different times, such as the hip abduction angle measured on the right and left hip of individuals. Repeated measures experiments are often done without replication, although they could be done with replication.^{[16],[18]}
Reasons why to use repeated measures analysis of variance
 When you collect data from the same participants over a period of time, individual differences (a source of between group differences) are reduced or eliminated
 Testing is more powerful because the sample size is not divided between groups
 The test can be economical, as you are using the same participants.
Assumptions for repeated measures analysis of variance
The results from your repeated measures ANOVA will be valid only if the following assumptions have not been violated:
 There must be one independent variable and one dependent variable
 The dependent variable must be a continuous variable, either on an interval scale or a ratio scale
 The independent variable must be categorical, either on the nominal scale or ordinal scale
 Ideally, levels of dependence between pairs of groups is equal (“sphericity”).^{[14]}
Test procedure for designs with replication
Set up hypotheses and determine level of significance
In addition to the two hypotheses for without repetition design, it has a third hypothesis for interaction.
Preparation of twoway analysis of variance table without replication
In addition to the calculation SS_{t}, SS_{c}, SS_{r} and SS_{e}, it involves the calculation of sum of squares within samples (SS_{w}). d.f. are worked out as under:
 d.f. for total variance = Total number of items in all the samples (N) − 1.
 d.f. for variance between columns = c − 1
 d.f. for variance between rows = r − 1
 d.f. for variance within samples = N − k where k = number of samples.
 d.f. for residual variance= (N − 1) − [(c − 1) + (r − 1) + (N − k)] = N − 1 – c + 1 – r + 1 – N + k = k – c – r + 1
After making all these computations, ANOVA table [Table 10] can be set up for drawing inferences.  Table 10: General format of twoway analysis of variance table for designs with replication
Click here to view 
We have also illustrated the test procedure for designs with replication with Example 4.
Example 4
Suppose that the same clinical trial in Example 3 is replicated in another with 12 patients and the following data [Table 11] are observed.
Solution
Hypotheses
The following three null hypotheses can be formulated:
 H_{01}: There is significant difference between the brands
 H_{02:} There is no significant difference between the symptoms
 H_{03:} There is no significant interaction between the brand and symptoms.
Preparation of analysis of variance table by manual methods
To minimize the calculation work, we make use of the coding method. We subtract three from all the given figures and obtain coded values as shown in [Table 12].
T = 27, N = 24, correction factor = = 135.375
SS_{t} = (2^{2} + 3^{2} + 2^{2} + 2^{2} + 3^{2} + 4^{2} + 1^{2} + 0^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 0^{2} + 1^{2} + 0^{2} + 0^{2} + 2^{2} + 3^{2} + 0^{2} + 0^{2} + 4^{2} + 5^{2} + 5^{2} + 6^{2}) − 135.375 = 217 − 135.375 = 81.625
SS_{b}, i.e., between brands, = 135.375 = 152.833 – 135.375 = 17.458
SS_{r}, i.e., between symptoms 135.375 = 174.625 − 135.375 = 39.25
SS_{w} = (2 − 2.5)^{2} + (3 − 2.5)^{2} + (2 − 2)^{2} + (2 − 2)^{2} + (3 − 3.5)^{2} + (4 – 3.5)^{2} + (1 − 0.5)^{2} + (0 − 0.5)^{2} + (2 − 2.5)^{2} + (3 − 2.5)^{2} + (4 − 4.5)^{2} + (5 − 4.5)^{2} + (0 − 0.5)^{2} + (1 − 0.5)^{2} + (0 − 0)^{2} + (0 − 0)^{2} + (2 − 2.5)^{2} + (3 − 2.5)^{2} + (0 − 0)^{2} + (0 − 0)^{2} + (4 − 4.5)^{2} + (5 − 4.5)^{2} + (5 − 5.5)^{2} + (6 − 5.5)^{2} = 4.5
SS_{e} = 81.625 – (17.458 + 39.25 + 4.5) = 20.417.
The information is summarized in [Table 13].
Preparation of analysis of variance table using online calculator
For userfriendly online calculator, you may visit this uniform locator https://home.ubalt.edu/ntsbarsh/Businessstat/otherapplets/ANOVA2Rep.htm. Enter your uptotwo replications for each blockandtreatment pairs [Table 14].
Click the calculate button to generate the results.
The output from the online calculator is summarized in [Table 15].
Decision
Since all the calculated F statistics are more than the corresponding critical values (15.518 > 3.49, 52.333 > 3.89, and 9.074 > 3.11), therefore, we reject the three null hypotheses [Table 13]. Similarly, from [Table 15], since the three P values are less than α (0.00069 < 0.05, 0.005 < 0.05, and 0.00119 < 0.05), we also reject the three null hypotheses.
Graphical method of analyzing interaction in a twoway design
Interaction is said to exist when the difference in the response variable on the levels of one factor depends on the level of the other factor.^{[17],[19]} To visually look for possible interactions, we construct a type of graph known as an interaction plot. To construct an interaction plot, we first compute the treatment mean (average) for all r × c possible treatment combinations. The vertical axis of the plot is the response variable. The horizontal axis represents levels of one of the factors. The profiles represent the levels of the other factor. If the line segments are parallel (or nearly parallel), there is no significant interaction. If the profiles are clearly not parallel, then an interaction exists. For Example 4, averages are shown in [Table 16] and interaction plot as [Figure 4].
[Figure 4] indicates that there is a significant interaction because the different connecting lines for the symptoms do cross over each other. The shortest time to pain relief for mild, severe, and severe and very severe symptoms are caused by brand D, A, and C, respectively.
Conceptual Framework for Conducting TwoWay Analysis of Variance   
The above concepts of conducting twoway ANOVA are summarized by a conceptual framework in [Figure 5].  Figure 5: Conceptual framework on conducting twoway analysis of variance.
Click here to view 
What Does TwoWay Analysis of Variance Tell and What It Does Not?   
When we are interested in the effects of two factors, a twoway design offers great advantages over two singlefactor studies. Twoway ANOVA measures the variation between columns, rows, and interaction. The means can also be used in an interaction plot to aid in the interpretation of the results. When there is interaction, main effects can be meaningful and important, but this is not always the case. Compared with using multiple ttests, twoway ANOVA requires fewer measurements to discover significant effects (i.e., the tests are said to have more power). This is one reason why ANOVA is used frequently when analyzing data from statistically designed experiments. Other ANOVA and multivariate ANOVA methods exist for more complex experimental situations, but a description of these is beyond the scope of this article.
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Conflicts of interest
There are no conflicts of interest.
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[Figure 1], [Figure 2], [Figure 3], [Figure 4], [Figure 5]
[Table 1], [Table 2], [Table 3], [Table 4], [Table 5], [Table 6], [Table 7], [Table 8], [Table 9], [Table 10], [Table 11], [Table 12], [Table 13], [Table 14], [Table 15], [Table 16]
