14B Preview

Preparing for the next class

In the next in-class activity, you will need to make connections between values  presented in an ANOVA table and describe the shape of the F Distribution. You will also  need to understand how the P-value is represented by an area under an F Distribution  curve and describe how the F-statistic is used in hypothesis testing for a one-way  ANOVA. Finally, you will need to use the P-value to reach a conclusion.

Suppose a researcher wants to investigate the effect of the amount of fertilizer on the  height of a common houseplant. More specifically, the researcher is interested in  determining if there is a difference in the mean height of plants between those receiving  one of following three different fertilizer levels: high, medium, and low.

The following data are the simulated results of this controlled experiment. (Note that this  small dataset is used introduce the concept and make calculations easier. When  conducting an ANOVA, larger sample sizes are usually needed to meet assumptions.)

Fertilizer Level Height of Plant (inches)
Low 23.2, 20.9, 21.5, 25.3
Medium 24.6, 27.7, 22.5, 30.1
High 29.2, 30.2, 31.1, 33.6

As we saw in the previous in-class activity, conducting a one-way ANOVA involves  comparing the variation within each of the groups to the variation between each of the  groups. When the variation between each of the groups is significantly larger than the  variation within each of the groups, we might conclude that there is a statistically significant difference among the means.

In an ANOVA table, the calculation illustrating the total variation within the groups of  interest is known as the error sum of squares (SSError). The calculation illustrating  the total variation between the groups is known as the group sum of squares (SSGroup).

Two other essential columns found in an ANOVA table are the degrees of freedom (df) and the mean square.

The following table illustrates how these values are calculated for each of the given  sources: Group or Error (i.e., between and within).

When calculating these values, it is important to know that [latex]k[/latex] represents the number of  groups being considered and [latex]N[/latex] represents the total number of data values among all  groups.

Source Degrees of Freedom (df) Sum of  Squares Mean  Square F-Statistic
Group [latex]k-1[/latex]

(The number of groups  minus 1)

 SSGroup SSGroup

[latex]k-1[/latex]

 MSGroup

MSError
Error [latex]N-k[/latex]

(The total number of data  points minus the number of  groups)

 SSError SSError

[latex]N-k[/latex]
Total [latex]N-1[/latex]

(The total number of data  points minus 1)

 SSGroup +  SSError

Question 1

1) Given the information in the previous table, let’s calculate the degrees of freedom  associated with the Group and Error sources for the fertilizer and plant height  scenario.

Part A: In the fertilizer and plant height scenario, what is the value for � (i.e., the  number of groups)?

Part B: In the fertilizer and plant height scenario, what is the value for � (i.e., the  total number of data values in the experiment)?

Part C: Given your previous responses, complete the Degrees of Freedom column  for the fertilizer and plant height scenario in the following table.

Source Degrees of

Freedom (df)
Group k − 1 =
Error N − k =
Total N − 1 =

When performing a formal hypothesis test for a one-way ANOVA, the mean square  values are used to calculate the value of our test statistic; thus, they impact the P-value  we get.  

As noted in the previous table, the mean square for error and mean square for group are calculated by taking each of the sum of square values and dividing them by the  degrees of freedom associated with the respective source (i.e., Group or Error).

[latex]Mean\;Square\;for\;Error\;(MSError)=\frac{Error\;sum\;of\;squares}{degrees\;of\;freedom\;(Error)}=\frac{SSE}{N-k}[/latex]

[latex]Mean\;Square\;for\;Group\;(MSGroup)=\frac{Group\;sum\;of\;squares}{degrees\;of\;freedom\;(Group)}=\frac{SSG}{k-1}[/latex]

Question 2

2) We are given the sum of squares and degrees of freedom for the fertilizer and plant  height scenario. Use these values to calculate the mean square for error and mean  square for group. Enter the values in the following table.

Source Degrees of

Freedom (df)

 Sum of Squares Mean Square
Group 2 55.9134
Error 9 140.0108
Total 11 195.9242

Question 3

3) The test statistic that we use to complete the appropriate hypothesis test for a one way ANOVA is calculated with the ratio below:

F-Statistic = MSGroup

MSError = Variation BETWEEN groups

Variation WITHIN groups

Use the MSGroup and MSError values you calculated to find the F-statistic for this  situation:

F-Statistic = MSGroup

MSError = Variation BETWEEN groups

Variation WITHIN groups = 

Question 4

4) Recall from previous in-class activities that hypothesis testing for two means is  based on the t Distribution, and we calculate the test statistic, t. ANOVA is based on  the F Distribution, so we will be calculating the F-statistic.

Go to the DCMP F Distribution tool at https://dcmathpathways.shinyapps.io/FDist/.

Using the data analysis tool, graph the F Distribution from the example by entering  the degrees of freedom from Question 1, Part C or by selecting varying values.

Part A: The F Distribution is:

a) Skewed left

b) Symmetric

c) Skewed right

Part B: The t Distribution is symmetric, centered at the mean 0. Thus, when  conducting a t-test, we have positive t-values and negative t-values. Adjust  the numerator and denominator degrees of freedom in the F Distribution tool.  What do you notice about the values in the F Distribution (the values on the  horizontal axis)?

a) The F values are always negative.

b) The F values are always positive.

c) The F values can be negative or positive.

