anova hypothesis test hypotheses

Hypothesis Testing - Analysis of Variance (ANOVA)

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The ANOVA Approach

Test statistic for anova.

All Modules

Table of F-Statistic Values

Consider an example with four independent groups and a continuous outcome measure. The independent groups might be defined by a particular characteristic of the participants such as BMI (e.g., underweight, normal weight, overweight, obese) or by the investigator (e.g., randomizing participants to one of four competing treatments, call them A, B, C and D). Suppose that the outcome is systolic blood pressure, and we wish to test whether there is a statistically significant difference in mean systolic blood pressures among the four groups. The sample data are organized as follows:

The hypotheses of interest in an ANOVA are as follows:

• H 0 : μ 1 = μ 2 = μ 3 ... = μ k
• H 1 : Means are not all equal.

where k = the number of independent comparison groups.

In this example, the hypotheses are:

• H 0 : μ 1 = μ 2 = μ 3 = μ 4
• H 1 : The means are not all equal.

The null hypothesis in ANOVA is always that there is no difference in means. The research or alternative hypothesis is always that the means are not all equal and is usually written in words rather than in mathematical symbols. The research hypothesis captures any difference in means and includes, for example, the situation where all four means are unequal, where one is different from the other three, where two are different, and so on. The alternative hypothesis, as shown above, capture all possible situations other than equality of all means specified in the null hypothesis.

The test statistic for testing H 0 : μ 1 = μ 2 = ... =   μ k is:

and the critical value is found in a table of probability values for the F distribution with (degrees of freedom) df 1 = k-1, df 2 =N-k. The table can be found in "Other Resources" on the left side of the pages.

NOTE: The test statistic F assumes equal variability in the k populations (i.e., the population variances are equal, or s 1 2 = s 2 2 = ... = s k 2 ). This means that the outcome is equally variable in each of the comparison populations. This assumption is the same as that assumed for appropriate use of the test statistic to test equality of two independent means. It is possible to assess the likelihood that the assumption of equal variances is true and the test can be conducted in most statistical computing packages. If the variability in the k comparison groups is not similar, then alternative techniques must be used.

The F statistic is computed by taking the ratio of what is called the "between treatment" variability to the "residual or error" variability. This is where the name of the procedure originates. In analysis of variance we are testing for a difference in means (H 0 : means are all equal versus H 1 : means are not all equal) by evaluating variability in the data. The numerator captures between treatment variability (i.e., differences among the sample means) and the denominator contains an estimate of the variability in the outcome. The test statistic is a measure that allows us to assess whether the differences among the sample means (numerator) are more than would be expected by chance if the null hypothesis is true. Recall in the two independent sample test, the test statistic was computed by taking the ratio of the difference in sample means (numerator) to the variability in the outcome (estimated by Sp).  

The decision rule 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 significance and the degrees of freedom. The F statistic has two degrees of freedom. These are denoted df 1 and df 2 , and called the numerator and denominator degrees of freedom, respectively. The degrees of freedom are defined as follows:

df 1 = k-1 and df 2 =N-k,

where k is the number of comparison groups and N is the total number of observations in the analysis.   If the null hypothesis is true, the between treatment variation (numerator) will not exceed the residual or error variation (denominator) and the F statistic will small. If the null hypothesis is false, then the F statistic will be large. The rejection region for the F test is always in the upper (right-hand) tail of the distribution as shown below.

Rejection Region for F   Test with a =0.05, df 1 =3 and df 2 =36 (k=4, N=40)

For the scenario depicted here, the decision rule is: Reject H 0 if F > 2.87.

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Statistics and probability

Course: statistics and probability   >   unit 16.

• ANOVA 1: Calculating SST (total sum of squares)
• ANOVA 2: Calculating SSW and SSB (total sum of squares within and between)

ANOVA 3: Hypothesis test with F-statistic

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Hypothesis Testing and ANOVA

• First Online: 20 July 2018

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• Marko Sarstedt 3 &
• Erik Mooi 4  

Part of the book series: Springer Texts in Business and Economics ((STBE))

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We first describe the essentials of hypothesis testing and how testing helps make critical business decisions of statistical and practical significance. Without using difficult mathematical formulas, we discuss the steps involved in hypothesis testing, the types of errors that may occur, and provide strategies on how to best deal with these errors. We also discuss common types of test statistics and explain how to determine which type you should use in which specific situation. We explain that the test selection depends on the testing situation, the nature of the samples, the choice of test, and the region of rejection. Drawing on a case study, we show how to link hypothesis testing logic to empirics in SPSS. The case study touches upon different test situations and helps you interpret the tables and graphics in a quick and meaningful way.

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In experimental studies, if respondents were paired with others (as in a matched case control sample), each person would be sampled once, but it still would be a paired sample.

The exact calculation of this test is shown on https://www.ibm.com/support/knowledgecenter/en/SSLVMB_20.0.0/com.ibm.spss.statistics.help/alg_npar_tests_mannwhitney.htm

The fundamental difference between the z - and t -distributions is that the t -distribution is dependent on sample size n (which the z -distribution is not). The distributions become more similar with larger values of n .

To obtain the critical value, you can also use the TINV function provided in Microsoft Excel, whose general form is “TINV( α, df ).” Here, α represents the desired Type I error rate and df the degrees of freedom. To carry out this computation, open a new Excel spreadsheet and type in “ = TINV(2*0.025,9).” Note that we have to specify “2*0.025” (or, directly 0.05) under α , because we are applying a two-tailed instead of a one-tailed test.

Unfortunately, there is some confusion about the difference between the α and p -value. See Hubbard and Bayarri ( 2003 ) for a discussion.

Note that this is convention and most textbooks discuss hypothesis testing in this way. Originally, two testing procedures were developed, one by Neyman and Pearson and another by Fisher (for more details, see Lehmann 1993 ). Agresti and Finlay ( 2014 ) explain the differences between the convention and the two original procedures.

Note that this rule doesn't always apply such as for exact tests of probabilities.

We don’t have to conduct manual calculations and tables when working with SPSS. However, we can calculate the p -value using the TDIST function in Microsoft Excel. The function has the general form “TDIST( t, df , tails)”, where t describes the test value, df the degrees of freedom, and tails specifies whether it’s a one-tailed test (tails = 1) or two-tailed test (tails = 2). Just open a new spreadsheet for our example and type in “ = TDIST(2.274,9,1)”. Likewise, there are several webpages with Java-based modules (e.g., https://graphpad.com/quickcalcs/pvalue1.cfm ) that calculate p -values and test statistic values.

The number of pairwise comparisons is calculated as follows : k· ( k − 1)/2, with k the number of groups to compare.

In fact, these two assumptions are interrelated, since unequal group sample sizes result in a greater probability that we will violate the homogeneity assumption.

SS is an abbreviation of “sum of squares,” because the variation is calculated using the squared differences between different types of values.

Note that the group-specific sample size in this example is too small to draw conclusions and is only used to show the calculation of the statistics.

