Another ANOVA Example
Calcium is an essential mineral that regulates the heart, is important for blood clotting and for building healthy bones. The National Osteoporosis Foundation recommends a daily calcium intake of 10001200 mg/day for adult men and women. While calcium is contained in some foods, most adults do not get enough calcium in their diets and take supplements. Unfortunately some of the supplements have side effects such as gastric distress, making them difficult for some patients to take on a regular basis.
A study is designed to test whether there is a difference in mean daily calcium intake in adults with normal bone density, adults with osteopenia (a low bone density which may lead to osteoporosis) and adults with osteoporosis. Adults 60 years of age with normal bone density, osteopenia and osteoporosis are selected at random from hospital records and invited to participate in the study. Each participant's daily calcium intake is measured based on reported food intake and supplements. The data are shown below.
Normal Bone Density 
Osteopenia 
Osteoporosis 

1200 
1000 
890 
1000 
1100 
650 
980 
700 
1100 
900 
800 
900 
750 
500 
400 
800 
700 
350 
Is there a statistically significant difference in mean calcium intake in patients with normal bone density as compared to patients with osteopenia and osteoporosis? We will run the ANOVA using the fivestep approach.
 Step 1. Set up hypotheses and determine level of significance
H_{0}: μ_{1} = μ_{2} = μ_{3} H_{1}: Means are not all equal α=0.05
 Step 2. Select the appropriate test statistic.
The test statistic is the F statistic for ANOVA, F=MSB/MSE.
 Step 3. Set up decision rule.
In order to determine the critical value of F we need degrees of freedom, df_{1}=k1 and df_{2}=Nk. In this example, df_{1}=k1=31=2 and df_{2}=Nk=183=15. The critical value is 3.68 and the decision rule is as follows: Reject H_{0} if F > 3.68.
 Step 4. Compute the test statistic.
To organize our computations we will complete the ANOVA table. In order to compute the sums of squares we must first compute the sample means for each group and the overall mean.
Normal Bone Density 
Osteopenia 
Osteoporosis 

n_{1}=6 
n_{2}=6 
n_{3}=6 



If we pool all N=18 observations, the overall mean is 817.8.
We can now compute:
Substituting:
Finally,
Next,
SSE requires computing the squared differences between each observation and its group mean. We will compute SSE in parts. For the participants with normal bone density:
Normal Bone Density 
(X  938.3) 
(X  938.3333)^{2} 

1200 
261.6667 
68,486.9 
1000 
61.6667 
3,806.9 
980 
41.6667 
1,738.9 
900 
38.3333 
1,466.9 
750 
188.333 
35,456.9 
800 
138.333 
19,126.9 
Total 
0 
130,083.3 
Thus,
For participants with osteopenia:
Osteopenia 
(X  800.0) 
(X  800.0)^{2} 

1000 
200 
40,000 
1100 
300 
90,000 
700 
100 
10,000 
800 
0 
0 
500 
300 
90,000 
700 
100 
10,000 
Total 
0 
240,000 
Thus,
For participants with osteoporosis:
Osteoporosis 
(X  715.0) 
(X  715.0)^{2} 

890 
175 
30,625 
650 
65 
4,225 
1100 
385 
148,225 
900 
185 
34,225 
400 
315 
99,225 
350 
365 
133,225 
Total 
0 
449,750 
Thus,
We can now construct the ANOVA table.
Source of Variation 
Sums of Squares (SS) 
Degrees of freedom (df) 
Mean Squares (MS) 
F 

Between Treatments 
152,477.7 
2 
76,238.6 
1.395 
Error or Residual 
819,833.3 
15 
54,655.5 

Total 
972,311.0 
17 
 Step 5. Conclusion.
We do not reject H_{0} because 1.395 < 3.68. We do not have statistically significant evidence at a =0.05 to show that there is a difference in mean calcium intake in patients with normal bone density as compared to osteopenia and osterporosis. Are the differences in mean calcium intake clinically meaningful? If so, what might account for the lack of statistical significance?
OneWay ANOVA in R
The video below by Mike Marin demonstrates how to perform analysis of variance in R. It also covers some other statistical issues, but the initial part of the video will be useful to you.