Now you are studying a habitat in west Texas which contains 6 different species

1. Use a one-factor ANOVA to test the null hypothesis that population abundance is equal among these 6 species. Be sure to interpret your results and explain your reasons for declaring them significant or not significant.

The obtained output is given below,

Oneway

The above output indicates that the p-value of the test is 0.014 which is smaller than the significance level 0.05 thus we are rejecting the null hypothesis concluding that the result in significant i.e. the group means are not equal.

2. Use either Bartlett’s test or Levene’s test to determine whether your groups have equal variances.

The obtained output is given below,

The p-value of the test is small. A p-value of 0.020 is indicating that the result is significant i.e. there is sufficient evidence that the groups do not have equal variances.

3. After determining that they do not, use a data transformation to help make your data obey the assumption of equality of variances. You can use any data transformation you would like, although you may have to try several to find one that works!

I used a natural logarithmic data transformation by taking natural log of the dependent variable Population. The result indicated that the groups have equal variances now.

4. Redo your one-factor ANOVA analysis, this time using the transformed data. Again, be sure to interpret your results and explain your reasons for declaring them significant or not significant. Did the results differ between your analysis here and your analysis in part A?

The obtained output is given below,

Oneway

Here also the p-value is small so we are rejecting the null hypothesis thus the conclusion in this part and in part A are exactly same.