11 Analysis of Variance
11.01: One-Way ANOVA
The \(z\)- and \(t\)-tests can be used to test the equality between two population means \(\mu_{1}\) and \(\mu_{2}\). When we have more than two groups, we would inflate the probability of making a type I error if we were to compare just two at a time and make a conclusion about all the groups together. To account for this P(Type I Error) inflation, we instead will do an analysis of variance (ANOVA) to test the equality between 3 or more population means \(\mu_{1}, \mu_{2}, \mu_{3}, \ldots, \mu_{k}\).
The F-test (for ANOVA) is a statistical test for testing the equality of \(k\) population means.
The one-way ANOVA F-test is a statistical test for testing the equality of \(k\) population means from 3 or more groups within one variable or factor. There are many different types of ANOVA; for now, we are going to start with what is commonly referred to as a one-way ANOVA, which has one main effect or factor that is split up into three or more independent treatment levels. In more advanced courses you would learn about dependent groups or two or more factors.
Assumptions:
- The populations are normally distributed continuous variables with equal variances.
- The observations are independent.
The hypotheses for testing the equality of \(k\) population means (ANOVA) are set up with all the means equal to one another in the null hypothesis and at least one mean is different in the alternative hypothesis.
\(H_{0}: \mu_{1} = \mu_{2} = \mu_{3} = \ldots = \mu_{k}\)
\(H_{1}:\) At least one mean is different.
Even though there is equality in \(H_{0}\), the ANOVA test is testing if the variance between groups is significantly greater than the variance within groups; hence, this will always be set up as a right-tailed test.
We will be using abbreviations for many of the numbers found in this section.
B = Between, W = Within
MS = Mean Square (This is a variance)
MSB = Mean Square (Variance) Between groups.
MSW = Mean Square (Variance) Within groups.
The formula for the F-test statistic is \(F = \frac{MSB}{MSW}\).
Use the F-distribution with degrees of freedom from the between and within groups. The numerator degrees of freedom are equal to the number of groups minus one, that is numerator degrees of freedom are \(df_{B} = k - 1\). The denominator degrees of freedom are equal to the total of all the sample sizes minus the number of groups, that is denominator degrees of freedom are \(df_{W} = N - k\).
The sum of squares, degrees of freedom and mean squares are organized in a table called an ANOVA table. Figure 11-1 below is a template for an ANOVA table.
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Where:
\(\bar{\chi}_{i}\) = sample mean from the \(i^{th}\) group
\(s_{i}^{2}\) = sample variance from the \(i^{th}\) group
\(n_{i}\) = sample size from the \(i^{th}\) group
\(k\) = number of groups
\(N = n_{1} + n_{2} + \cdots + n_{k}\) = sum of the individual sample sizes for groups
Grand mean from all groups = \(\bar{\chi}_{GM} = \frac{\sum \chi_{i}}{N}\)
Sum of squares between groups = SSB = \(\sum n_{i} \left(\bar{\chi}_{i} - \bar{\chi}_{GM}\right)^{2}\)
Sum of squares within groups = SSW = \(\sum \left(n_{i} - 1\right) s_{i}^{2}\)
Mean squares between groups (or the between-groups variance \(s_{B}^{2}\)) = \(MSB = \frac{SSB}{k-1}\)
Mean squares within-group (or error within-groups variance \(s_{W}^{2}\)) = \(MSW = \frac{SSW}{N-k}\)
\(F = \frac{MSB}{MSW}\) is the test statistic.
These calculations can be time-consuming to do by hand, so use technology to find the ANOVA table values, critical value and/or p-value.
Different textbooks and computer software programs use different labels in the ANOVA tables.
- The TI-calculators use the word Factor for Between Groups and Error for Within Groups.
- Some software packages use Treatment instead of Between groups. You may see different notation depending on which textbook, software or video you are using.
- For between groups SSB = SSB = SSTR = SST = SSF and for within groups SSW = SSW = SSE.
- One thing that is consistent within the ANOVA table is that the between = factor = treatment always appears on the first row of the ANOVA table, and the within = error always is in the second row of the ANOVA table.
- The \(df\) column usually is in the second column since we divide the sum of squares by the \(df\) to find the mean squares. However, some software packages will put the \(df\) column before the sum of squares column.
- The test statistic is under the column labeled F.
- Many software packages will give an extra column for the p-value and some software packages give a critical value too.
Assumption: The population we are sampling from must be approximately normal with equal variances. If these assumptions are not met there are more advanced statistical methods that should be used.
An educator wants to see if there is a difference in the average grades given to students for the 4 different instructors who teach intro statistics courses. They randomly choose courses that each of the 4 instructors taught over the last few years and perform an ANOVA test. What would be the correct hypotheses for this test?
Solution
There are 4 groups (the 4 instructors) so there will be 4 means in the null hypothesis, \(H_{0}: \mu_{1} = \mu_{2} = \mu_{3} = mu_{4}\).
It is tempting to write \(\mu_{1} \neq \mu_{2} \neq \mu_{3} \neq mu_{4}\) for the alternative hypothesis, but the test is testing the opposite of all equal is that at least one mean is different. We could for instance have just the mean for group 2 be different and groups 1, 3 and 4 have equal means. If we wanted to write out all the way to have unequal groups, you would have a combination problem with \({}_{4} C_{3} + {}_{4} C_{2} + {}_{4} C_{1}\) ways of getting unequal means. Instead of all of these possibilities, we just write a sentence: “at least one mean is different.”
The hypotheses are:
\(H_{0}: \mu_{1} = \mu_{2} = \mu_{3} = mu_{4}\)
\(H_{1}:\) At least one mean is different.
A researcher claims that there is a difference in the average age of assistant professors, associate professors, and full professors at her university. Faculty members are selected randomly and their ages are recorded. Assume faculty ages are normally distributed. Test the claim at the \(\alpha\) = 0.01 significance level. The data are listed below.
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Solution
The claim is that there is a difference in the average age of assistant professors \((\mu_{1})\), associate professors \((\mu_{2})\), and full professors \((\mu_{3})\) at her university.
The correct hypotheses are:
\(H_{0}: \mu_{1} = \mu_{2} = \mu_{3}\)
\(H_{1}:\) At least one mean differs.
We need to compute all of the necessary parts for the ANOVA table and F-test.
Compute the descriptive stats for each group with your calculator using 1–Var Stats L1.