Note that when initiating the analysis by going to ► Analyze ► General Linear Model ► Univariate, we can request these statistics under Options ( Estimates of effect size ).

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Further Readings

Kanji, G. K. (2006). 100 statistical tests (3rd ed.). London: Sage.

Van Belle, G. (2011). Statistical rules of thumb (2nd ed.). Hoboken, N.J.: John Wiley & Sons.

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Home » ANOVA (Analysis of variance) – Formulas, Types, and Examples

ANOVA (Analysis of variance) – Formulas, Types, and Examples

Table of Contents

Analysis of Variance (ANOVA)

Analysis of Variance (ANOVA) is a statistical method used to test differences between two or more means. It is similar to the t-test, but the t-test is generally used for comparing two means, while ANOVA is used when you have more than two means to compare.

ANOVA is based on comparing the variance (or variation) between the data samples to the variation within each particular sample. If the between-group variance is high and the within-group variance is low, this provides evidence that the means of the groups are significantly different.

ANOVA Terminology

When discussing ANOVA, there are several key terms to understand:

• Factor : This is another term for the independent variable in your analysis. In a one-way ANOVA, there is one factor, while in a two-way ANOVA, there are two factors.
• Levels : These are the different groups or categories within a factor. For example, if the factor is ‘diet’ the levels might be ‘low fat’, ‘medium fat’, and ‘high fat’.
• Response Variable : This is the dependent variable or the outcome that you are measuring.
• Within-group Variance : This is the variance or spread of scores within each level of your factor.
• Between-group Variance : This is the variance or spread of scores between the different levels of your factor.
• Grand Mean : This is the overall mean when you consider all the data together, regardless of the factor level.
• Treatment Sums of Squares (SS) : This represents the between-group variability. It is the sum of the squared differences between the group means and the grand mean.
• Error Sums of Squares (SS) : This represents the within-group variability. It’s the sum of the squared differences between each observation and its group mean.
• Total Sums of Squares (SS) : This is the sum of the Treatment SS and the Error SS. It represents the total variability in the data.
• Degrees of Freedom (df) : The degrees of freedom are the number of values that have the freedom to vary when computing a statistic. For example, if you have ‘n’ observations in one group, then the degrees of freedom for that group is ‘n-1’.
• Mean Square (MS) : Mean Square is the average squared deviation and is calculated by dividing the sum of squares by the corresponding degrees of freedom.
• F-Ratio : This is the test statistic for ANOVAs, and it’s the ratio of the between-group variance to the within-group variance. If the between-group variance is significantly larger than the within-group variance, the F-ratio will be large and likely significant.
• Null Hypothesis (H0) : This is the hypothesis that there is no difference between the group means.
• Alternative Hypothesis (H1) : This is the hypothesis that there is a difference between at least two of the group means.
• p-value : This is the probability of obtaining a test statistic as extreme as the one that was actually observed, assuming that the null hypothesis is true. If the p-value is less than the significance level (usually 0.05), then the null hypothesis is rejected in favor of the alternative hypothesis.
• Post-hoc tests : These are follow-up tests conducted after an ANOVA when the null hypothesis is rejected, to determine which specific groups’ means (levels) are different from each other. Examples include Tukey’s HSD, Scheffe, Bonferroni, among others.

Types of ANOVA

Types of ANOVA are as follows:

One-way (or one-factor) ANOVA

This is the simplest type of ANOVA, which involves one independent variable . For example, comparing the effect of different types of diet (vegetarian, pescatarian, omnivore) on cholesterol level.

Two-way (or two-factor) ANOVA

This involves two independent variables. This allows for testing the effect of each independent variable on the dependent variable , as well as testing if there’s an interaction effect between the independent variables on the dependent variable.

Repeated Measures ANOVA

This is used when the same subjects are measured multiple times under different conditions, or at different points in time. This type of ANOVA is often used in longitudinal studies.

Mixed Design ANOVA

This combines features of both between-subjects (independent groups) and within-subjects (repeated measures) designs. In this model, one factor is a between-subjects variable and the other is a within-subjects variable.

Multivariate Analysis of Variance (MANOVA)

This is used when there are two or more dependent variables. It tests whether changes in the independent variable(s) correspond to changes in the dependent variables.

Analysis of Covariance (ANCOVA)

This combines ANOVA and regression. ANCOVA tests whether certain factors have an effect on the outcome variable after removing the variance for which quantitative covariates (interval variables) account. This allows the comparison of one variable outcome between groups, while statistically controlling for the effect of other continuous variables that are not of primary interest.

Nested ANOVA

This model is used when the groups can be clustered into categories. For example, if you were comparing students’ performance from different classrooms and different schools, “classroom” could be nested within “school.”

ANOVA Formulas

ANOVA Formulas are as follows:

Sum of Squares Total (SST)

This represents the total variability in the data. It is the sum of the squared differences between each observation and the overall mean.

• yi represents each individual data point
• y_mean represents the grand mean (mean of all observations)

Sum of Squares Within (SSW)

This represents the variability within each group or factor level. It is the sum of the squared differences between each observation and its group mean.

• yij represents each individual data point within a group
• y_meani represents the mean of the ith group

Sum of Squares Between (SSB)

This represents the variability between the groups. It is the sum of the squared differences between the group means and the grand mean, multiplied by the number of observations in each group.

• ni represents the number of observations in each group
• y_mean represents the grand mean

Degrees of Freedom

The degrees of freedom are the number of values that have the freedom to vary when calculating a statistic.

For within groups (dfW):

For between groups (dfB):

For total (dfT):

• N represents the total number of observations
• k represents the number of groups

Mean Squares

Mean squares are the sum of squares divided by the respective degrees of freedom.

Mean Squares Between (MSB):

Mean Squares Within (MSW):

F-Statistic

The F-statistic is used to test whether the variability between the groups is significantly greater than the variability within the groups.

If the F-statistic is significantly higher than what would be expected by chance, we reject the null hypothesis that all group means are equal.

Examples of ANOVA

Examples 1:

Suppose a psychologist wants to test the effect of three different types of exercise (yoga, aerobic exercise, and weight training) on stress reduction. The dependent variable is the stress level, which can be measured using a stress rating scale.

Here are hypothetical stress ratings for a group of participants after they followed each of the exercise regimes for a period:

• Yoga: [3, 2, 2, 1, 2, 2, 3, 2, 1, 2]
• Aerobic Exercise: [2, 3, 3, 2, 3, 2, 3, 3, 2, 2]
• Weight Training: [4, 4, 5, 5, 4, 5, 4, 5, 4, 5]

The psychologist wants to determine if there is a statistically significant difference in stress levels between these different types of exercise.

To conduct the ANOVA:

1. State the hypotheses:

• Null Hypothesis (H0): There is no difference in mean stress levels between the three types of exercise.
• Alternative Hypothesis (H1): There is a difference in mean stress levels between at least two of the types of exercise.

