Good Example Of Anova Essay

Mathematics

The company’s pilots have been exposed to three types of weight loss programs: diet alone, exercise alone, and a combination of diet and exercise. Random samples of pilots’ weight lost amongst the three groups have been obtained.
ANOVA (Analysis of variance) is a hypothesis testing technique to test the equality of population means by examining the variances of samples. ANOVA would be able to determine whether the e=differences between the samples are simply due to random error or whether there are systemic effects because of which the means differ.

The assumptions while carrying out ANOVA are:-

The populations of weight loss data of the three techniques follow a normal distribution.
All populations have the same variance.
The samples taken are randomly selected and individually independent of each other.
The data given is as shown below in Table 1:
Null and Alternate Hypothesis
H0 : μ1 = μ2= μ3. ie Mean weight loss is statistically equal across all three types of weight loss programs.
Since the null hypothesis assumes all the means are equal, we could reject the null hypothesis if only one mean is not equal.

Ha : At least one mean weight loss is not statistically equal.

Test Statistic
The test statistic in ANOVA is the ratio of the between and within variation in the data. It follows an F Distribution.
Total Sum of Squares (SST). It indicates the total variation in the data, and is the sum of the between and within variation.
SST = i=1rj=1c(Xij-X)2 , where r is the number of rows in the table, c is the number of columns, X is the grand mean, and Xij is the ith observation in the jth column.

Using the data in Table 1,

X = grand mean = (19+82+92)/ 30 = 193/30 = 6.43
SST = (2-6.43)2 + (3-6.43)2 + (3-6.43)2 + (2-6.43)2 + (9-6.43)2 =363.36

Treatment Sum of Squares (SSTR).

Between Sum of Squares ( or Treatment Sum of Squares) is the variation in the data between the different samples.
SSTR = rjXj-X2 , where rj is the number of rows in the jth treatment and Xj is the mean of the jth treatment.

Using the data in Table 1,

SSTR = {10 x (1.9-6.43)2 } + { 10 x (8.2- 6.43)2 } + {10 x (9.2-6.43)2} = 313.26
Within Variation (Error Sum of Squares) SSE. Error sum of squares (SSE) is the variation in the data from each individual treatment.
Error sum of squares (SSE) = (Xij- Xj)2

The next step is to determine the average sources of variation in the data using SST, SSR and SSE

Total Mean Squares (MST)
MST = SST /( N-1) , ie the ‘average total variation in the data’ (where N is the number of observations), = 363.36/29 = 12.52

Mean Square Treatment (MSTR)

MSTR = SSTR / (c-1), is the ‘average between variation’, where c is the number of columns in the data, = 313.26/(3-1) = 313.26/2 = 156.62

Mean Square Error (MSE)

MSE = SSE/ (N-c), is the ‘average within variation’, = 50.1/ (30-3) = 1.67

Test Statistic F Distribution

The test statistic, F Distribution, for one way ANOVA is the ratio between MSTR and MSE, ie the ration of the ‘average between variation’ to the ‘average within variation’.
F = MSTR/ MSE = 156.62/ 1.67 = 93.79

Critical Value

FCV has df1 and df2 degrees of freedom, where df1 is the numerator degrees of freedom equal to c-1 and df2 is the denominator degrees of freedom equal to N-c. We specify a significance level α of 5%
For our data, df1 = 3-2=2.
df2=30-3=27.
We need to find F2,27CV . In Excel, the Critical value is found by the function F.INV.RT function.
Using Excel, we find F.INV.RT(0.05, 2,27) = Critical value = 3.35

Decision Rule

We reject the Null Hypothesis if: F (Observed value) > FCV. For our data, 93.79> 3.35. Hence, we reject the Null Hypothesis.

Interpretation

As we have rejected the Null Hypothesis, we are 95% confident (1-α) that the mean weight loss is not statistically the same for the three different approaches to weight-loss, ie diet, diet and exercise and exercise only.

Thus, the ANOVA test tells us that at least one mean is different.

An additional test would be required to determine which mean/s is/are different.
Least Significant Difference
Least Significant Difference for a balanced sample: 2 × MSE × F1,N-cr , where MSE is the Mean Square error and r is the number of rows in each treatment.
F1,N-c = F1,27 = 4.21
For our data, LSD = 2 × 1.67 × 4.2110 = 1.185
Now, if the absolute value of the difference between any two treatments is greater than the LSD, we may conclude that they are not statistically equal.

Diet vs Exercise

| (1.9 – 8.2) | = 6.3 . As 6.3>1.185, we may conclude that mean weight loss is statistically different when doing diet as opposed to exercise.

Diet vs Diet cum Exercise

| (1.9 – 9.2) | = 7.3 . As 7.3>1.185, we may conclude that mean weight loss is statistically different when doing diet as opposed to diet cum exercise.

Exercise vs Exercise cum Diet

| (8.2 – 9.2) | = 1 . As 1>1.185, we may conclude that mean weight loss is statistically the same when following the exercise regimen as opposed to diet cum exercise.

Recommendation

Using One-Way ANOVA, we have established that the three weight loss regimens are not statistically the same. Using the Least Significant Difference method, we find that diet is statistically different from exercise and exercise cum diet. As the mean weight loss while doing diet is lesser, we eliminate diet from the recommendation for pilots.

Statistically, there is no difference in weight loss while doing exercise as opposed ot exercise cum diet.

Therefore, pilots would be recommended to follow either of the two options – exercise, or exercise cum diet.

Reference

Seltman, H. (2014). Experimental design and analysis. [Online Book]. Carnegie Melon University. Retrieved February 21, 2015, from http://www.stat.cmu.edu/~hseltman/309/Book/Book.pdf