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

							Mea n
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. Which antipyretics are most effective at reducing fever?
   1. Answer your research question using the means in the Data Table and the ANOVA results.

F: 4.78

P Value = 0.298

One- way ANOVA is Body temperature verses Treatment

Treatment = 2, 4.079, 2.039, 7.11, 0.007

Total = 17, 8.382

P Value= The P Value is below 0.05 and with this we reject the H_o and can say there is at least one treatment that has a different effect than the others

Individual CL = 97.97%

Error= 4.303, 02.87

• Which treatment(s) would you recommend to reduce a fever for this population? The treatment I would recommend is acetaminophen. This is because the mean effects of aspirin and ibuprofen are insignificant; acetaminophen is significantly different from the other two groups.
  • 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)? The Tukey’s HSD method can be used to compare each treatment group to the other two at a time with out inflating the type 1 error (a)
  • Why is it important to make sure you do not increase the type I error? If we increase the type 1 error, we increase the chance of rejecting a true null hypothesis. Doing this can lead to an incorrect hypothesis.

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

• Evidence of nonrandom differences between group means occurs when the variance between

groups is _different than         the variance within groups.

• Why are scatterplots helpful when investigating the association between quantitative variables? Scatter plots are important when investigating the association between quantitative variables because it helps to be able to determine the relationship between two variables. From this we can observe if there is a negative or positive skew. We can also find out the R value which will let us know the direction and power of a linear relationship between the two variables.

• The Pearson correlation statistic, r, is always greater than or equal to     -1       and less than or

equal to _1

. Absolutely no association is present when r =     . Between r = -0.56 and r =

+0.46, the stronger correlation is _-0.56         .

• A Pearson correlation statistic is only valid when the relationship between the two quantitative

(continuous) variables is linear or above 0.07                           .

• Explain why it is true that the slope of a line is related to the Pearson correlation statistic, r.

It is a true that the slope of a line is related to the Pearson correlation statistic because it shows us that the data in the Pearson Correlation coefficient is +1. All units in a variable do not have an increase, however, it means there is no variation between data and the line.

• Create a scatterplot to investigate the association between the amount of fluoride in domestic water (ppm) and the number of dental caries in permanent teeth per 100 children for 21 cities. The data are below.
• Create the scatterplot

• Describe the association you see in your scatterplot.

Looking at the scatterplot we can see that the relationship between fluoride and caries is not linear.

c.     The value of r is -0.86, -0.36, 0.36, or 0.86?

The value of R is -0.86

CityID	FLUORIDE (ppm)	CARIES
1	1.9	236
2	2.6	246
3	1.8	252
4	1.2	258
5	1.2	281
6	1.2	303
7	1.3	323
8	0.9	343
9	0.6	412
10	0.5	444
11	0.4	556
12	0.3	652
13	0.0	673
14	0.2	703
15	0.1	706
16	0.0	722
17	0.2	733
18	0.1	772
19	0.0	810
20	0.1	823
21	0.1	1037

Data from Dean, H.T., Arnold, F. A., Jr., & Elvove, E. (1942). Domestic water and dental caries. Public Health Reports, 57, 1155-1179.

1. State the research question that this ANOVA answers.

Which medication treatment is the most effective in reducing the temperature in patients with flu diagnoses?

• Answer your research question using the means in the Data Table and the ANOVA results.

1.     H0: μ1 = μ2 = μ3; Ha: at least one μ is different where μ1 = true mean effect of aspirin, μ2 = true mean effect of ibuprofen, and μ3 = true mean effect of Acetaminophen

• Since p-value=0.007<0.05, we reject H0 and conclude that at least one treatment has different effects. For finding the treatments which are responsible for this rejection, we perform Tukey’s HSD test.

• Treatment N Mean Grouping Aspirin 5 1.2600 A Ibuprofen 6 1.2017 A Acetaminophen 7 0.2529 B
• Tukey 95% Simultaneous Confidence Intervals

All Pairwise Comparisons among Levels of Treatment

  Individual confidence level = 97.97%

• The simultaneous confidence intervals for treatments differences are: i. Aspirin-Acetaminophen: (0.1933, 1.8210)
  • Ibuprofen-Acetaminophen: (0.1755, 1.7221)
  • Ibuprofen-Aspirin: (-0.9000, 0.7833)

Answers 1-5 under section B of question 1

1. Which treatment(s) would you recommend to reduce a fever for this population?

CI for aspirin-acetaminophen and Ibuprofen-acetaminophen contain only positive values and CI for Ibuprofen and aspirin have 0. The difference in means shows significant in Ibuprofen and Aspirin versus Acetaminophen. I recommend Acetaminophen to reduce fever. I am on the fence with this answer. The alternative I came out with was                                                           Group 1: Asprin and Group 2: Ibuprofen

• 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)?

ANOVA and post-hoc rests or multiple comparison tests can be used, but in this test, the Tukey method was used

• Why is it important to make sure you do not increase the type I error?

Increasing the type I error would increase the chance that you reject a true null hypothesis. In this case (if the null hypothesis were true) that would increase the chance that we reject the true hypothesis that the means do not differ, which would lead to not recommending a treatment that could help reduce the body temperature.

• Why are scatterplots helpful when investigating the association between quantitative variables? They help understand the relationship behavior between the two variables, and they can help us see the form, direction, strength, and outliers

• The Pearson correlation statistic, r, is always greater than or equal to _-1      and less than or equal to  1_. Absolutely no association is present when r = -1                               . Between r = -0.56 and r =

+0.46, the stronger correlation is _-0.56.       .

• A Pearson correlation statistic is only valid when the relationship between the two quantitative (continuous) variables is _linear                   .

• Explain why it is true that the slope of a line is related to the Pearson correlation statistic, r. Slope of regression line shows how the response variable changes when the predictor variable changes. The correlation coefficient represents the linear relation between two variables, so if two variables have a positive linear relationship, then the slope line is positive. Two variables with a negative correlation

have a slope line that is negative. A slope regression line is related to Pearson correlation statistic. Or in simple terms. If the slope of a line is positive to the Pearson correlation statistic, r is positive. If the slope of a line is negative, to the Pearson correlation statistic, r is negative.