2. Calculate the ANOVA statistics:

• Compute the Sum of Squares Between (SSB), Sum of Squares Within (SSW), and Sum of Squares Total (SST).
• Calculate the Degrees of Freedom (dfB, dfW, dfT).
• Calculate the Mean Squares Between (MSB) and Mean Squares Within (MSW).
• Compute the F-statistic (F = MSB / MSW).

3. Check the p-value associated with the calculated F-statistic.

• If the p-value is less than the chosen significance level (often 0.05), then we reject the null hypothesis in favor of the alternative hypothesis. This suggests there is a statistically significant difference in mean stress levels between the three exercise types.

4. Post-hoc tests

• If we reject the null hypothesis, we conduct a post-hoc test to determine which specific groups’ means (exercise types) are different from each other.

Examples 2:

Suppose an agricultural scientist wants to compare the yield of three varieties of wheat. The scientist randomly selects four fields for each variety and plants them. After harvest, the yield from each field is measured in bushels. Here are the hypothetical yields:

The scientist wants to know if the differences in yields are due to the different varieties or just random variation.

Here’s how to apply the one-way ANOVA to this situation:

• Null Hypothesis (H0): The means of the three populations are equal.
• Alternative Hypothesis (H1): At least one population mean is different.
• Calculate the Degrees of Freedom (dfB for between groups, dfW for within groups, dfT for total).
• If the p-value is less than the chosen significance level (often 0.05), then we reject the null hypothesis in favor of the alternative hypothesis. This would suggest there is a statistically significant difference in mean yields among the three varieties.
• If we reject the null hypothesis, we conduct a post-hoc test to determine which specific groups’ means (wheat varieties) are different from each other.

How to Conduct ANOVA

Conducting an Analysis of Variance (ANOVA) involves several steps. Here’s a general guideline on how to perform it:

• Null Hypothesis (H0): The means of all groups are equal.
• Alternative Hypothesis (H1): At least one group mean is different from the others.
• The significance level (often denoted as α) is usually set at 0.05. This implies that you are willing to accept a 5% chance that you are wrong in rejecting the null hypothesis.
• Data should be collected for each group under study. Make sure that the data meet the assumptions of an ANOVA: normality, independence, and homogeneity of variances.
• Calculate the Degrees of Freedom (df) for each sum of squares (dfB, dfW, dfT).
• Compute the Mean Squares Between (MSB) and Mean Squares Within (MSW) by dividing the sum of squares by the corresponding degrees of freedom.
• Compute the F-statistic as the ratio of MSB to MSW.
• Determine the critical F-value from the F-distribution table using dfB and dfW.
• If the calculated F-statistic is greater than the critical F-value, reject the null hypothesis.
• If the p-value associated with the calculated F-statistic is smaller than the significance level (0.05 typically), you reject the null hypothesis.
• If you rejected the null hypothesis, you can conduct post-hoc tests (like Tukey’s HSD) to determine which specific groups’ means (if you have more than two groups) are different from each other.
• Regardless of the result, report your findings in a clear, understandable manner. This typically includes reporting the test statistic, p-value, and whether the null hypothesis was rejected.

When to use ANOVA

ANOVA (Analysis of Variance) is used when you have three or more groups and you want to compare their means to see if they are significantly different from each other. It is a statistical method that is used in a variety of research scenarios. Here are some examples of when you might use ANOVA:

• Comparing Groups : If you want to compare the performance of more than two groups, for example, testing the effectiveness of different teaching methods on student performance.
• Evaluating Interactions : In a two-way or factorial ANOVA, you can test for an interaction effect. This means you are not only interested in the effect of each individual factor, but also whether the effect of one factor depends on the level of another factor.
• Repeated Measures : If you have measured the same subjects under different conditions or at different time points, you can use repeated measures ANOVA to compare the means of these repeated measures while accounting for the correlation between measures from the same subject.
• Experimental Designs : ANOVA is often used in experimental research designs when subjects are randomly assigned to different conditions and the goal is to compare the means of the conditions.

Here are the assumptions that must be met to use ANOVA:

• Normality : The data should be approximately normally distributed.
• Homogeneity of Variances : The variances of the groups you are comparing should be roughly equal. This assumption can be tested using Levene’s test or Bartlett’s test.
• Independence : The observations should be independent of each other. This assumption is met if the data is collected appropriately with no related groups (e.g., twins, matched pairs, repeated measures).

Applications of ANOVA

The Analysis of Variance (ANOVA) is a powerful statistical technique that is used widely across various fields and industries. Here are some of its key applications:

Agriculture

ANOVA is commonly used in agricultural research to compare the effectiveness of different types of fertilizers, crop varieties, or farming methods. For example, an agricultural researcher could use ANOVA to determine if there are significant differences in the yields of several varieties of wheat under the same conditions.

Manufacturing and Quality Control

ANOVA is used to determine if different manufacturing processes or machines produce different levels of product quality. For instance, an engineer might use it to test whether there are differences in the strength of a product based on the machine that produced it.

Marketing Research

Marketers often use ANOVA to test the effectiveness of different advertising strategies. For example, a marketer could use ANOVA to determine whether different marketing messages have a significant impact on consumer purchase intentions.

Healthcare and Medicine

In medical research, ANOVA can be used to compare the effectiveness of different treatments or drugs. For example, a medical researcher could use ANOVA to test whether there are significant differences in recovery times for patients who receive different types of therapy.

ANOVA is used in educational research to compare the effectiveness of different teaching methods or educational interventions. For example, an educator could use it to test whether students perform significantly differently when taught with different teaching methods.

Psychology and Social Sciences

Psychologists and social scientists use ANOVA to compare group means on various psychological and social variables. For example, a psychologist could use it to determine if there are significant differences in stress levels among individuals in different occupations.

Biology and Environmental Sciences

Biologists and environmental scientists use ANOVA to compare different biological and environmental conditions. For example, an environmental scientist could use it to determine if there are significant differences in the levels of a pollutant in different bodies of water.

Advantages of ANOVA

Here are some advantages of using ANOVA:

Comparing Multiple Groups: One of the key advantages of ANOVA is the ability to compare the means of three or more groups. This makes it more powerful and flexible than the t-test, which is limited to comparing only two groups.

Control of Type I Error: When comparing multiple groups, the chances of making a Type I error (false positive) increases. One of the strengths of ANOVA is that it controls the Type I error rate across all comparisons. This is in contrast to performing multiple pairwise t-tests which can inflate the Type I error rate.

Testing Interactions: In factorial ANOVA, you can test not only the main effect of each factor, but also the interaction effect between factors. This can provide valuable insights into how different factors or variables interact with each other.

Handling Continuous and Categorical Variables: ANOVA can handle both continuous and categorical variables . The dependent variable is continuous and the independent variables are categorical.

Robustness: ANOVA is considered robust to violations of normality assumption when group sizes are equal. This means that even if your data do not perfectly meet the normality assumption, you might still get valid results.

Provides Detailed Analysis: ANOVA provides a detailed breakdown of variances and interactions between variables which can be useful in understanding the underlying factors affecting the outcome.