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Record the sample size, sample mean, sum of \(\chi\), and sample variances. Take the standard deviation \(s_{\chi}\) and square it to find the variance \(s_{\chi}^{2}\) for each group.
| Assistant Prof | \(n_{1} = 7\) | \(\bar{\chi}_{1} = 37\) | \(\sum \chi_{1} = 259\) | \(s_{1}^{2} = 53\) |
| Associate Prof | \(n_{2} = 7\) | \(\bar{\chi}_{2} = 52\) | \(\sum \chi_{2} = 364\) | \(s_{2}^{2} = 55.6667\) |
| Prof | \(n_{3} = 7\) | \(\bar{\chi}_{1} = 54\) | \(\sum \chi_{3} = 378\) | \(s_{3}^{2} = 35\) |
Compute the grand mean: \(N = n_{1} + n_{2} + n_{3} = 7 + 7 + 7 = 21\), \(\bar{\chi}_{GM} = \frac{\sum \chi_{i}}{N} = \frac{(259 + 364 + 378)}{21} = 47.66667\).
Compute the sum of squares for between groups: SSB = \(\sum n_{i} \left(\bar{\chi}_{i} - \bar{\chi}_{GM}\right)^{2} = n_{1} \left(\bar{\chi}_{1} - \bar{\chi}_{GM}\right)^{2} + n_{2} \left(\bar{\chi}_{2} - \bar{\chi}_{GM}\right)^{2} + n_{3} \left(\bar{\chi}_{3} - \bar{\chi}_{GM}\right)^{2} = 7(37 - 47.66667)^{2} + 7(52 − 47.66667)^{2} + 7(54 − 47.66667)^{2} = 1208.66667\).
Compute the sum of squares within groups: SSW = \(\sum \left(n_{i} - 1\right) s_{i}^{2} = \left(n_{1} - 1\right) s_{1}^{2} + \left(n_{2} - 1\right) s_{2}^{2} + \left(n_{3} - 1\right) s_{3}^{2} = 6 \cdot 53 + 6 \cdot 55.66667 + 6 \cdot 35 = 862\).
Place the sum of squares into your ANOVA table and add them up to get the total.
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Next, find the degrees of freedom: \(k = 3\) since there are 3 groups, so \(df_{B} = k-1 = 2\); \(df_{W} = N-k = 21-3 = 18\).
Add the degrees of freedom to the table to get the total \(df\).
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Compute the mean squares by dividing the sum of squares by their corresponding \(df\), then add these numbers to the table. \(MSB = \frac{SSB}{k-1} = \frac{1208.66667}{2} = 604.3333 \quad MSW = \frac{SSW}{N-k} = \frac{862}{18} = 47.8889\)
The test statistic is the ratio of these two mean squares: \(F = \frac{MSB}{MSW} = \frac{604.3333}{47.8889} = 12.6195\). Add the test statistic to the table under F.
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All ANOVA tests are right-tailed tests, so the critical value for a right-tailed F-test is found with the F-distribution. Use \(\alpha\) = 0.01 area in the right-tail. The degrees of freedom are \(df_{N} = 2\), and \(df_{D} = 18\). The critical value is 6.0129; see the sampling distribution curve in Figure 11-2.
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Note that the F-distribution starts at zero, and is skewed to the right. The test statistic of 12.6295 is larger than the critical value of 6.0129 so the decision would be to reject the null hypothesis.
Decision: Reject \(H_{0}\).
Summary: There is enough evidence to support the claim that there is a difference in the average age of assistant professors, associate professors, and full professors at her university.
If we were using the p-value method, then we would use the calculator or computer for a right-tailed test. The p-value = 0.0003755 is less than \(\alpha\) = 0.01, which leads to the same decision of rejecting \(H_{0}\) that we found using the critical value method.
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Alternatively, use technology to compute the ANOVA table and p-value.
TI-84: ANOVA, hypothesis test for the equality of k population means. Note you have to have the actual raw data to do this test on the calculator. Press the [STAT] key and then the [EDIT] function, type the three lists of data into list one, two and three. Press the [STAT] key, arrow over to the [TESTS] menu, arrow down to the option [F:ANOVA(] and press the [ENTER] key. This brings you back to the regular screen where you should now see ANOVA(. Now hit the [2nd] [L1] [,] [2nd] [L2] [,][2nd] [L3][)] keys in that order. You should now see ANOVA(L1,L2,L3); if you had 4 lists you would then have an additional list.
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Press the [ENTER] key. The calculator returns the F-test statistic, the p-value, Factor (Between) \(df\), SS and MS, Error (Within) \(df\), SS and MS. The last value, Sxp is the square root of the MSE.
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TI-89: ANOVA, hypothesis test for the equality of \(k\) population means. Go to the [Apps] Stat/List Editor, then type in the data for each group into a separate list (or if you don’t have the raw data, enter the sample size, sample mean and sample variance for group 1 into list1 in that order, repeat for list2, etc.). Press [2nd] then F6 [Tests], then select C:ANOVA. Select the input method data or stats. Select the number of groups. Press the [ENTER] key to calculate. The calculator returns the F-test statistic, the p-value, Factor (Between) df, SS and MS, Error (Within) df, SS and MS. The last value, Sxp, is the square root of the MSE.
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Excel: Type the labels and data into adjacent columns (it is important not to have any blank columns, which would be an additional group counted as zeros). Select the Data tab, then Data Analysis, ANOVA: Single-Factor, then OK.
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Next, select all three columns of data at once for the input range. Check the box that says Labels in first row (only select this if you actually selected the labels in the input range). Change your value of alpha and output range will be one cell reference where you want your output to start, see below.
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You get the following output:
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Excel gives both the p-value and critical value so you can use either method when making your decision, but make sure you are comfortable with both.
Summary
The ANOVA test gives evidence that there is a difference between three or more means. The null hypothesis will always have the means equal to one another versus the alternative hypothesis that at least one mean is different. The F-test results are about the difference in means, but the test is actually testing if the variation between the groups is larger than the variation within the groups. If this between group variation is significantly larger than the within groups then we can say there is a statistically significant difference in the population means. Hence, we are always performing a right-tailed F-test for ANOVA. Make sure to only compare the p-value with \(\alpha\) and the test statistic to the critical value.
11.02: Pairwise Comparisons of Means (Post-Hoc Tests)
If you do in fact reject \(H_{0}\), then you know that at least two of the means are different. The ANOVA test does not tell which of those means are different, only that a difference exists. Most likely your sample means will be different from each other, but how different do they need to be for there to be a statistically significant difference?
To determine which means are significantly different, you need to conduct further tests. These post-hoc tests include the range test, multiple comparison tests, Duncan test, Student-Newman-Keuls test, Tukey test, Scheffé test, Dunnett test, Fisher’s least significant different test, and the Bonferroni test, to name a few. There are more options, and there is no consensus on which test to use. These tests are available in statistical software packages such as R, Minitab and SPSS.
One should never use two-sample \(t\)-tests from the previous chapter. This would inflate the type I error.
