One- way ANOVA(Analysis of Variance) with example


One - way ANOVA(Analysis of Variance):

One-way ANOVA (Analysis of Variance) is a statistical test used to determine if there is a significant difference between the means of three or more groups. This test assumes that the population variances are equal.

The term "one-way" refers to the presence of only one independent variable (or factor) that is being tested with multiple levels (or categories). For instance, in a study investigating the effectiveness of various pain medications, the independent variable would be the medication type (e.g., aspirin, ibuprofen, acetaminophen), and the levels would be the different types of medication.

The null hypothesis for a one-way ANOVA states that there is no difference between the means of the groups, while the alternative hypothesis suggests that at least one group mean is different from the others.

To conduct a one-way ANOVA, the F-statistic is calculated by dividing the between-group variance by the within-group variance. A higher F-statistic implies a higher probability of a significant difference between the group means.

If the F-statistic is significant (e.g., p < 0.05), the null hypothesis is rejected, indicating that there is a significant difference between at least two group means. Post-hoc tests such as Tukey's HSD or the Bonferroni correction can be performed to determine which groups are significantly different from each other.

In conclusion, the one-way ANOVA is a widely used statistical test for comparing the means of multiple groups. It is applicable to various research questions and study designs.


Example:

Suppose a farmer wants to determine if three different fertilizer brands (A, B, and C) have a significant effect on the crop yield of a particular field. The farmer randomly selects 20 sections of the field and applies each of the three fertilizer brands to an equal number of sections. After the crop has matured, the farmer records the crop yield (in kg) for each of the 60 sections.

The crop yield data for the three fertilizer brands are summarized in the following table:

Fertilizer brand

Sample size (n)

Mean crop yield (kg)

Sample variance

A

20

175

125

B

20

160

144

C

20

180

100


The null hypothesis is that the means of the crop yields for the three fertilizer brands are equal. The alternative hypothesis is that at least one of the means is different.

Step 1: Calculate the sum of squares within groups (SSW)

To calculate SSW, we first need to calculate the total variance of the crop yields.

Total variance = ∑(xi - x̄)² / (N - 1)

                        = (125 + 144 + 100) / (20 + 20 + 20 - 1)

                        = 241600 / 57

                        = 4231.58

For Brand A:

SSWA = (nA - 1)s²A

            = 19(125)

            = 2375


For Brand B:

SSWB = (nB - 1)s²B

            = 19(144)

            = 2736


For Brand C:

SSWC = (nC - 1)s²C

            = 19(100)

            = 1900


SSW = SSWA + SSWB + SSWC

         = 2375 + 2736 + 1900

          = 7011


Step 2: Calculate the sum of squares between groups (SSB)

To calculate SSB, we need to find the total mean crop yield and the sum of squares for each of the three fertilizer brands.

Total mean crop yield = (mean yield for Brand A + mean yield for Brand B + mean yield for Brand C) / 3

         = (175 + 160 + 180) / 3

         = 171.67


For Brand A:

(xi - x̄)² = (175 - 171.67)² = 11.2225

∑(xi - x̄)² = 20(175 - 171.67)² = 4445


For Brand B:

(xi - x̄)² = (160 - 171.67)² = 134.9556

∑(xi - x̄)² = 20(160 - 171.67)² = 26991.12


For Brand C:

(xi - x̄)² = (180 - 171.67)² = 70.7556

∑(xi - x̄)² = 20(180 - 171.67)² = 14151.12


SSB = ∑(xi - x̄)²

        = 4445 + 26991.12 + 14151.12

        = 45587.24


Step 3: Calculate the total sum of squares (SST)

SST = SSW + SSB

        = 7011 + 45587.24

        = 52698.24


Step 4: Calculate the degrees of freedom


DFW = N - k = 60 - 3 = 57 (where N is the total sample size and k is the number of groups)

DFB = k - 1 = 3 - 1 = 2


Step 5: Calculate the mean square values


MSW = SSW / DFW = 7011 / 57 = 122.79

MSB = SSB / DFB = 45587.24 / 2 = 22793.62


Step 6: Calculate the F-value


F = MSB / MSW = 22793.62 / 122.79 = 185.59


Step 7: Determine the critical value and p-value


Using a significance level of α = 0.05, the critical value for F with 2 and 57 degrees of freedom is 3.17. Since the calculated F-value (185.59) is much larger than the critical value, we can reject the null hypothesis and conclude that at least one of the means is significantly different.

To find the p-value, we can use an F-distribution table or a statistical software. The p-value is the probability of obtaining an F-value as extreme or more extreme than the calculated F-value, assuming the null hypothesis is true. In this case, the p-value is less than 0.0001, indicating strong evidence against the null hypothesis.

Therefore, we can conclude that there is a significant difference in crop yield among the three fertilizer brands, and further analysis (such as post-hoc tests) can be conducted to determine which brands are significantly different from each other.


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