Capability to Handle Complex Experimental Designs: Advanced types of ANOVA (like repeated measures ANOVA, MANOVA, etc.) can handle more complex experimental designs, including those where measurements are taken on the same subjects over time, or when you want to analyze multiple dependent variables at once.

Disadvantages of ANOVA

Some limitations or disadvantages that are important to consider:

Assumptions: ANOVA relies on several assumptions including normality (the data follows a normal distribution), independence (the observations are independent of each other), and homogeneity of variances (the variances of the groups are roughly equal). If these assumptions are violated, the results of the ANOVA may not be valid.

Sensitivity to Outliers: ANOVA can be sensitive to outliers. A single extreme value in one group can affect the sum of squares and consequently influence the F-statistic and the overall result of the test.

Dichotomous Variables: ANOVA is not suitable for dichotomous variables (variables that can take only two values, like yes/no or male/female). It is used to compare the means of groups for a continuous dependent variable.

Lack of Specificity: Although ANOVA can tell you that there is a significant difference between groups, it doesn’t tell you which specific groups are significantly different from each other. You need to carry out further post-hoc tests (like Tukey’s HSD or Bonferroni) for these pairwise comparisons.

Complexity with Multiple Factors: When dealing with multiple factors and interactions in factorial ANOVA, interpretation can become complex. The presence of interaction effects can make main effects difficult to interpret.

Requires Larger Sample Sizes: To detect an effect of a certain size, ANOVA generally requires larger sample sizes than a t-test.

Equal Group Sizes: While not always a strict requirement, ANOVA is most powerful and its assumptions are most likely to be met when groups are of equal or similar sizes.

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Statistics Made Easy

Understanding the Null Hypothesis for ANOVA Models

A one-way ANOVA is used to determine if there is a statistically significant difference between the mean of three or more independent groups.

A one-way ANOVA uses the following null and alternative hypotheses:

• H 0 :  μ 1  = μ 2  = μ 3  = … = μ k  (all of the group means are equal)
• H A : At least one group mean is different   from the rest

To decide if we should reject or fail to reject the null hypothesis, we must refer to the p-value in the output of the ANOVA table.

If the p-value is less than some significance level (e.g. 0.05) then we can reject the null hypothesis and conclude that not all group means are equal.

A two-way ANOVA is used to determine whether or not there is a statistically significant difference between the means of three or more independent groups that have been split on two variables (sometimes called “factors”).

A two-way ANOVA tests three null hypotheses at the same time:

• All group means are equal at each level of the first variable
• All group means are equal at each level of the second variable
• There is no interaction effect between the two variables

To decide if we should reject or fail to reject each null hypothesis, we must refer to the p-values in the output of the two-way ANOVA table.

The following examples show how to decide to reject or fail to reject the null hypothesis in both a one-way ANOVA and two-way ANOVA.

Example 1: One-Way ANOVA

Suppose we want to know whether or not three different exam prep programs lead to different mean scores on a certain exam. To test this, we recruit 30 students to participate in a study and split them into three groups.

The students in each group are randomly assigned to use one of the three exam prep programs for the next three weeks to prepare for an exam. At the end of the three weeks, all of the students take the same exam. 

The exam scores for each group are shown below:

When we enter these values into the One-Way ANOVA Calculator , we receive the following ANOVA table as the output:

Notice that the p-value is 0.11385 .

For this particular example, we would use the following null and alternative hypotheses:

• H 0 :  μ 1  = μ 2  = μ 3 (the mean exam score for each group is equal)

Since the p-value from the ANOVA table is not less than 0.05, we fail to reject the null hypothesis.

This means we don’t have sufficient evidence to say that there is a statistically significant difference between the mean exam scores of the three groups.

Example 2: Two-Way ANOVA

Suppose a botanist wants to know whether or not plant growth is influenced by sunlight exposure and watering frequency.

She plants 40 seeds and lets them grow for two months under different conditions for sunlight exposure and watering frequency. After two months, she records the height of each plant. The results are shown below:

In the table above, we see that there were five plants grown under each combination of conditions.

For example, there were five plants grown with daily watering and no sunlight and their heights after two months were 4.8 inches, 4.4 inches, 3.2 inches, 3.9 inches, and 4.4 inches:

She performs a two-way ANOVA in Excel and ends up with the following output:

We can see the following p-values in the output of the two-way ANOVA table:

• The p-value for watering frequency is 0.975975 . This is not statistically significant at a significance level of 0.05.
• The p-value for sunlight exposure is 3.9E-8 (0.000000039) . This is statistically significant at a significance level of 0.05.
• The p-value for the interaction between watering  frequency and sunlight exposure is 0.310898 . This is not statistically significant at a significance level of 0.05.

These results indicate that sunlight exposure is the only factor that has a statistically significant effect on plant height.

And because there is no interaction effect, the effect of sunlight exposure is consistent across each level of watering frequency.

That is, whether a plant is watered daily or weekly has no impact on how sunlight exposure affects a plant.

Additional Resources

The following tutorials provide additional information about ANOVA models:

How to Interpret the F-Value and P-Value in ANOVA How to Calculate Sum of Squares in ANOVA What Does a High F Value Mean in ANOVA?

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2 Replies to “Understanding the Null Hypothesis for ANOVA Models”

Hi, I’m a student at Stellenbosch University majoring in Conservation Ecology and Entomology and we are currently busy doing stats. I am still at a very entry level of stats understanding, so pages like these are of huge help. I wanted to ask, why is the sum of squares (treatment) for the one way ANOVA so high? I calculated it by hand and got a much lower number, could you please help point out if and where I went wrong?

As I understand it, SSB (treatment) is calculated by finding the mean of each group and the grand mean, and then calculating the sum of squares like this: GM = 85.5 x1 = 83.4 x2 = 89.3 x3 = 84.7

SSB = (85.5 – 83.4)^2 + (85.5 – 89.3)^2 + (85.5 – 84.7)^2 = 18.65 DF = 2

I would appreciate any help, thank you so much!

Hi Theo…Certainly! Here are the equations rewritten as they would be typed in Python:

### Sum of Squares Between Groups (SSB)

In a one-way ANOVA, the sum of squares between groups (SSB) measures the variation due to the interaction between the groups. It is calculated as follows:

1. **Calculate the group means**: “`python mean_group1 = 83.4 mean_group2 = 89.3 mean_group3 = 84.7 “`

2. **Calculate the grand mean**: “`python grand_mean = 85.5 “`

3. **Calculate the sum of squares between groups (SSB)**: Assuming each group has `n` observations: “`python n = 10 # Number of observations in each group

ssb = n * ((mean_group1 – grand_mean)**2 + (mean_group2 – grand_mean)**2 + (mean_group3 – grand_mean)**2) “`

### Example Calculation

For simplicity, let’s assume each group has 10 observations: “`python n = 10

ssb = n * ((83.4 – 85.5)**2 + (89.3 – 85.5)**2 + (84.7 – 85.5)**2) “`

Now calculate each term: “`python term1 = (83.4 – 85.5)**2 # term1 = (-2.1)**2 = 4.41 term2 = (89.3 – 85.5)**2 # term2 = (3.8)**2 = 14.44 term3 = (84.7 – 85.5)**2 # term3 = (-0.8)**2 = 0.64 “`

Sum these squared differences: “`python sum_of_squared_diffs = term1 + term2 + term3 # sum_of_squared_diffs = 4.41 + 14.44 + 0.64 = 19.49 ssb = n * sum_of_squared_diffs # ssb = 10 * 19.49 = 194.9 “`

So, the sum of squares between groups (SSB) is 194.9, assuming each group has 10 observations.