The probability of at least one type I error increases exponentially with the number of groups you are comparing. Let us assume that \(\alpha = 0.05\), then the probability that an observed difference between two groups that does not occur by chance is \(1 - \alpha = 0.95\). If two comparisons are made, the probability that the observed difference is true is no longer 0.95. The probability is \((1 - \alpha)^{2} = 0.9025\), and the P(Type I Error) = \(1 - 0.9025 = 0.0975\). Therefore, the P(Type I Error) occurs if \(m\) comparisons are made is \(1 - (1 - \alpha)m\).
For instance, if we are comparing the means of four groups: There would be \(m = {}_4 C_{2} = 6\) different ways to compare the 4 groups: groups (1,2), (1,3), (1,4), (2,3), (2,4), and (3,4). The P(Type I Error) = \(1 - (1 - \alpha)6 = 0.2649\). This is why a researcher should use ANOVA for comparing means, instead of independent \(t\)-tests.
There are many different methods to use. Many require special tables or software. We could actually just start with post-hoc tests, but they are a lot of work. If we run an ANOVA and we fail to reject the null hypothesis, then there is no need for further testing and it will save time if you were doing these steps by hand. Most statistical software packages give you the ANOVA table followed by the pairwise comparisons with just a change in the options menu. Keep in mind that Excel is not a statistical software and does not give pairwise comparisons.
We will use the Bonferroni Test, named after the mathematician Carlo Bonferroni. The Bonferroni Test uses the t-distribution table and is similar to previous t-tests that we have used, but adjusts \(\alpha\) to the number of comparisons being made.
The Bonferroni test is a statistical test for testing the difference between two population means (only done after an ANOVA test shows not all means are equal).
The formula for the Bonferroni test statistic is \(t = \dfrac{\bar{x}_{i} - \bar{x}_{j}}{\sqrt{\left( MSW \left(\frac{1}{n_{i}} + \frac{1}{n_{j}}\right) \right)}}\).
where \(\bar{x}_{i}\) and \(\bar{x}_{j}\) are the means of the samples being compared, \(n_{i}\) and \(n_{j}\) are the sample sizes, and \(MSW\) is the within-group variance from the ANOVA table.
The Bonferroni test critical value or p-value is found by using the t-distribution with within degrees of freedom \(df_{W} = N-k\), using an adjusted \(\frac{\alpha}{m}\) two-tail area under the t-distribution, where \(k\) = number of groups and \(m = {}_{k} C_{2}\), all the combinations of pairs out of \(k\) groups.
Critical Value Method
According to the ANOVA test that we previously performed, there does appear to be a difference in the average age of assistant professors \((\mu_{1})\), associate professors \((\mu_{2})\), and full professors \((\mu_{3})\) at this university.
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The hypotheses were:
\(H_{0}: \mu_{1} = \mu_{2} = \mu_{3}\)
\(H_{1}:\) At least one mean differs.
The decision was to reject \(H_{0}\), which means there is a significant difference in the mean age. The ANOVA test does not tell us, though, where the differences are. Determine which of the difference between each pair of means is significant. That is, test if \(\mu_{1} \neq \mu_{2}\), if \(\mu_{1} \neq \mu_{3}\), and if \(\mu_{2} \neq \mu_{3}\).
Solution
The alternative hypothesis for the ANOVA was “at least one mean is different.” There will be \({}_{3} C_{2} = 3\) subsequent hypothesis tests to compare all the combinations of pairs (Group 1 vs. Group 2, Group 1 vs. Group 3, and Group 2 vs. Group 3). Note that if you have 4 groups then you would have to do \({}_{4} C_{2} = 6\) comparisons, etc.
Use the t-distribution to find the critical value for the Bonferroni test. The total of all the individual sample sizes \(N = 21\) and \(k = 3\), and \(m = {}_{3} C_{2} = 3\), then the area for both tails would be \(\frac{\alpha}{m} = \frac{0.01}{m} = 0.003333\).
This is a two-tailed test so the area in one tail is \(\frac{0.003333}{2}\) with \(df_{W} = N-k = 21-3 = 18\) gives \(\text{C.V.} = \pm 3.3804\). The critical values are really far out in the tail so it is hard to see the shaded area. See Figure 11-3.
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Compare \(\mu_{1}\) and \(\mu_{2}\):
\(H_{0}: \mu_{1} = \mu_{2}\)
\(H_{1}: \mu_{1} \neq \mu_{2}\)
The test statistic is \(t = \frac{\bar{x}_{1} - \bar{x}_{2}}{\sqrt{\left( MSW \left(\frac{1}{n_{1}} + \frac{1}{n_{2}}\right) \right)}} = \frac{37 - 52}{\sqrt{\left(47.8889 \left(\frac{1}{7} + \frac{1}{7}\right) \right)}} = -4.0552\).
Compare the test statistic to the critical value. Since the test statistic \(-4.0552 < \text{critical value} = -3.3804\), we reject \(H_{0}\).
There is enough evidence to conclude that there is a difference in the average age of assistant and associate professors.
Compare \(\mu_{1}\) and \(\mu_{3}\):
\(H_{0}: \mu_{1} = \mu_{3}\)
\(H_{1}: \mu_{1} \neq \mu_{3}\)
The test statistic is \(t = \frac{\bar{x}_{1} - \bar{x}_{3}}{\sqrt{\left( MSW \left(\frac{1}{n_{1}} + \frac{1}{n_{3}}\right) \right)}} = \frac{37 - 54}{\sqrt{\left(47.8889 \left(\frac{1}{7} + \frac{1}{7}\right) \right)}} = -4.5958\).
Compare the test statistic to the critical value. Since the test statistic \(-4.5958 < \text{critical value} = -3.3804\), we reject \(H_{0}\).
Reject \(H_{0}\), since the test statistic is in the lower tail. There is enough evidence to conclude that there is a difference in the average age of assistant and full professors.
Compare \(\mu_{2}\) and \(\mu_{3}\):
\(H_{0}: \mu_{2} = \mu_{3}\)
\(H_{1}: \mu_{2} \neq \mu_{3}\)
The test statistic is \(t = \frac{\bar{x}_{2} - \bar{x}_{3}}{\sqrt{\left( MSW \left(\frac{1}{n_{2}} + \frac{1}{n_{3}}\right) \right)}} = \frac{52 - 54}{\sqrt{\left(47.8889 \left(\frac{1}{7} + \frac{1}{7}\right) \right)}} = -0.5407\)
Compare the test statistic to the critical value. Since the test statistic is between the critical values \(-3.3804 < -0.5407 < 3.3804\), we fail to reject \(H_{0}\).
Do not reject \(H_{0}\), since the test statistic is between the two critical values. There is enough evidence to conclude that there is not a difference in the average age of associate and full professors.
Note: you should get at least one group that has a reject \(H_{0}\), since you only do the Bonferroni test if you reject \(H_{0}\) for the ANOVA. Also, note that the transitive property does not apply. It could be that group 1 = group 2 and group 2 = group 3; this does not mean that group 1 = group 3.
