RESULTS AND CONCLUSIONS FOR ANOVA
WHEN SHOULD YOU USE THE ANOVA TEST?
Quantitative data
Comparing the MEAN of MORE THAN TWO groups
Sample data
Effect of Stress on the Heights of Phaseolus vulgaris After 30 Days
Height of plants stressed once daily (cm) / Height of plants stressed twice daily (cm) / Height of unstressed plants (cm)55.21 / 75.33 / 48.04
65.33 / 82.18 / 64.56
50.50 / 76.22 / 59.11
57.12 / 76.89 / 57.25
59.14 / 79.43 / 51.66
73.00 / 80.02 / 63.22
57.82 / 83.64 / 64.78
54.24 / 83.55 / 58.36
61.92 / 97.46 / 44.27
67.03 / 54.10 / 49.99
Step 1: State the null hypothesis.
The mean heights of all plant groups are not significantly different.
Step 2: Establish the level of significance (0.05)
probability of error in rejecting null hypothesis is 5/100
Step 3: Calculate ANOVA using Excel or the graphing calculator (see end of this document)
Results from Excel:
Anova: Single FactorSUMMARY
Groups / Count / Sum / Average / Variance
Height of plants stressed once daily (cm) / 10 / 601 / 60.131 / 45.61141
Height of plants stressed twice daily (cm) / 10 / 789 / 78.882 / 115.5401733
Height of unstressed plants (cm) / 10 / 561 / 56.124 / 52.78344889
ANOVA
Source of Variation / SS / df / MS / F / P-value / F crit
Between Groups / 2951.9 / 2 / 1475.971023 / 20.69746607 / 3.55294E-06 / 3.354130829
Within Groups / 1925.4 / 27 / 71.31167741
Total / 4877.4 / 29
Results from TI-84:
One-way ANOVA
F = 20.69746607 Critical F
p = 3.5529399E -6 (3.55 X 10-6) p value
Factor Variance between groups
df = 2 Number of groups minus 1
SS = 2951.94205
MS = 1475.97102 Mean Squares (variance)
Error Variance within groups
df = 27 Sample size minus number of groups
SS = 1925.41529
MS = 71.3116774 Mean Squares (variance)
Step 4: Compare the calculated value for F to the critical value for F.
TI-84: Look at a table of critical values of F
Excel: Look at Fcrit
Step 5: Decide to reject or not reject the null hypothesis
Calculated F < critical F → null hypothesis not rejected
Calculated F critical F → null hypothesis is rejected
Critical F at 0.05 level.= 3.35; calculated F of 20.70 > 3.35.
The null hypothesis is rejected.
Step 6: Determine whether the statistical findings support the research hypothesis.
IF Null hypothesis was rejected = research hypothesis was supported
IF Null hypothesis not rejected = research hypothesis was not supported
Because the null hypothesis was rejected, the research hypothesis that there would be a difference in the mean heights of Phaseolus vulgaris plants exposed to two levels of stress and no stress was supported.
Step 7: Construct a data table that communicates all statistics
Table A: Effect of Stress on the Mean Height of Phaseolus vulgaris After 30 Days
Plants Stressed Twice Daily / Plants Stressed Once Daily / Unstressed groupMean
Standard deviation
1SD
2SD
Number / 78.88 cm
10.75 cm
68.13 – 89.63 cm
57.38 – 100.38 cm
10 / 60.13 cm
6.75 cm
53.38 – 66.88 cm
46.63 – 73.63 cm
10 / 56.12 cm
7.27 cm
48.85 – 63.39 cm
41.58 – 70.66 cm
10
F: 20.70
Numerator df: 2 (between groups)
Denominator df: 27 (within groups)
/ F of 20.70 > 3.35p = 3.55 X 10 -6
Step 8: Write a paragraph describing results
¨ Write a topic sentence stating the independent and dependent variables, and a reference to
tables and graphs.
Effects of stress on the height of Phaseolus vulgaris plants are summarized in Table A.
¨ Write sentences comparing the means and standard deviation of the groups.
