Math 17, Section 2 Spring 2011 Lab 9

Name _____ SOLUTIONS ____ Math 17, Section 2 Spring 2011 Lab 9 Goal: To gain experience with an ANOVA test, and use Rcmdr to run ANOVA tests as well as scatterplots. Part 1 – ANOVA A pharmaceutical company tested three formulations of a pain relief medicine for migraine headache sufferers. For the experiment, 27 volunteers were selected and 9 were randomly assigned to one of three drug formulations. The subjects were instructed to take the drug during their next migraine headache episode and to report their pain on a scale of 1 = no pain to 10 = extreme pain 30 minutes after taking the drug. A: 4 5 4 3 2 4 3 4 4 s
 3.6667 0.8660 B: 6 8 4 5 4 6 5 8 6 5.7778 C: 6 7 6 6 7 5 6 5 5 5.8889 0.7817 Overall (all data pooled together) 5.1111 1.4814 1.4763 Use Rcmdr to compute the overall sample average and standard deviation, as well as the average and standard deviation for each of the three drugs. You can import the data into Rcmdr using the Ch28_Analgesics.xls file on the course CMS website (see the Lab 9 assignment). Q1] State the null and alternative hypotheses. Perform an ANOVA by Hand. Just to see where the number in our computer outcome come from, we’ll do the ANOVA by hand and using the computer and compare the two. We need to fill in the table below. Analysis of Variance Sum of Source DF Squares Treatment Mean Square 2 28.2223 14.1111 Error 24 28.4444 1.1852 Total 26 56.6667 F-ratio 11.91 P-value < 0.01 We have a total of 27 observations, so the total degrees of freedom is n – 1 = 26. There are 2 degrees of freedom for DRUG (k – 1) and 24 degrees of freedom for error. Total Sum of Squares 
 
 Drug Pain Error Sum of Squares 
 
  A 4 
 
  4  5.1111  1.2345 A 5 0.0123 1.7778 A 4 1.2346 0.1111 A 3 4.4568 0.4444 A 2 9.6790 2.7778 A 4 1.2346 0.1111 A 3 4.4568 0.4444 A 4 1.2346 0.1111 A 4 1.2346 0.1111 B 6 0.7901 0.0494 B 8 8.3457 4.9383 B 4 1.2346 3.1605 B 5 0.0123 0.6049 B 4 1.2346 3.1605 B 6 0.7901 0.0494 B 5 0.0123 0.6049 B 8 8.3457 4.9383 B 6 0.7901 0.0494 C 6 0.7901 0.0123 C 7 3.5679 1.2346 C 6 0.7901 0.0123 C 6 0.7901 0.0123 C 7 3.5679 1.2346 C 5 0.0123 0.7901 C 6 0.7901 0.0123 C 5 0.0123 0.7901 C 5 0.0123 0.7901 
 