P-Value Method
A research organization tested microwave ovens. At \(\alpha\) = 0.10, is there a significant difference in the average prices of the three types of oven?
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Solution
The ANOVA was run in Excel.
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To test if there is a significant difference in the average prices of the three types of oven, the hypotheses are:
\(H_{0}: \mu_{1} = \mu_{2} = \mu_{3}\)
\(H_{1}:\) At least one mean differs.
Use the Excel output to find the p-value in the ANOVA table of 0.001019, which is less than \(\alpha\) so reject \(H_{0}\); there is at least one mean that is different in the average oven prices.
There is a statistically significant difference in the average prices of the three types of oven. Use the Bonferroni test p-value method to see where the differences are.
Compare \(\mu_{1}\) and \(\mu_{2}\):
\(H_{0}: \mu_{1} = \mu_{2}\)
\(H_{1}: \mu_{1} \neq \mu_{2}\)
\(t = \frac{\bar{x}_{1} - \bar{x}_{2}}{\sqrt{\left( MSW \left(\frac{1}{n_{1}} + \frac{1}{n_{2}}\right) \right)}} = \frac{233.3333-203.125}{\sqrt{\left((1073.794 \left(\frac{1}{6} + \frac{1}{8}\right) \right)}} = 1.7070\)
To find the p-value, find the area in both tails and multiply this area by \(m\). The area to the right of \(t = 1.707\), using \(df_{W} = 19\), is 0.0520563. Remember these are always two-tail tests, so multiply this area by 2, to get both tail areas of 0.104113.
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Then multiply this area by \(m = {}_{3} C_{2} = 3\) to get a p-value = 0.3123.
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Since the p-value = \(0.3123 > \alpha = 0.10\), we do not reject \(H_{0}\). There is a statistically significant difference in the average price of the 1,000- and 900-watt ovens.
Compare \(\mu_{1}\) and \(\mu_{3}\):
\(H_{0}: \mu_{1} = \mu_{3}\)
\(H_{1}: \mu_{1} \neq \mu_{3}\)
\(t = \frac{\bar{x}_{1} - \bar{x}_{3}}{\sqrt{\left( MSW \left(\frac{1}{n_{1}} + \frac{1}{n_{3}}\right) \right)}} = \frac{233.3333-155.625}{\sqrt{\left((1073.794 \left(\frac{1}{6} + \frac{1}{8}\right) \right)}} = 4.3910\)
Use \(df_{W}\) = 19 to find the p-value.
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Since the p-value = (tail areas)*3 = \(0.00094 < \alpha = 0.10\), we reject \(H_{0}\). There is a statistically significant difference in the average price of the 1,000- and 800-watt ovens.
Compare \(\mu_{2}\) and \(\mu_{3}\):
\(H_{0}: \mu_{2} = \mu_{3}\)
\(H_{1}: \mu_{2} \neq \mu_{3}\)
\(t = \frac{\bar{x}_{2} - \bar{x}_{3}}{\sqrt{\left( MSW \left(\frac{1}{n_{2}} + \frac{1}{n_{3}}\right) \right)}} = \frac{203.125-155.625}{\sqrt{\left((1073.794 \left(\frac{1}{8} + \frac{1}{8}\right) \right)}} = 2.8991\)
Use \(df_{W} = 19\) to find the p-value (remember that these are always two-tail tests).
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Since the p-value = \(0.0276 < \alpha = 0.10\), we reject \(H_{0}\). There is a statistically significant difference in the average price of the 900- and 800-watt ovens.
There is a chance that after we multiply the area by the number of comparisons, the p-value would be greater than one. However, since the p-value is a probability we would cap the probability at one.
This is a lot of math! The calculators and Excel do not have post-hoc pairwise comparisons shortcuts, but we can use the statistical software called SPSS to get the following results. We will look specifically at interpreting the SPSS output for Example 11-4.
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The first table, labeled "Descriptives", gives descriptive statistics; the second table is the ANOVA table, and note that the p-value is in the column labeled Sig. The Multiple Comparisons table is where we want to look. There are repetitive pairs in the last table, just in a different order.
The first two rows in Figure 11-4 are comparing group 1 with groups 2 and 3. If we follow the first row across under the Sig. column, this gives the p-value = 0.312 for comparing the 1,000- and 900-watt ovens.
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The second row in Figure 11-4 compares the 1,000- and 800-watt ovens, p-value = 0.001.
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The third row in Figure 11-4 compares the 900- and 1000-watt ovens in the reverse order as the first row; note that the difference in the means is negative but the p-value is the same.
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The fourth row in Figure 11-4 compares the 900- and 800-watt ovens, p-value = 0.028.
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The last set of rows in Figure 11-4 are again repetitive and give the 800-watt oven compared to the 900- and 1000-watt ovens.
Keep in mind that post-hoc is defined as occurring after an event. A post-hoc test is done after an ANOVA test shows that there is a statistically significant difference. You should get at least one group that has a result of "reject \(H_{0}\)", since you only do the Bonferroni test if you reject \(H_{0}\) for the ANOVA.
11.03: Two-Way ANOVA (Factorial Design)
Two-way analysis of variance (two-way ANOVA) is an extension of one-way ANOVA. It can be used to compare the means of two independent variables or factors from two or more populations. It can also be used to test for interaction between the two independent variables.
We will not be doing the sum of squares calculations by hand. These numbers will be given to you in a partially filled out ANOVA table or an Excel output will be given in the problem.
There are three sets of hypotheses for testing the equality of \(k\) population means from two independent variables, and to test for interaction between the two variables (two-way ANOVA):
| Row Effect (Factor A): |
\(H_{0}:\) The row variable has no effect on the average ___________________. \(H_{1}:\) The row variable has an effect on the average ___________________. |
| Column Effect (Factor B): |
\(H_{0}\): The column variable has no effect on the average ___________________. \(H_{1}\): The column variable has an effect on the average ___________________. |
| Interaction Effect (A×B): |
\(H_{0}:\) There is no interaction effect between row variable and column variable on the average ___________________. \(H_{1}:\) There is an interaction effect between row variable and column variable on the average ___________________. |
These ANOVA tests are always right-tailed F-tests.
The F-test (for two-way ANOVA) is a statistical test for testing the equality of k independent quantitative population means from two nominal variables, called factors. The two-way ANOVA also tests for interaction between the two factors.
Assumptions:
- The populations are normal.
- The observations are independent.
- The variances from each population are equal.
- The groups must have equal sample sizes.
