Analysis of Variance

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Variance Analysis (ANOVA)

Problem: The mean income level of three states, Alaska, Michigan, and Seattle, is used to investigate the occurrence of any paygrade differences. Five companies with branch offices in all three states were chosen, and the median yearly income of these companies was collected, as shown below:

The Company’s Name

Alaska

Michigan

Seattle

33.568 48.273 57.37 Company A

30.505 48.1 55.66 Company B

33.156 48.6 61.42 Company C

31.205 46.2 56.246 Company D

E Company 28.9 35.056 51.245

*Figures are in Thousands

Null Hypothesis:

There is no significant difference among the median salaries in these states.

Mathematically, H0: μ1 = μ2 = μ3

Alternative Hypothesis:

At least one state has a different paygrade among these three states.

Mathematically, H0: μ1 ≠μ2 ≠ μ3

Let us perform an ANOVA test to testify the hypothesis by analyzing the variances in each category.

Step 1: The first step consists of computing variance for each group. It can be done by using excel function VAR.S () as follows:

Step 2: The Mean Square Error (MSE) is given by the mean of all variances above.

MSE = = 16.79

Step 3: Now compute the variance of three variances computed above. The variance is:

228.24

The Mean Square Between (MSB) value can be obtained by multiplying the above variance by the number of companies, which is 5 in this case. MSB = 1141.21

Step 4: F-value may be obtained from the quotient of MSB and MSE.

F-value = = = 13.59

Step 5: degrees of freedom, df1 and df2 may be computed as follows. We have three different categories (states) with 5 number of companies in each. This gives us,

df1 = 3-1 = 2 and

df2 = 3× (5-1) = 3×4 =12

Step 6: p-value may be obtained from Excel function F. DIST.RT(F-value, df1, df2). Here the degree of freedoms, df1 and df2 are 2 and 12 respectively.

p-value = 0.00033329

Step 7: Let us consider a 5% level of significance for which α = 0.05. If the p-value is less than the level of significance then we reject the null hypothesis and accept otherwise. Here, p-value of 0.00033329 is much less than 0.05; we reject the null hypothesis.

Step 8: Decision: The ANOVA test confirms the presence of significant amount of difference among the income level of these states.