  4  3.6667  0.1111 Compute the total sum of squares by taking each data point, subtracting off the overall mean, squaring the difference, and summing them all up. 56.6667 To get the error sum of squares, do the same thing, but instead subtract off the mean of the group each particular data point was in. 28.4444 The treatment sum of squares is the total sum of squares minus the error sum of squares. 28.2223 Fill in these quantities on the ANOVA table on the previous page. Compute the mean square treatment (MSTr) by taking the treatment sum of squares and dividing by its degrees of freedom. Computer the mean square error (MSE) by taking the error sum of squares and dividing by its degrees of freedom. Compute the F-statistic as the ratio of the two above quantities, and fill them in on the ANOVA table above. P-values can be computed by hand using an F-table. To use it, note the degrees of freedom for the numerator (MSTr) and the denominator (MSE). Here, we have 2 numerator and 24 denominator degrees of freedom. Go to the column corresponding to 2, and the row corresponding to 24. We have an F-value of 5.61. Note that the table says α = 0.01 in the upper left corner. If our particular F-statistic is larger than 5.61, then it has a p-value < 0.01. Q2] Based on your F-statistic and p-value, what do you conclude? I conclude that we have evidence to reject the null hypothesis, and conclude that at least one of the means is different. Further testing could be done using multiple comparisons. An ANOVA using Rcmdr • Import the Ch28_Analgesics.xls file into R commander. • Go to Statistics
Means
One-way ANOVA • Select DRUG for the groups and PAIN for the response • Click OK Q3] Does the ANOVA table from R commander match the one you made by hand? It should! Yes it does! > summary(AnovaModel.1) Df Sum Sq Mean Sq F value Pr(>F) Drug 2 28.222 14.1111 11.906 0.0002559 *** Residuals 24 28.444 1.1852 --Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Q4] Do the assumptions needed for ANOVA seem to be satisfied? Not really. There’s an outlier, and some of the boxes look very strange due to skewness. Part 2 – Scatterplots In this part of the lab, we’ll work with a scatterplot. Which mountain bike should we buy? Is there a relationship between the weight of a mountain bike and its price? A lighter bike is often preferred but do lighter bikes tend to be more expensive? The following table gives data on price and weight for 12 brands. Brand and Model Trek VRX200 Cannondale Super V400 GT XCR-4000 Specialized FSR Trek 6500 Specialized Rockhopper Haro Escape 7.1 Giant Yukon SE Mongoose SX 6.5 Diamondback Sorrento Motiv Rockbridge Huffy Anorak 36789 Price ($) 1000 1100 940 1100 700 600 440 450 550 340 180 140 Weight(lbs) 32 31 34 30 29 28 29 29 30 33 34 37 We are interested in whether and how weight affects price. Q5] Which variable is the logical choice for the explanatory variable? Weight should be the explanatory variable. Which variable is the logical choice for the response variable? Price is the logical choice for the response variable. Q6] Create a scatterplot using Rcmdr. To do this, import the bike data file from the CMS page. The file is Lab9_Bikedata.xls. Then, go to Graphs
Scatterplot. Select the x- and yvariables, and uncheck all the option checkboxes. Click on OK. Does the relationship look linear? No it does not. In what way does it deviate from linearity? It seems almost arced in shape. Part 3 –Correlation In this part of the lab, we’ll work with correlation. Suppose you want to study how well eyesight predicts your reaction time. You measure 10 people, measuring their eyesight in diopters and their reaction time in seconds. The data are shown below: Subject 1 2 3 4 5 6 7 8 9 10 Eyesight 0.20 1.30 2.20 2.70 3.30 4.80 5.10 6.40 8.10 8.21 Reaction Time 0.11 0.16 0.12 0.23 0.27 0.28 0.31 0.36 0.32 0.42 These data may be found on the CMS site as Lab9_Eyesightdata.xls. Q7] Which variable is the explanatory variable? Which is the response variable? Eyesight is the explanatory variable and reaction time is the response. Q8] Use Rcmdr to draw a scatterplot of the data is given above. Do you think r will be close to –1, 0, or 1? I think it will be close to 1. Q9] Describe the direction, form, and strength of the association between eyesight and reaction time? We have a positive, linear association that is quite strong. Q10] Calculate and interpret r? Go to Statistics
Summaries
Correlation Matrix. Select both variables. You can hold down the shift key or ctrl key to select multiple variables. Leave the “Type of Correlations” button set on “Pearson product-moment.” > cor(Eyesight[,c("Eyesight","ReactionTime")], use="complete.obs") Eyesight ReactionTime Eyesight 1.000000 0.922299 ReactionTime 0.922299 1.000000 We have r = 0.92, indicating a strong, positive linear association. Q11] Suppose we have an eleventh data point with an observation of 7.1 diopters for eyesight and a reaction time 0.15 seconds. What is the correlation (r) with this new observation included? You can click the “Edit data set” box in Rcmdr and type in this new observation. Then you can re-run the scatterplot and correlation computations. This should illustrate the impact of outliers on the correlation. We have r = 0.73. The outlier changed things quite a bit! > cor(Eyesight[,c("Eyesight","ReactionTime")], use="complete.obs") Eyesight ReactionTime Eyesight 1.0000000 0.7316096 ReactionTime 0.7316096 1.0000000