The formulas for the F-test statistics are:
| Factor 1: | \(F_{A} = \frac{MS_{A}}{MSE}\) with \(df_{A} = a-1\) and \(df_{\text{E}} = ab(n-1)\) |
| Factor 2: | \(F_{B} = \frac{MS_{B}}{MSE}\) with \(df_{B} = b-1\) and \(df_{\text{E}} = ab(n-1)\) |
| Interaction: | \(F_{A \times B} = \frac{MS_{A \times B}}{MSE}\) with \(df_{A \times B} = (a-1)(b-1)\) and \(df_{\text{E}} = ab(n-1)\) |
Where:
\(SS_{\text{A}}\) = sum of squares for factor A, the row variable
\(SS_{\text{B}}\) = sum of squares for factor B, the column variable
\(SS_{\text{A} \times \text{B}}\) = sum of squares for interaction between factor A and B
\(SSE\) = sum of squares of error, also called sum of squares within groups
\(a\) = number of levels of factor A
\(b\) = number of levels of factor B
\(n\) = number of subjects in each group
It will be helpful to make a table. Figure 11-5 is called a two-way ANOVA table.
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Since the computations for the two-way ANOVA are tedious, this text will not cover performing the calculations by hand. Instead, we will concentrate on completing and interpreting the two-way ANOVA tables.
A farmer wants to see if there is a difference in the average height for two new strains of hemp plants. They believe there also may be some interaction with different soil types so they plant 5 hemp plants of each strain in 4 types of soil: sandy, clay, loam and silt. At \(\alpha\) = 0.01, analyze the data shown, using a two-way ANOVA as started below in Figure 11-6. See below for raw data.
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Solution
First, find the missing values for the ANOVA table in Figure 11-6. The sum of squares should add up to the total in the SS column of 5108.
Subtract the 3 sum of squares that were already given in the table from the total sum of squares to find the missing error sum of squares: \(5108 - 12.1 - 2501 - 268.1 = 2326.8\).
Find the degrees of freedom for each category.
There are 2 types of hemp strains so \(a = 2\) and the \(df_{\text{A}} = 2-1 = 1\).
There are 4 soil types so \(b = 4\) and the \(df_{\text{B}} = 4-1 = 3\).
The interaction \(df_{\text{A} \times \text{B}} = (2-1)(4-1) = 3\).
There are 5 plants within each group so \(n = 5\), and \(df_{\text{E}} = 2 \cdot 4(5-1) = 32\).
The total \(df\) should add up to \(N-1\). There are 8 groups of 5 so there are 40 data points all together; \(df_{\text{Total}} = 39\). Fill in your \(SS\) and \(df\) to the ANOVA table and then divide the sum of squares by \(df\) to get the mean squares. See below.
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Next, find the test statistic for by dividing the \(MS\) for each category by the \(MSE\). For the row effect \(F_{\text{A}} = 12.1/72.7125 = 0.1664\); for the column effect \(F_{\text{B}} = 833.6667/72.7125 = 11.4652\); and for the interaction effect \(F_{\text{A} \times \text{B}} = 89.3667/72/7125 = 1.2290\). Add these to complete the ANOVA table.
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There are three hypothesis tests that we are performing: the strain of hemp plan has an effect on the mean plant height (row effect), the type of soil has an effect on the mean plant height (column effect), and the interaction effect between strain of hemp plan and soil type on the mean plant height.
Row Effect-Factor A:
\(H_{0}:\) The strain of hemp plant has no effect on the mean plant height.
\(H_{1}:\) The strain of hemp plant has an effect on the mean plant height.
Test Statistic: \(F_{\text{A}} = 0.1664\).
Critical Value: Use an F-distribution with \(\alpha = 0.01\) right-tail area, \(df_{\text{A}} = 1\) and \(df_{\text{E}} = 32\).
The critical value is 7.4993. See Figure 11-7.
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The test statistic of 0.1664 is less than the critical value and not in the shaded tail; therefore, do not reject \(H_{0}\). There is no significant difference in the mean plant height between the two hemp strains.
Column Effect-Factor B:
\(H_{0}:\) The type of soil has no effect on the mean plant height.
\(H_{1}:\) The type of soil has an effect on the mean plant height.
Test Statistic: \(F_{\text{B}} = 11.4652\)
Critical Value: Use an F-distribution with \(\alpha\) = 0.01 right-tail area, \(df_{\text{B}} = 3\) and \(df_{\text{E}} = 32\). The critical value is 4.4594.
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The test statistic is greater than the critical value and is in the shaded tail; therefore, reject \(H_{0}\). The type of soil has a significant effect on the mean plant height.
Interaction Effect-Factor A \(\times\) B
\(H_{0}:\) There is no interaction effect between the strain of hemp and the soil type on the mean plant height.
\(H_{1}:\) There is an interaction effect between the strain of hemp and the soil type on the mean plant height.
Test Statistic: \(F_{\text{A} \times \text{B}} = 1.229\)
Critical Value: Use an F-distribution with \(\alpha\) = 0.01 right-tail area, \(df_{\text{A} \times \text{B}} = 3\) and \(df_{\text{E}} = 32\). The critical value is 4.4594.
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The test statistic is not in the shaded tail; therefore, do not reject \(H_{0}\). There is no significant interaction effect between the hemp strain and the soil type on the mean plant height.
Rough drawings from memory were futile. He didn't even know how long it had been, beyond Ford Prefect's rough guess at the time that it was "a couple of million years" and he simply didn't have the maths. Still, in the end he worked out a method which would at least produce a result. He decided not to mind the fact that with the extraordinary jumble of rules of thumb, wild approximations and arcane guesswork he was using he would be lucky to hit the right galaxy, he just went ahead and got a result. He would call it the right result. Who would know?
As it happened, through the myriad and unfathomable chances of fate, he got it exactly right, though he of course would never know that. He just went up to London and knocked on the appropriate door. "Oh. I thought you were going to phone me first."
(Adams, 2002)
11.04: Chapter 11 Formulas
One-Way ANOVA
\(H_{0}: \mu_{1} = \mu_{2} = \mu_{3} = \ldots = \mu_{k}\)
f\(H_{1}:\) At least one mean is different.
| Source | \(SS\) = Sum of Squares | \(df\) | \(MS\) = Mean Square | F |
|---|---|---|---|---|
| Between (Factor) | \(\sum n_{i} \left(\bar{x}_{i} - \bar{x}_{GM}\right)^{2}\) | \(k-1\) | \(MSB = \frac{SSB}{k-1}\) | \(F = \frac{MSB}{MSW}\) |
| Within (Error) | \(\sum \left(n_{i} - 1\right) s_{i}^{2}\) | \(N-k\) | \(MSW = \frac{SSW}{N-k}\) | |
| Total | SST | \(N-1\) |
\(\bar{x}_{i}\) = sample mean from the \(i^{th}\) group
\(n_{i}\) = sample size of the \(i^{th}\) group
\(k\) = number of groups
\(s_{i}^{2}\) = sample variance from the \(i^{th}\) group
\(N = n_{1} + n_{2} + \ldots + n_{k}\)
\(\bar{x}_{GM} = \frac{\sum x_{i}}{N}\)
Bonferroni Test
\(H_{0}: \mu_{i} = \mu_{j}\)
\(H_{1}: \mu_{i} \neq \mu_{j}\)
Bonferroni test statistic: \(t = \dfrac{\bar{x}_{i} - \bar{x}_{j}}{\sqrt{ \left(MSW \left(\frac{1}{n_{i}} + \frac{1}{n_{j}}\right) \right)}}\)
Multiply p-value by \(m = {}_k C_{2}\), divide area for critical value by \(m = {}_k C_{2}\).
