A research laboratory was developing a new compound for therelief of severe cases of hay fever. In an experiment with 36volunteers, the amounts of the two active ingredients (A & B)in the compound were varied at three levels each. Randomization wasused in assigning four volunteers to each of the nine treatments.The data on hours of relief can be found in the following .csvfile: Fever.csv
State the Null and Alternate Hypothesis for conducting one-wayANOVA for both the variables ‘A’ and ‘B’ individually.
1.2) Perform one-way ANOVA for variable ‘A’ with respect to thevariable ‘Relief’. State whether the Null Hypothesis is accepted orrejected based on the ANOVA results.
1.3) Perform one-way ANOVA for variable ‘B’ with respect to thevariable ‘Relief’. State whether the Null Hypothesis is accepted orrejected based on the ANOVA results.
1.4) Analyse the effects of one variable on another with thehelp of an interaction plot.
What is an interaction between two treatments?
[hint: use the ‘pointplot’ function from the ‘seaborn’function]
1.5) Perform a two-way ANOVA based on the different ingredients(variable ‘A’ & ‘B’) with the variable 'Relief' and state yourresults.
1.6) Mention the business implications of performing ANOVA forthis particular case study.
| A | B | Volunteer | Relief |
| 1 | 1 | 1 | 2.4 |
| 1 | 1 | 2 | 2.7 |
| 1 | 1 | 3 | 2.3 |
| 1 | 1 | 4 | 2.5 |
| 1 | 2 | 1 | 4.6 |
| 1 | 2 | 2 | 4.2 |
| 1 | 2 | 3 | 4.9 |
| 1 | 2 | 4 | 4.7 |
| 1 | 3 | 1 | 4.8 |
| 1 | 3 | 2 | 4.5 |
| 1 | 3 | 3 | 4.4 |
| 1 | 3 | 4 | 4.6 |
| 2 | 1 | 1 | 5.8 |
| 2 | 1 | 2 | 5.2 |
| 2 | 1 | 3 | 5.5 |
| 2 | 1 | 4 | 5.3 |
| 2 | 2 | 1 | 8.9 |
| 2 | 2 | 2 | 9.1 |
| 2 | 2 | 3 | 8.7 |
| 2 | 2 | 4 | 9 |
| 2 | 3 | 1 | 9.1 |
| 2 | 3 | 2 | 9.3 |
| 2 | 3 | 3 | 8.7 |
| 2 | 3 | 4 | 9.4 |
| 3 | 1 | 1 | 6.1 |
| 3 | 1 | 2 | 5.7 |
| 3 | 1 | 3 | 5.9 |
| 3 | 1 | 4 | 6.2 |
| 3 | 2 | 1 | 9.9 |
| 3 | 2 | 2 | 10.5 |
| 3 | 2 | 3 | 10.6 |
| 3 | 2 | 4 | 10.1 |
| 3 | 3 | 1 | 13.5 |
| 3 | 3 | 2 | 13 |
| 3 | 3 | 3 | 13.3 |
| 3 | 3 | 4 | 13.2 |