As we just saw, the F-statistic is the ratio of the variation between groups (MSGroup) to the variation within groups (MSError). Larger values of the F-statistic (greater than 1)  would imply that the variation between groups is larger than the variation within groups.

Question 5

5) As the variation between groups gets significantly larger than the variation within  groups, we will get:

  1. a) A larger F-statistic that corresponds to a greater P-value and less evidence to  support the null hypothesis
  2. b) A larger F-statistic that corresponds to a smaller P-value and more evidence to  support the alternative hypothesis
  3. c) A P-value that does not change as the F-statistic changes

When there is a greater difference among the group means, the F-statistic will be larger;  when there is a smaller difference among the group means, the F-statistic will be  smaller.

Question 6

6) Which of the following plots will likely have the largest F-statistic? Explain.  Note that the means for each group are exactly the same in Plot A and Plot B.

Plot A

A table and a box plot. The table is titled “Descriptive Statistics” and has the columns “Group,” “Sample Size,” “Mean,” “Standard Deviation,” and “Standard Error.” The first row reads “Drug 1,” 6, 204, 11.5, 4.70, the second row reads “Drug 2,” 6, 191, 12.5, 5.10, and the third row reads “Drug 3, 6, 176, 12.9, 5.28. Beneath this is a box plot labeled “Total Cholesterol (mg/dL)” on the horizontal axis. For Drug 1, the low point is at approximately 185 and the high point is at approximately 220. The low end of the box is at approximately 198, the high end is at approximately 209, and the middle line is at approximately 205. For Drug 2, the low point is at approximately 175 and the high point is at approximately 210. The low end of the box is at approximately 182, the high end is at approximately 196, and the middle line is at approximately 193. For Drug 3, the low point is at approximately 160 and the high end is at approximately 190. The low end of the box is at approximately 167, the high end is at approximately 188, and the middle line is at approximately 180. There are points on the horizontal axis “y bar sub 1” at approximately 204, “y bar sub 2” at approximately 191, and “y bar sub 3” at approximately 178.

Plot B

A table and a box plot. The table is titled “Descriptive Statistics” and has the columns “Group,” “Sample Size,” “Mean,” “Standard Deviation,” and “Standard Error.” The first row reads “Drug 1,” 5, 204, 1.58, 0.707, the second row reads “Drug 2,” 5, 191, 1.58, 0.707, and the third row reads “Drug 3, 5, 178, 1.58, 0.707. Beneath this is a box plot labeled “Total Cholesterol (mg/dL)” on the horizontal axis. For Drug 1, the low point is at approximately 202 and the high point is at approximately 206. The low end of the box is at approximately 203, the high end is at approximately 205, and the middle line is at approximately 204. For Drug 2, the low point is at approximately 188 and the high point is at approximately 193. The low end of the box is at approximately 190, the high end is at approximately 192, and the middle line is at approximately 191. For Drug 3, the low point is at approximately 176 and the high end is at approximately 180. The low end of the box is at approximately 177, the high end is at approximately 179, and the middle line is at approximately 178. There are points on the horizontal axis “y bar sub 1” at approximately 204, “y bar sub 2” at approximately 191, and “y bar sub 3” at approximately 178.

Remember that in hypothesis testing, the P-value is our statistical evidence to support  our conclusions. When the P-value is less than our significance level, α, we reject the  null hypothesis and have sufficient evidence to support the alternative hypothesis.  Otherwise, we fail to reject the null hypothesis and do not have sufficient evidence to  support the alternative hypothesis.

Question 7

7) Suppose we ran an ANOVA at the alpha (α) = 0.05 significance level to answer the following question: “Is there a difference in the mean hours of exercise per week  among people in the U.S. regions of the Northeast, South, West, and Midwest?”

Part A: If the test resulted in a P-value of 0.0245, what should you do? a) Reject the null hypothesis.

  1. b) Fail to reject the null hypothesis.
  2. c) Accept the null hypothesis.

Part B: Based on your answer to Part A, what would be your conclusion?

  1. a) There is convincing evidence to suggest that there is a difference in the  mean hours of exercise between at least two regions.
  2. b) There is convincing evidence to suggest that there is a difference in the  mean hours of exercise between all four regions.
  3. c) There is not convincing evidence to suggest that there is a difference in the mean hours of exercise between the four regions.

Question 8

8) Suppose we ran an ANOVA at the alpha (α) = 0.05 significance level for the  following question: “Is there is a difference in the mean reduction of blood pressure  for the following three different techniques: diet, exercise, and medication.”

Part A: If the test resulted in a P-value of 0.3214, what should you do? a) Reject the null hypothesis.

  1. b) Fail to reject the null hypothesis.
  2. c) Accept the null hypothesis.

Part B: Based on your answer to Part A, what would be your conclusion?

  1. a) There is convincing evidence that there is a difference in the mean blood  pressure reduction between at least two techniques.
  2. b) There is convincing evidence that there is a difference in the mean blood  pressure reduction between all three techniques.
  3. c) There is not convincing evidence that there is a difference in the mean  blood pressure reduction between the three techniques.