Two-Way ANOVA
| Row Effect (Factor A): |
\(H_{0}:\) The row variable has no effect on the average ___________________. \(H_{1}:\) The row variable has an effect on the average ___________________. |
| Column Effect (Factor B): |
\(H_{0}\): The column variable has no effect on the average ___________________. \(H_{1}\): The column variable has an effect on the average ___________________. |
| Interaction Effect (A×B): |
\(H_{0}:\) There is no interaction effect between row variable and column variable on the average ___________________. \(H_{1}:\) There is an interaction effect between row variable and column variable on the average ___________________. |
| Source | \(SS\) | \(df\) | \(MS\) | F |
|---|---|---|---|---|
| \(A\) (row factor) | \(SS_{A}\) | \(a-1\) | \(MS_{A} = \frac{SS_{A}}{df_{A}}\) | \(F_{A} = \frac{MS_{A}}{MSE}\) |
| \(B\) (column factor) | \(SS_{B}\) | \(b-1\) | \(MS_{B} = \frac{SS_{B}}{df_{B}}\) | \(F_{B} = \frac{MS_{B}}{MSE}\) |
| \(A \times B\) (interaction) | \(SS_{A \times B}\) | \((a-1)(b-1)\) | \(MS_{A \times B} = \frac{SS_{A \times B}}{df_{A \times B}}\) | \(F_{A \times B} = \frac{MS_{A \times B}}{MSE}\) |
| Error (within) | \(SSE\) | \(ab(n-1)\) | \(MSE = \frac{SSE}{df_{E}}\) | |
| Total | \(SST\) | \(N-1\) |
11.05: Chapter 11 Exercises
1. What does the acronym ANOVA stand for?
a) Analysis of Variance
b) Analysis of Means
c) Analyzing Various Means
d) Analysis of Variance
e) Anticipatory Nausea and Vomiting
f) Average Noise Variance
2. What would the test statistic equal if MSB = MSW?
a) -1
b) 0
c) 1
d) 4
e) 1.96
3. A researcher would like to test to see if there is a difference in the average profit between 5 different stores. Which are the correct hypotheses for an ANOVA?
a) \(H_{0}: \mu_{1} = \mu_{2} = \mu_{3}\) \(H_{1}:\) At least one mean is different.
b) \(H_{0}: \mu_{1} = \mu_{2} = \mu_{3} = \mu_{4} = \mu_{5}\) \(H_{1}: \mu_{1} \neq \mu_{2} \neq \mu_{3} \neq \mu_{4} \neq \mu_{5}\)
c) \(H_{0}: \mu_{1} \neq \mu_{2} \neq \mu_{3} \neq \mu_{4} \neq \mu_{5}\) \(H_{1}: \mu_{1} = \mu_{2} = \mu_{3} = \mu_{4} = \mu_{5}\)
d) \(H_{0}: \sigma_{B}^{2} \neq \sigma_{W}^{2}\) \(H_{1}: \sigma_{B}^{2} = \sigma_{W}^{2}\)
e) \(H_{0}: \mu_{1} = \mu_{2} = \mu_{3} = \mu_{4} = \mu_{5}\) \(H_{1}:\) At least one mean is different
4. An ANOVA was run to test to see if there was a significant difference in the average cost between three different brands of snow skis. Random samples for each of the three brands were collected from different stores. Assume the costs are normally distributed. At \(\alpha\) = 0.05, test to see if there is a difference in the means. State the hypotheses, fill in the ANOVA table to find the test statistic, compute the p-value, state the decision and summary.
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5. An ANOVA test was run for the per-pupil costs for private school tuition for three counties in the Portland, Oregon, metro area. Assume tuition costs are normally distributed. At \(\alpha\) = 0.05, test to see if there is a difference in the means.
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a) State the hypotheses.
b) Fill out the ANOVA table to find the test statistic.
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c) Compute the p-value.
d) State the correct decision and summary.
6. Cancer is a terrible disease. Surviving may depend on the type of cancer the person has. To see if the mean survival time for several types of cancer are different, data was collected on the survival time in days of patients with one of these cancers in advanced stage. The data is from "Cancer survival story," 2013. (Please realize that this data is from 1978. There have been many advances in cancer treatment, so do not use this data as an indication of survival rates from these cancers.) Does the data indicate that there is a difference in the mean survival time for these types of cancer? Use a 1% significance level.
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a) State the hypotheses.
b) Fill in an ANOVA table to find the test statistic.
c) Compute the p-value.
d) State the correct decision and summary.
7. What does the Bonferroni comparison test for?
a) The analysis of between and within variance.
b) The difference between all the means at once.
c) The difference between two pairs of mean.
d) The sample size between the groups.
8. True or false: The Bonferroni test should only be done when you reject the null hypothesis F-test?
9. A manufacturing company wants to see if there is a significant difference in three types of plastic for a new product. They randomly sample prices for each of the three types of plastic and run an ANOVA. Use \(\alpha\) = 0.05 to see if there is a statistically significant difference in the mean prices. Part of the computer output is shown below.
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a) State the hypotheses.
b) Fill in the ANOVA table to find the test statistic.
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c) Compute the critical value.
d) State the decision and summary.
e) Which group(s) are significantly different based on the Bonferroni test?
10. A manager on an assembly line wants to see if they can speed up production by implementing a new switch for their conveyor belts. There are four switches to choose from and replacing all the switches along the assembly line will be quite costly. They test out each of the four designs and record assembly times. Use \(\alpha\) = 0.05 to see if there is a statistically significant difference in the mean times.
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a) State the hypotheses.
b) Fill in the ANOVA table to find the test statistic.
c) Compute the critical value.
d) State the correct decision and summary.
e) Should a post-hoc Bonferroni test be done? Why?
i. No, since the p-value > \(\alpha\) there is no difference in the means.
ii. Yes, we should always perform a post-hoc test after an ANOVA
iii. No, since we already know that there is a difference in the means.
iv. Yes, since the p-value < \(\alpha\) we need to see where the differences are.
f) All four new switches are significantly faster than the current switch method. Of the four new types of switches, switch 3 cost the least amount to implement. Which of the 4, if any, should the manager choose?
i. The manager should stay with the old switch method since we failed to reject the null hypothesis.
ii. The manager should switch to any of the four new switches since we rejected the null hypothesis.
iii. The manager should randomly pick from switch types 1, 2 or 4.
iv. Since there is no statistically significant difference in the mean time they should choose switch 3 since it is the least expensive.