### Degrees of Freedom (DF)

The degrees of freedom for SSB is calculated as: “`python df_between = k – 1 “` where `k` is the number of groups.

For three groups: “`python k = 3 df_between = k – 1 # df_between = 3 – 1 = 2 “`

### Summary

– **SSB** should consider the number of observations in each group. – **DF** is the number of groups minus one.

By ensuring you include the number of observations per group in your SSB calculation, you can get the correct SSB value.

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11.4 One-Way ANOVA and Hypothesis Tests for Three or More Population Means

Learning objectives.

• Conduct and interpret hypothesis tests for three or more population means using one-way ANOVA.

The purpose of a one-way ANOVA (analysis of variance) test is to determine the existence of a statistically significant difference among the means of three or more populations.  The test actually uses variances to help determine if the population means are equal or not.

Throughout this section, we will use subscripts to identify the values for the means, sample sizes, and standard deviations for the populations:

[latex]k[/latex] is the number of populations under study, [latex]n[/latex] is the total number of observations in all of the samples combined, and [latex]\overline{\overline{x}}[/latex] is the mean of the sample means.

[latex]\begin{eqnarray*} n & = & n_1+n_2+\cdots+n_k \\ \\ \overline{\overline{x}} & = & \frac{n_1 \times \overline{x}_1 +n_2 \times \overline{x}_2 +\cdots+n_k \times \overline{x}_k}{n} \end{eqnarray*}[/latex]

One-Way ANOVA

A predictor variable is called a factor or independent variable .  For example age, temperature, and gender are factors.  The groups or samples are often referred to as treatments .  This terminology comes from the use of ANOVA procedures in medical and psychological research to determine if there is a difference in the effects of different treatments.

A local college wants to compare the mean GPA for players on four of its sports teams:  basketball, baseball, hockey, and lacrosse.  A random sample of players was taken from each team and their GPA recorded in the table below.

In this example, the factor is the sports team.

[latex]\begin{eqnarray*} k & = & 4 \\ \\ n & = & n_1+n_2+n_3+n_4 \\ & = & 5+5+5+5 \\ & = & 20 \\ \\ \overline{\overline{x}} & = & \frac{n_1 \times \overline{x}_1+n_2 \times \overline{x}_2+n_3 \times \overline{x}_3+n_4 \times \overline{x}_4}{n} \\ & = & \frac{5 \times 3.22+5 \times 3.02+5 \times 3+5 \times 2.94}{20}  \\& = & 3.045 \end{eqnarray*}[/latex]

The following assumptions are required to use a one-way ANOVA test:

• Each population from which a sample is taken is normally distributed.
• All samples are randomly selected and independently taken from the populations.
• The populations are assumed to have equal variances.
• The population data is numerical (interval or ratio level).

The logic behind one-way ANOVA is to compare population means based on two independent estimates of the (assumed) equal variance [latex]\sigma^2[/latex] between the populations:

• One estimate of the equal variance [latex]\sigma^2[/latex] is based on the variability among the sample means themselves (called the between-groups estimate of population variance).
• One estimate of the equal variance [latex]\sigma^2[/latex] is based on the variability of the data within each sample (called the within-groups estimate of population variance).

The one-way ANOVA procedure compares these two estimates of the population variance [latex]\sigma^2[/latex] to determine if the population means are equal or if there is a difference in the population means.  Because ANOVA involves the comparison of two estimates of variance, an [latex]F[/latex]-distribution is used to conduct the ANOVA test.  The test statistic is an [latex]F[/latex]-score that is the ratio of the two estimates of population variance:

[latex]\displaystyle{F=\frac{\mbox{variance between groups}}{\mbox{variance within groups}}}[/latex]

The degrees of freedom for the [latex]F[/latex]-distribution are [latex]df_1=k-1[/latex] and [latex]df_2=n-k[/latex] where [latex]k[/latex] is the number of populations and [latex]n[/latex] is the total number of observations in all of the samples combined.

The variance between groups estimate of the population variance is called the mean square due to treatment , [latex]MST[/latex].  The [latex]MST[/latex] is the estimate of the population variance determined by the variance of the sample means from the overall sample mean [latex]\overline{\overline{x}}[/latex].  When the population means are equal, [latex]MST[/latex] provides an unbiased estimate of the population variance.  When the population means are not equal, [latex]MST[/latex] provides an overestimate of the population variance.

[latex]\begin{eqnarray*} SST & = & n_1 \times (\overline{x}_1-\overline{\overline{x}})^2+n_2\times (\overline{x}_2-\overline{\overline{x}})^2+ \cdots +n_k \times (\overline{x}_k-\overline{\overline{x}})^2 \\  \\ MST & =& \frac{SST}{k-1} \end{eqnarray*}[/latex]

The variance within groups estimate of the population variance is called the mean square due to error , [latex]MSE[/latex].  The [latex]MSE[/latex] is the pooled estimate of the population variance using the sample variances as estimates for the population variance.  The [latex]MSE[/latex] always provides an unbiased estimate of the population variance because it is not affected by whether or not the population means are equal.

[latex]\begin{eqnarray*} SSE & = & (n_1-1) \times s_1^2+ (n_2-1) \times s_2^2+ \cdots + (n_k-1) \times s_k^2\\  \\ MSE & =& \frac{SSE}{n -k} \end{eqnarray*}[/latex]

The one-way ANOVA test depends on the fact that the variance between groups [latex]MST[/latex] is influenced by differences between the population means, which results in [latex]MST[/latex] being either an unbiased or overestimate of the population variance.  Because the variance within groups [latex]MSE[/latex] compares values of each group to its own group mean, [latex]MSE[/latex] is not affected by differences between the population means and is always an unbiased estimate of the population variance.

The null hypothesis in a one-way ANOVA test is that the population means are all equal and the alternative hypothesis is that there is a difference in the population means.  The [latex]F[/latex]-score for the one-way ANOVA test is [latex]\displaystyle{F=\frac{MST}{MSE}}[/latex] with [latex]df_1=k-1[/latex] and [latex]df_2=n-k[/latex].  The p -value for the test is the area in the right tail of the [latex]F[/latex]-distribution, to the right of the [latex]F[/latex]-score.

