Question

1. A trial evaluated the fever-inducing effects of three substances. Study subjects were adults seen in an emergency room with diagnoses of the flu and body temperatures between 100.0 and 100.9ºF. The three treatments (aspirin, ibuprofen and acetaminophen) were assigned randomly to study subjects. Body temperatures were reevaluated 2 hours after administration of treatments. The below table lists the data.

Data Table: Decreases in body temperature (degrees Fahrenheit) for each patient                            

							Mean
Group 1 (aspirin)	0.95	1.48	1.33	1.28			1.26
Group 2 (ibuprofen)	0.39	0.44	1.31	2.48	1.39		1.20
Group 3 (acetaminophen)	0.19	1.02	0.07	0.01	0.62	-0.39	0.25

The ANOVA table that corresponds to this data is below.

1. State the research question that this ANOVA answers.
2. Answer your research question using the means in the Data Table and the ANOVA results.
3. Which treatment(s) would you recommend to reduce a fever for this population?
4. What type of tests could you conduct that would allow you to compare each treatment group to the other (2 at a time) without inflating the type I error (α)?
5. Why is it important to make sure you do not increase the type I error?

ANOVA Table:

Fev_red	Sum of Squares	df	Mean Square	F	Sig.
Between groups	3.426	2	1.713	4.777	0.030
Within groups	4.303	12	0.359
Total	7.729	14

Answer 1

a. Effect of three treatments (aspirin, ibuprofen and acetaminophen) are equal or not.

b. ANOVA table:

One-way ANOVA: body temperature versus Group

Source DF     SS     MS     F      P
Group    2 3.426 1.713 4.78 0.030
Error   12 4.303 0.359
Total   14 7.729

Since p-value<0.05 so there is enough evidence to conclude that three treatments are not equally effective.

c. Since Mean temperature decreased for aspirin is maximum so we recommend aspirin to reduce a fever for this population. d. We perform multiple comparison methods to compare each treatment group to the other (2 at a time) without inflating the type I error (α). There are several multiple comparison methods. For this problem we generally use Tukey's post hoc test.

Minitab output:

Grouping Information Using Tukey Method

Group N    Mean Grouping
1      4 1.2600 A
2      5 1.2020 A
3      6 0.2533 A

Means that do not share a letter are significantly different.

Tukey 95% Simultaneous Confidence Intervals
All Pairwise Comparisons among Levels of Group

Individual confidence level = 97.94%

Group = 1 subtracted from:

Group    Lower   Center   Upper     +---------+---------+---------+---------
2      -1.1288 -0.0580 1.0128              (---------*----------)
3      -2.0371 -1.0067 0.0238     (---------*---------)
                                    +---------+---------+---------+---------
                                 -2.0      -1.0       0.0       1.0

Group = 2 subtracted from:

Group    Lower   Center   Upper     +---------+---------+---------+---------
3      -1.9153 -0.9487 0.0180      (---------*--------)
                                    +---------+---------+---------+---------

                                 -2.0      -1.0       0.0       1.0

From the above confidence intervals we see that all three treatments are equally effective.

e. When we perform a large number of statistical tests, P-values of some of them is less than our critical value or level of significance and we reject corresponding null hypotheses. But These decisions may be wrong; the corresponding null hypotheses might be true, and the significant results might be due to chance. This is occurred because overall Type I error is increased. So we need to control the overall type I error and this is done by using some post hoc tests.