For exercises 11-16, Assume that all distributions are normal with equal population standard deviations, and the data was collected independently and randomly. Show all 5 steps for hypothesis testing. If there is a significant difference is found, run a Bonferroni test to see which means are different.
a) State the hypotheses.
b) Compute the test statistic.
c) Compute the critical value or p-value.
d) State the decision.
e) Write a summary.
11. Is a statistics class's delivery type a factor in how well students do on the final exam? The table below shows the average percent on final exams from several randomly selected classes that used the different delivery types. Use a level of significance of \(\alpha\) = 0.10.
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12. The dependent variable is the number of times a photo gets a like on social media. The independent variable is the subject matter, selfie or people, landscape, meme, or a cute animal. The researcher is exploring whether the type of photo makes a difference on the mean number of likes. A random sample of photos were taken from social media. Test to see if there is a significant difference in the means using \(\alpha\) = 0.05.
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13. The dependent variable is movie ticket prices, and the groups are the geographical regions where the theaters are located (suburban, rural, urban). A random sample of ticket prices were taken from randomly chosen states. Test to see if there is a significant difference in the means using \(\alpha\) = 0.05.
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14. Recent research indicates that the effectiveness of antidepressant medication is directly related to the severity of the depression (Khan, Brodhead, Kolts & Brown, 2005). Based on pre-treatment depression scores, patients were divided into four groups based on their level of depression. After receiving the antidepressant medication, depression scores were measured again and the amount of improvement was recorded for each patient. The following data are similar to the results of the study. Use a significance level of \(\alpha\) = 0.05. Test to see if there is a difference in the mean scores.
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15. An ANOVA was run to test to see if there was a significant difference in the average cost between three different types of fabric for a new clothing company. Random samples for each of the three fabric types was collected from different manufacturers. At \(\alpha\) = 0.10, run an ANOVA test to see if there is a difference in the means.
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16. Three students, Linda, Tuan, and Javier, are given laboratory rats for a nutritional experiment. Each rat's weight is recorded in grams. Linda feeds her 9 rats Formula A, Tuan feeds his 9 rats Formula B, and Javier feeds his 9 rats Formula C. At the end of a specified time-period, each rat is weighed again, and the net gain in grams is recorded. Using a significance level of 0.10, test to see if there is a difference in the mean weight gain for the three formulas.
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17. For a two-way ANOVA, a row factor has 3 different levels, a column factor has 4 different levels. There are 15 data values in each group. Find the following.
a) The degrees of freedom for the row effect.
b) The degrees of freedom for the column effect.
c) The degrees of freedom for the interaction effect.
18. Fill out the following two-way ANOVA table.
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19. Fill out the following two-way ANOVA table.
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20. Fill out the following two-way ANOVA table.
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21. Fill out the following two-way ANOVA table.
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22. Fill out the following two-way ANOVA table.
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23. A professor is curious if class size and format for which homework is administered has an impact on students’ test grades. In a particular semester, she samples 4 students in each category below and records their grade on the department-wide final exam. The data are recorded below. Assume the variables are normally distributed. Run a two-way ANOVA using \(\alpha\) = 0.05.
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24. A study was conducted to observe the impact of young adults eating a diet that is high in healthy fats. A random sample of young adults was instructed to eat a particular menu for a month. They were then tested to check a combination of recall skills, reflexes, and physical fitness and scored from 1-10 based on performance. They were divided into two groups, one eating a menu that is high in healthy fats, and the other low in healthy fats. They were also divided based on age. The data are recorded below. Assume the variables are normally distributed. Run a two-way ANOVA using \(\alpha\) = 0.05.
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25. A door-to-door sales company sells three types of vacuums. The company manager is interested to find out if the type of vacuum sold has an effect on whether a sale is made, as well as what time of day the sale is made. She samples 36 sales representatives and divides them into the following categories, then records their sales (in hundreds of dollars) for a week. Assume the variables are normally distributed. Run a two-way ANOVA using \(\alpha\) = 0.05.
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26. A customer shopping for a used car is curious if the price of a vehicle varies based on type of vehicle and location of used car dealership. She samples 5 vehicles in each category below and records the price of each vehicle. Each vehicle is in similar shape in regards to age, mileage, and condition. A two-way ANOVA test was run and the information from the test is summarized in the table below. State all 3 hypotheses, critical values, decisions and summaries using \(\alpha\) = 0.05.
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27. A sample of patients are tested for cholesterol level and divided into categories by age and by location of residence in the United States. The data are recorded below. Assume the variables are normally distributed. A two-way ANOVA test was run and the information from the test is summarized in the table below. State all 3 hypotheses, critical values, decisions and summaries using \(\alpha\) = 0.05.
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28. The employees at a local nursery swear by a certain variety of tomato seed and a certain variety of fertilizer. To test their instincts, they take a sample of 3 varieties of tomato seed and 4 varieties of fertilizer and find the following yield of tomatoes from each tomato plant. Assume the variables are normally distributed. Run a two-way ANOVA using \(\alpha\) = 0.05.
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29. An obstetrician feels that her patients who are taller and leaner before becoming pregnant typically have quicker deliveries. She samples 3 women in each of the following categories of Height and Body Mass Index and records the time they spent in the pushing phase of labor in minutes. All women in the sample had a natural vaginal delivery and it was their first childbirth. The data are recorded below. Assume the variables are normally distributed. Run a two-way ANOVA using \(\alpha\) = 0.05.
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- Solutions to Odd-Numbered Exercises
-
1. a
3. e
5. \(H_{0}: \mu_{1} = \mu_{2} = \mu_{3}\); \(H_{1}:\) At least one mean is different.
F = 0.5902; p-value = 0.5605.
Do not reject \(H_{0}\). There is not enough evidence to support the claim that there is a difference in the mean per-pupil costs for private school tuition for three counties in the Portland, Oregon, metro area.7. c
9. a) \(H_{0}: \mu_{A} = \mu_{B} = \mu_{C}\); \(H_{1}:\) At least one mean is different.
b) \(F = 10.64046\)
c) \(F_{\alpha} = 3.0781\)
d) Reject \(H_{0}\). There is enough evidence to support the claim that there is a difference in the mean price of the three types of plastic.
e) \(H_{0}: \mu_{A} = \mu_{B}\); \(H_{1} = \mu_{A} \neq \mu_{B}\); p-value=0; reject \(H_{0}\). There is significant difference in price between plastics A and B. \(H_{0}: \mu_{A} = \mu_{C}\); \(H_{1} = \mu_{A} \neq \mu_{C}\); p-value = 1; do not reject \(H_{0}\). There is not a significant difference in price between plastics A and C. \(H_{0}: \mu_{B} = \mu_{C}\); \(H_{1} = \mu_{B} \neq \mu_{C}\); p-value = 0.003; reject \(H_{0}\). There is significant difference in price between plastics B and C.11. \(H_{0}: \mu_{1} = \mu_{2} = \mu_{3}\); \(H_{1}:\) At least one mean is different. F = 2.5459; p-value = 0.0904; do not reject \(H_{0}\). There is not enough evidence to support the claim that there is a difference in the mean movie ticket prices by geographical regions.