• When the variance between groups [latex]MST[/latex] and variance within groups [latex]MSE[/latex] are close in value, the [latex]F[/latex]-score is close to 1 and results in a large p -value.  In this case, the conclusion is that the population means are equal.
• When the variance between groups [latex]MST[/latex] is significantly larger than the variability within groups [latex]MSE[/latex], the [latex]F[/latex]-score is large and results in a small p -value.  In this case, the conclusion is that there is a difference in the population means.

Steps to Conduct a Hypothesis Test for Three or More Population Means

• Verify that the one-way ANOVA assumptions are met.

[latex]\begin{eqnarray*} \\ H_0: &  &  \mu_1=\mu_2=\cdots=\mu_k\end{eqnarray*}[/latex].

[latex]\begin{eqnarray*} \\ H_a: &  & \mbox{at least one population mean is different from the others} \\ \\ \end{eqnarray*}[/latex]

• Collect the sample information for the test and identify the significance level [latex]\alpha[/latex].

[latex]\begin{eqnarray*}F & = & \frac{MST}{MSE} \\ \\ df_1 & = & k-1 \\ \\ df_2 &  = & n-k \\ \\ \end{eqnarray*}[/latex]

• The results of the sample data are significant.  There is sufficient evidence to conclude that the null hypothesis [latex]H_0[/latex] is an incorrect belief and that the alternative hypothesis [latex]H_a[/latex] is most likely correct.
• The results of the sample data are not significant.  There is not sufficient evidence to conclude that the alternative hypothesis [latex]H_a[/latex] may be correct.
• Write down a concluding sentence specific to the context of the question.

Assume the populations are normally distributed and have equal variances.  At the 5% significance level, is there a difference in the average GPA between the sports team.

Let basketball be population 1, let baseball be population 2, let hockey be population 3, and let lacrosse be population 4. From the question we have the following information:

Previously, we found [latex]k=4[/latex], [latex]n=20[/latex], and [latex]\overline{\overline{x}}=3.045[/latex].

Hypotheses:

[latex]\begin{eqnarray*} H_0: & & \mu_1=\mu_2=\mu_3=\mu_4 \\   H_a: & & \mbox{at least one population mean is different from the others} \end{eqnarray*}[/latex]

To calculate out the [latex]F[/latex]-score, we need to find [latex]MST[/latex] and [latex]MSE[/latex].

[latex]\begin{eqnarray*} SST & = & n_1 \times (\overline{x}_1-\overline{\overline{x}})^2+n_2\times (\overline{x}_2-\overline{\overline{x}})^2+n_3 \times (\overline{x}_3-\overline{\overline{x}})^2  +n_4 \times (\overline{x}_4-\overline{\overline{x}})^2\\  & = & 5 \times (3.22-3.045)^2+5 \times (3.02-3.045)^2+5 \times (3-3.045)^2 \\ &  & +5 \times (2.94 -3.045)^2 \\ & = & 0.2215 \\ \\ MST & = & \frac{SST}{k-1} \\ & = & \frac{0.2215 }{4-1} \\ & = & 0.0738...\\ \\  SSE & = & (n_1-1) \times s_1^2+ (n_2-1) \times s_2^2+  (n_3-1) \times s_3^2+ (n_4-1) \times s_4^2\\  & = &( 5-1) \times 0.277+(5-1) \times 0.487+(5-1) \times 0.56 +(5-1)\times 0.623 \\ & = & 7.788 \\ \\ MSE & = & \frac{SSE}{n-k} \\ & = & \frac{7.788 }{20-4} \\ & = & 0.48675\end{eqnarray*}[/latex]

The p -value is the area in the right tail of the [latex]F[/latex]-distribution.  To use the f.dist.rt  function, we need to calculate out the [latex]F[/latex]-score and the degrees of freedom:

[latex]\begin{eqnarray*} F & = &\frac{MST}{MSE} \\ & = & \frac{0.0738...}{0.48675} \\ & = & 0.15168... \\ \\ df_1 & = & k-1 \\ & = & 4-1 \\ & = & 3 \\ \\df_2 & = & n-k \\ & = & 20-4 \\ & = & 16\end{eqnarray*}[/latex]

So the p -value[latex]=0.9271[/latex].

Conclusion:

Because p -value[latex]=0.9271 \gt 0.05=\alpha[/latex], we do not reject the null hypothesis.  At the 5% significance level there is  enough evidence to suggest that the mean GPA for the sports teams are the same.

• The null hypothesis [latex]\mu_1=\mu_2=\mu_3=\mu_4[/latex] is the claim that the mean GPA for the sports teams are all equal.
• The alternative hypothesis is the claim that at least one of the population means is not equal to the others.  The alternative hypothesis does not say that all of the population means are not equal, only that at least one of them is not equal to the others.
• The function is f.dist.rt because we are finding the area in the right tail of an [latex]F[/latex]-distribution.
• Field 1 is the value of [latex]F[/latex].
• Field 2 is the value of [latex]df_1[/latex].
• Field 3 is the value of [latex]df_2[/latex].
• The p -value of 0.9271 is a large probability compared to the significance level, and so is likely to happen assuming the null hypothesis is true.  This suggests that the assumption that the null hypothesis is true is most likely correct, and so the conclusion of the test is to not reject the null hypothesis.  In other words, the population means are all equal.

ANOVA Summary Tables

The calculation of the [latex]MST[/latex], [latex]MSE[/latex], and the [latex]F[/latex]-score for a one-way ANOVA test can be time consuming, even with the help of software like Excel.  However, Excel has a built-in one-way ANOVA summary table that not only generates the averages, variances, [latex]MST[/latex] and [latex]MSE[/latex], but also calculates the required [latex]F[/latex]-score and p -value for the test.

USING EXCEL TO CREATE A ONE-WAY ANOVA SUMMARY TABLE

In order to create a one-way ANOVA summary table, we need to use the Analysis ToolPak.  Follow these instructions to add the Analysis ToolPak.

• Enter the data into an Excel worksheet.
• Go to the Data tab and click on Data Analysis .  If you do not see Data Analysis in the Data tab, you will need to install the Analysis ToolPak.
• In the Data Analysis window, select Anova:  Single Factor .  Click OK .
• In the Inpu t range, enter the cell range for the data.
• In the Grouped By box, select rows if your data is entered as rows (the default is columns).
• Click on Labels in first row if the you included the column headings in the input range.
• In the Alpha box, enter the significance level for the test.
• From the Output Options , select the location where you want the output to appear.

This website provides additional information on using Excel to create a one-way ANOVA summary table.

Because we are using the p -value approach to hypothesis testing, it is not crucial that we enter the actual significance level we are using for the test.  The p -value (the area in the right tail of the [latex]F[/latex]-distribution) is not affected by significance level.  For the critical-value approach to hypothesis testing, we must enter the correct significance level for the test because the critical value does depend on the significance level.

Let basketball be population 1, let baseball be population 2, let hockey be population 3, and let lacrosse be population 4.

The ANOVA summary table generated by Excel is shown below:

The p -value for the test is in the P -value column of the between groups row .  So the p -value[latex]=0.9271[/latex].