Plants stressed twice daily exhibited a greater mean height (78.88 cm) than plants stressed once daily (60.13 cm) and non-stressed plants (56.12 cm). Variation within the group of plants stressed twice daily was greater than the other two groups of plants. This is shown by a standard deviation of 10.75 for the plants stressed twice daily compared to a standard deviation of 6.75 for plants stressed once daily and 7.27 for plants not stressed.
¨ Write sentences describing the statistical test, level of significance, and null hypothesis.
The ANOVA test was used to test the following null hypothesis at the 0.05 level of
significance: The mean heights of plants stressed twice daily, plants stressed once daily and plants not stressed are not significantly different.
¨ Write sentences comparing the calculated F value with the critical value and make a
statement about rejection of the null hypothesis.
The null hypothesis was rejected (F = 20.70 > 3.35; p =0.00)
¨ Write sentences stating support of the research hypothesis by the data.
The data did support the research hypothesis that the mean heights of Phaseolus vulgaris plants exposed to two stress treatments and those not exposed to stress would be different
Step 9: Construct a box-and-whiskers plot to illustrate the variation for each group.
Step 10: Write an appropriate conclusion.
· What was the purpose of the experiment?
The effect of stress on the growth of Phaseolus vulgaris plants was investigated by comparing the height of three groups of plants exposed to varying levels of stress. One group of ten plants was not stressed, a second group of ten plants was stressed once daily, and a third group of ten plants was stressed twice daily.
· What were the major findings? (Focus on results of the statistical test)
There was a significant difference between the mean heights of plants stressed once daily, plants stressed twice daily, and unstressed plants 30 days after transplanting.
· Was the research hypothesis supported by the data?
The research hypothesis that there would be a difference in the mean heights of Phaseolus vulgaris plants stressed once daily, twice daily and unstressed was supported. The mean height of plants stressed twice daily is much greater than the mean heights of the other two groups, but the mean height of plants stressed once daily is very similar to the mean height of unstressed plants. Further statistical tests are needed to determine if there is a significant difference between the plants stressed once daily and the unstressed plants.
· How did your findings compare (similarities and differences) with your preliminary research?
Similarly, Japanese farmers found that hitting and pulling rice plants were beneficial.(Osaki 57)
· What possible explanations can you offer for similarities and/or differences between your results and other researchers?
The type of stress (pulling of plants) used in this experiment was the same as in the Osaki experiment. The results were similar even though the Phaseolus vulgaris plant is a dicot and the rice plants used in the Osaki experiment are monocots.
· What recommendations do you have for further study and for improving the experiment?
Additional investigations using various sources of stress at more frequent intervals with
both monocots and dicots should be conducted. Improved experimental design techniques should be implemented, including a larger sample size, more frequent measurement, and a longer growing period.
NOTE: You should be able to write much more than I did. After all, you did an extensive literature review before experimentation and you are the “expert” for your topic.
How to do ANOVA with Excel or calculator
ANOVA with Excel
Statistical procedures are found in the Data Analysis Tools, which are included with Microsoft Excel.
You need to first verify that Data Analysis tools are installed:
· Open Microsoft Excel
· Select File ® Options ® Add-Ins.
· In the Add-Ins dialog box that appears, select the Analysis ToolPak and Analysis ToolPak – VBA check boxes from the Add-Ins Available list and click the OK button.
· Exit Microsoft Excel (to save the selections)
You should only need to do this once. (School computers are an exception)
1. Enter your data into columns on the spreadsheet. The first cell of each column should have your column heading.
2. Select Tools ® Data Analysis and in the dialog box select ANOVA: Single Factor and click OK.
3. In the ANOVA: Single Factor dialog box, click on the icon in the Input Range box, then highlight all of your data. Click on the icon again to return to the ANOVA: Single Factor dialog box.
4. Leave Alpha as 0.05 and check the Labels box to include your column headings.
5. Select the Output Range option and choose a cell away from your data (E1 or F1).
6. Click OK.
7. Results appear in the columns next to your data.
8. Change the width of the columns to view all contents.
ANOVA with the graphing calculator
1. Enter your data into lists by using STAT, EDIT (see calculator sheet for t-Test).
2. Choose STAT and select “TESTS”.
3. Select “H: ANOVA(“ and ENTER.
4. Enter the list names (L1, L2, L3 …) containing your data, separated by commas.
5. Press ENTER and your ANOVA values will appear.