13. \(H_{0}: \mu_{1} = \mu_{2} = \mu_{3}\); \(H_{1}:\) At least one mean is different. F = 2.7121; p-value = 0.0896; reject \(H_{0}\). There is sufficient evidence to support the claim that course delivery type is a factor in final exam score.
15. \(H_{0}: \mu_{A} = \mu_{B} = \mu_{C}\); \(H_{1}:\) At least one mean is different. F = 2.895; p-value = 0.06; reject \(H_{0}\). 0. There is sufficient evidence to support the claim that there is a difference in the mean cost between three different types of fabric.
\(H_{0}: \mu_{A} = \mu_{B}\); \(H_{1}: \mu_{A} \neq \mu_{B}\); p-value = 1; do not reject \(H_{0}\). There is not a significant difference in the mean cost of fabrics A and B. \(H_{0}: \mu_{A} = \mu_{C}\); \(H_{1}: \mu_{A} \neq \mu_{C}\); p-value = 0.222; do not reject \(H_{0}\). There is not a significant difference in the mean cost of fabrics A and C. \(H_{0}: \mu_{B} = \mu_{C}\); \(H_{1}: \mu_{B} \neq \mu_{C}\); p-value = 0.07; reject \(H_{0}\). There is significant difference in the mean cost of fabrics B and C.17. a) \(df_{\text{A}} = 2, df_{\text{E}} = 168\)
b) \(df_{\text{B}} = 3, df_{\text{E}} = 168\)
c) \(df_{\text{A} \times \text{B}} = 6, df_{\text{E}} = 168\)19.
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21.
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23. \(H_{0}:\) The format of the homework (paper vs. online) has no effect on the mean test grade. \(H_{1}:\) The format of the homework (paper vs. online) has an effect on the mean test grade. F = 0.7185; CV = F.INV.RT(0.05,1,12) = 4.7472; do not reject \(H_{0}\). There is not enough evidence to support the claim that the format of the homework (paper vs. online) has an effect on the mean test grade.
\(H_{0}:\) The class size has no effect on the mean test grade. \(H_{1}:\) The class size has an effect on the mean test grade. F = 5.7064; CV = F.INV.RT(0.05,1,12) = 4.7472; reject \(H_{0}\). There is enough evidence to support the claim that the class size has an effect on the mean test grade.
\(H_{0}:\) There is no interaction effect between the format of the homework (paper vs. online) and the class size on the mean test grade. \(H_{1}:\) There is an interaction effect between the format of the homework (paper vs. online) and the class size on the mean test grade. F = 0.8082; CV = F.INV.RT(0.05,1,12) = 4.7472; do not reject \(H_{0}\). There is not enough evidence to support the claim that there is an interaction effect between the format of the homework (paper vs. online) and the class size on the mean test grade.25. \(H_{0}:\) The time of day has no effect on the mean number of vacuum sales. \(H_{1}:\) The time of day has an effect on the mean number of vacuum sales. F = 4.4179; CV = F.INV.RT(0.05,2,27) = 3.3541; reject \(H_{0}\). There is enough evidence to support the claim that the time of day has an effect on the mean number of vacuum sales.
\(H_{0}:\) The type of vacuum has no effect on the mean number of vacuum sales. \(H_{1}:\) The type of vacuum has an effect on the mean number of vacuum sales. F = 27.4172; CV = F.INV.RT(0.05,2,27) = 3.3541; reject \(H_{0}\). There is enough evidence to support the claim that the type of vacuum has an effect on the mean number of vacuum sales.
\(H_{0}:\) There is no interaction effect between time of day and type of vacuum on the mean number of vacuum sales. \(H_{1}:\) There is an interaction effect between time of day and type of vacuum on the mean number of vacuum sales. F = 0.9021; CV = F.INV.RT(0.05,4,27) = 2.7278; do not reject \(H_{0}\). There is not enough evidence to support the claim that there is an interaction effect between time of day and type of vacuum on the mean number of vacuum sales.27. \(H_{0}:\) Age has no effect on the mean cholesterol level. \(H_{1}:\) Age has an effect on the mean cholesterol level. F = 7.863; CV = F.INV.RT(0.05,2,24) = 3.4028; reject \(H_{0}\). There is enough evidence to support the claim that age has an effect on the mean cholesterol level.
\(H_{0}:\) Location has no effect on the mean cholesterol level. \(H_{1}:\) Location has no effect on the mean cholesterol level. F = 5.709; CV = F.INV.RT(0.05,3,24) = 3.0087; reject \(H_{0}\). There is enough evidence to support the claim that the location has an effect on the mean cholesterol level.
\(H_{0}:\) There is no interaction effect between age and location on the mean cholesterol level. \(H_{1}:\) There is an interaction effect between age and location on the mean cholesterol level. F = 1.455; CV = F.INV.RT(0.05,6,24) = 2.5082; do not reject \(H_{0}\). There is not enough evidence to support the claim that there is an interaction effect between age and location on the mean cholesterol level.29. \(H_{0}:\) Height has no effect on the mean delivery time. \(H_{1}:\) Height has an effect on the mean delivery time. F = 3.2798; CV = F.INV.RT(0.05,2,19) = 3.5219; do not reject \(H_{0}\). There is not enough evidence to support the claim that height has an effect on the mean delivery time.
\(H_{0}:\) BMI has no effect on the mean delivery time. \(H_{1}:\) BMI has an effect on the mean delivery time. F = 1.3763; CV = F.INV.RT(0.05,2,19) = 3.5219; do not reject \(H_{0}\). There is not enough evidence to support the claim that BMI has an effect on the mean delivery time.
\(H_{0}:\) There is no interaction effect between the height and BMI on the mean delivery time. \(H_{1}:\) There is an interaction effect between the height and BMI on the mean delivery time. F = 0.1125; CV = F.INV.RT(0.05,4,19) = 2.8951; do not reject \(H_{0}\). There is not enough evidence to support the claim that there is an interaction effect between the height and BMI on the mean delivery time.
From Mostly Harmless Statistics by Rachel L. Webb, adapted from LibreTexts. Licensed CC BY-SA 4.0. XYZ Homework OER web edition, adapted with 7 verified corrections (see errata).