• In the top part of the ANOVA summary table (under the Summary heading), we have the averages and variances for each of the groups (basketball, baseball, hockey, and lacrosse).
• The value of [latex]SST[/latex] (in the SS column of the between groups row).
• The value of [latex]MST[/latex] (in the MS column of the between group s row).
• The value of [latex]SSE[/latex] (in the SS column of the within groups row).
• The value of [latex]MSE[/latex] (in the MS column of the within groups row).
• The value of the [latex]F[/latex]-score (in the F column of the between groups row).
• The p -value (in the p -value column of the between groups row).

A fourth grade class is studying the environment.  One of the assignments is to grow bean plants in different soils.  Tommy chose to grow his bean plants in soil found outside his classroom mixed with dryer lint.  Tara chose to grow her bean plants in potting soil bought at the local nursery.  Nick chose to grow his bean plants in soil from his mother’s garden.  No chemicals were used on the plants, only water.  They were grown inside the classroom next to a large window.  Each child grew five plants.  At the end of the growing period, each plant was measured, producing the data (in inches) in the table below.

Assume the heights of the plants are normally distribution and have equal variance.  At the 5% significance level, does it appear that the three media in which the bean plants were grown produced the same mean height?

Let Tommy’s plants be population 1, let Tara’s plants be population 2, and let Nick’s plants be population 3.

[latex]\begin{eqnarray*} H_0: & & \mu_1=\mu_2=\mu_3 \\   H_a: & & \mbox{at least one population mean is different from the others} \end{eqnarray*}[/latex]

So the p -value[latex]=0.8760[/latex].

Because p -value[latex]=0.8760 \gt 0.05=\alpha[/latex], we do not reject the null hypothesis.  At the 5% significance level there is  enough evidence to suggest that the mean heights of the plants grown in three media are the same.

• The null hypothesis [latex]\mu_1=\mu_2=\mu_3[/latex] is the claim that the mean heights of the plants grown in the three different media are all equal.
• The p -value of 0.8760 is a large probability compared to the significance level, and so is likely to happen assuming the null hypothesis is true.  This suggests that the assumption that the null hypothesis is true is most likely correct, and so the conclusion of the test is to not reject the null hypothesis.  In other words, the population means are all equal.

A statistics professor wants to study the average GPA of students in four different programs: marketing, management, accounting, and human resources.  The professor took a random sample of GPAs of students in those programs at the end of the past semester.  The data is recorded in the table below.

Assume the GPAs of the students are normally distributed and have equal variance.  At the 5% significance level, is there a difference in the average GPA of the students in the different programs?

Let marketing be population 1, let management be population 2, let accounting be population 3, and let human resources be population 4.

[latex]\begin{eqnarray*} H_0: & & \mu_1=\mu_2=\mu_3=\mu_4\\   H_a: & & \mbox{at least one population mean is different from the others} \end{eqnarray*}[/latex]

So the p -value[latex]=0.0462[/latex].

Because p -value[latex]=0.0462 \lt 0.05=\alpha[/latex], we reject the null hypothesis in favour of the alternative hypothesis.  At the 5% significance level there is enough evidence to suggest that there is a difference in the average GPA of the students in the different programs.

A manufacturing company runs three different production lines to produce one of its products.  The company wants to know if the average production rate is the same for the three lines.  For each production line, a sample of eight hour shifts was taken and the number of items produced during each shift was recorded in the table below.

Assume the numbers of items produced on each line during an eight hour shift are normally distributed and have equal variance.  At the 1% significance level, is there a difference in the average production rate for the three lines?

Let Line 1 be population 1, let Line 2 be population 2, and let Line 3 be population 3.

So the p -value[latex]=0.0073[/latex].

Because p -value[latex]=0.0073 \lt 0.01=\alpha[/latex], we reject the null hypothesis in favour of the alternative hypothesis.  At the 1% significance level there is enough evidence to suggest that there is a difference in the average production rate of the three lines.

Concept Review

A one-way ANOVA hypothesis test determines if several population means are equal.  In order to conduct a one-way ANOVA test, the following assumptions must be met:

• Each population from which a sample is taken is assumed to be normal.
• All samples are randomly selected and independent.

The analysis of variance procedure compares the variation between groups [latex]MST[/latex] to the variation within groups [latex]MSE[/latex]. The ratio of these two estimates of variance is the [latex]F[/latex]-score from an [latex]F[/latex]-distribution with [latex]df_1=k-1[/latex] and [latex]df_2=n-k[/latex].  The p -value for the test is the area in the right tail of the [latex]F[/latex]-distribution.  The statistics used in an ANOVA test are summarized in the ANOVA summary table generated by Excel.

The one-way ANOVA hypothesis test for three or more population means is a well established process:

• Write down the null and alternative hypotheses in terms of the population means.  The null hypothesis is the claim that the population means are all equal and the alternative hypothesis is the claim that at least one of the population means is different from the others.
• Collect the sample information for the test and identify the significance level.
• The p -value is the area in the right tail of the [latex]F[/latex]-distribution.  Use the ANOVA summary table generated by Excel to find the p -value.
• Compare the  p -value to the significance level and state the outcome of the test.

Attribution

“ 13.1   One-Way ANOVA “  and “ 13.2   The F Distribution and the F-Ratio “ in Introductory Statistics by OpenStax  is licensed under a  Creative Commons Attribution 4.0 International License .

Introduction to Statistics Copyright © 2022 by Valerie Watts is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

Module 13: F-Distribution and One-Way ANOVA

One-way anova, learning outcomes.

• Conduct and interpret one-way ANOVA

The purpose of a one-way ANOVA test is to determine the existence of a statistically significant difference among several group means. The test actually uses variances to help determine if the means are equal or not. In order to perform a one-way ANOVA test, there are five basic assumptions to be fulfilled:

• Each population from which a sample is taken is assumed to be normal.
• All samples are randomly selected and independent.
• The populations are assumed to have equal standard deviations (or variances) .
• The factor is a categorical variable.
• The response is a numerical variable.

The Null and Alternative Hypotheses

The null hypothesis is simply that all the group population means are the same. The alternative hypothesis is that at least one pair of means is different. For example, if there are k groups:

H 0 : μ 1 = μ 2 = μ 3 = … = μ k

H a : At least two of the group means μ 1 , μ 2 , μ 3 , …, μ k are not equal.

The graphs, a set of box plots representing the distribution of values with the group means indicated by a horizontal line through the box, help in the understanding of the hypothesis test. In the first graph (red box plots), H 0 : μ 1 = μ 2 = μ 3 and the three populations have the same distribution if the null hypothesis is true. The variance of the combined data is approximately the same as the variance of each of the populations.

If the null hypothesis is false, then the variance of the combined data is larger which is caused by the different means as shown in the second graph (green box plots).

(b) H 0 is not true. All means are not the same; the differences are too large to be due to random variation.

