null hypothesis one way anova

Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices.

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.