Concept Review

Analysis of variance extends the comparison of two groups to several, each a level of a categorical variable (factor). Samples from each group are independent, and must be randomly selected from normal populations with equal variances. We test the null hypothesis of equal means of the response in every group versus the alternative hypothesis of one or more group means being different from the others. A one-way ANOVA hypothesis test determines if several population means are equal. The distribution for the test is the F distribution with two different degrees of freedom.

Assumptions:

• The populations are assumed to have equal standard deviations (or variances).
• OpenStax, Statistics, One-Way ANOVA. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution
• Introductory Statistics . Authored by : Barbara Illowski, Susan Dean. Provided by : Open Stax. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]
• Completing a simple ANOVA table. Authored by : masterskills. Located at : https://youtu.be/OXA-bw9tGfo . License : All Rights Reserved . License Terms : Standard YouTube License

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3.1.4: Practice with Job Applicants

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• Page ID 22112

• Michelle Oja
• Taft College

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Let's practice our 4-step process using the ANOVA Summary Table to complete the calculations, but the Sum of Squares will be provided in this first example.

Our data come from three groups of 10 people each, all of whom applied for a single job opening: those with no college degree, those with a college degree that is not related to the job opening, and those with a college degree from a relevant field. We want to know if we can use this group membership to account for our observed variability in their scores on a test related to the job that they applied for, and, by doing so, see if there is a difference between our three group means.

To help understand what's going on in this scenario, let's answer some questions:

Exercise \(\PageIndex{1}\)

Answer the following questions to understand the variables and groups that we are working with.

• Who is the sample?
• Who do might be the population?
• What is the IV (groups being compared)?
• What is the DV (quantitative variable being measured)?
• The sample is the 30 people with different degrees (or no degree)?
• The population might be any job applicant?
• Related to job
• Unrelated to job
• The DV is the score on the test.

Identifying the IV levels and DV helps when constructing your hypotheses.

Step 1: State the Hypotheses

O ur hypotheses are concerned with the average score on the test for each of the groups based on education level, so you get to decide which groups you think will have a higher score, which groups will earn a lower average score, and which groups will have scores that are similar.

Exercise \(\PageIndex{2}\)

Determine the research hypothesis in words and symbols. You can fill in the following underlined spot with the symbols for greater than (>), less than (<), or equal signs. Just remember, at least one pair of means must be predicted to be different from each other.

• \( \overline{X}_{N} \) _____ \( \overline{X}_{R} \)
• \( \overline{X}_{N} \) _____ \(\overline{X}_{U} \)
• \( \overline{X}_{R} \) _____ \(\overline{X}_{U} \)

Here's a reasonable research hypothesis. However, without the group means to guide us, your research hypothesis might be slightly different. Just remember, at least one pair of means must be predicted to be different from each other.

• Research Hypothesis: Those with No Degree will have a lower average test score than those with a Related Degree, but will have a similar average test score to those with an unrelated degree. The average test score for those with a Related Degree will also have a higher average test score compared to those with an Unrelated Degree.
• \( \overline{X}_{N} < \overline{X}_{R} \)
• \( \overline{X}_{N} = \overline{X}_{U} \)
• \( \overline{X}_{R} > \overline{X}_{U} \)

What about the null hypothesis?

Exercise \(\PageIndex{3}\)

State the null hypothesis in words and symbols.

• Null Hypothesis: The average test score will be similar for each group; the degree does not affect the hiring rate.
• Symbols: \( \overline{X}_{N} = \overline{X}_{U} = \overline{X}_{R} \)

Step 2: Find the Critical Values

Our test statistic for ANOVA, as we saw above, is \(F\). Because we are using a new test statistic, we will get a new table: the \(F\) distribution table shown in the next section.

There are now two degrees of freedom we must use to find our critical value: Numerator and Denominator. These correspond to the numerator and denominator of our test statistic, which, if you look at the ANOVA table presented earlier, are our Between Groups and Within Groups rows, respectively. The \(df_B\) is the Degrees of Freedom: for the Numerator because it is the degrees of freedom value used to calculate the Mean Square for the Between Groups source, which in turn was the numerator of our \(F\) statistic. Likewise, the \(df_W\) is the Degrees of Freedome for the Denominator because it is the degrees of freedom value used to calculate the Mean Square for the Within Groups (sometimes called Error) source, which was our denominator for \(F\).

The formula for \(df_B\) is \(k – 1\), and remember that k is the number of groups we are assessing. In this example, \(k = 3\) so our \(df_B\) = 2. This tells us that we will use the fourth column, the one labeled 2, to find our critical value. To find the proper row, we calculate the \(df_W\), which was \(N – k\). The original prompt told us that we have “three groups of 10 people each,” so our total sample size is 30. This makes our value for \(df_W\) = 27. If we follow the fourth column down to the row for \(df_W\) = 27, then find the middle row for \(p\) = 0.05, we see that our critical value is 3.35. We use this critical value the same way as we did before: it is our criterion against which we will compare our calculated test statistic to determine statistical significance.

Step 3: Calculate the Test Statistic

Now that we have our hypotheses and the criterion we will use to test them, we can calculate our test statistic. To do this, we will fill in the ANOVA table. We will use the Sum of Squares values that we are provided in Table \(\PageIndex{1}\) .

These may seem like random numbers, but remember that they are based on the distances between the groups themselves and within each group. Figure \(\PageIndex{2}\) shows the plot of the data with the group means and total mean included. If we wanted to, we could use this information, combined with our earlier information that each group has 10 people, to calculate the Between Groups Sum of Squares by hand. However, doing so would take some time, and without the specific values of the data points, we would not be able to calculate our Within Groups Sum of Squares, so we are just trusting that these values are the correct ones.

Example \(\PageIndex{1}\)

Using the information provided in the scenario and Table \(\PageIndex{1}\), fill in the rest of the ANOVA Summary Table in Table \(\PageIndex{2}\)

Using the formulas that we learned about earlier, we can complete Table \(\PageIndex{3}\):

We leave those three empty cells blank; no information is needed from them. So, that leaves us with the final table looking like Table \(\PageIndex{4}\) showing calculated F-score of 36.86.

We can move on to comparing our calculated value to the critical value in step 4.

Step 4: Make the Decision

Our calculated test statistic was found to be \(F_{calc} = 36.86\) and our critical value was found to be \(F_{crit} = 3.35\). Our calculated statistic is larger than our critical value, so we can reject the null hypothesis.

(Critical \(<\) Calculated) \(=\) Reject null \(=\) At least one mean is different from at least one other mean. \(= p<.05\)

(Critical \(>\) Calculated) \(=\) Retain null \(=\) All of the means are similar. \(= p>.05\)

Based on our three groups of 10 people, we can conclude that average job test scores are statistically significantly different based on education level, \(F(2,27)=36.86,p<.05\). Notice that when we report \(F\), we include both degrees of freedom. We always report the numerator then the denominator, separated by a comma.

Because we were only testing for any difference, we cannot yet conclude which groups are different from the others or if our research hypothesis was supported, partially supported, or not supported . We will learn about pairwise comparisons next to answer these last questions!