Wilcoxon Rank-Sum Test. In Exercises 5–8, use the Wilcoxon rank-sum test.
7. Clinical Trials of Lipitor The sample data below are changes in LDL cholesterol levels in clinical trials of Lipitor (atorvastatin). It was claimed that Lipitor had an effect on LDL cholesterol. (The data are based on results given in a Parke-Davis memo from David G. Orloff, M.D., the medical team leader for clinical trials of Lipitor. Pfizer declined to provide the author with the original data values.) Negative values represent decreases in LDL cholesterol. Use a 0.05 significance level to test the claim that for those treated with 20 mg of Lipitor and those treated with 80 mg of Lipitor, changes in LDL cholesterol have the same
Want to see the full answer?
Check out a sample textbook solutionChapter 13 Solutions
MyLab Statistics with Pearson eText -- Standalone Access Card -- for Elementary Statistics
- Rental revenue (in millions of dollars) 4. Can movie rental revenue be predicted? A movie studio wishes to determine the relationship between the revenue from rental of comedies on streaming services and the revenue generated from the theatrical release of such movies. The studio has the following bivariate data from a sample of fifteen comedies released over the past five years. These data give the revenue x from theatrical release (in millions of dollars) and the revenue y from streaming service rentals (in millions of dollars) for each of the fifteen movies. Also shown are the scatter plot and the least- squares regression line for the data. The equation for this line is y=3.30 +0.15x. Theater revenue, X Rental revenue, y (in millions of dollars) (in millions of dollars) 61.1 16.5 44.8 6.1 个y 18- 36.8 12.1 16 61.4 9.6 14- 13.3 9.7 12- 25.8 9.0 10- ャ★ 20.2 4.6 X. 8 26.3 7.7 7.3 2.6 48.5 15.9 27.7 11.7 20 30 70 28.3 2.8 Theater revenue 66.3 10.1 (in millions of dollars) 29.9 6.3 22arrow_forwardYou perform an ANOVA to test that there are no main effects of factor A, no main effects of factor B, and no interaction between factors A and B. Some of the results are presented in the following ANOVA table. Source Between treatments Factor A Factor B A X B interaction Within treatments Total ANOVA Table SS 6.5667 0.0667 12.1667 df 5 29 MS 4.0333 1.2334 F 17.29 5.29 Work through the following steps to complete the preceding ANOVA table. 1. The main effect for factor A evaluates the mean differences between the levels of factor A. The main effect for factor B evaluates the mean differences between the levels of factor B. Select the correct values for the sums of squares for factors A and B in the ANOVA table. 2. Select the correct value for the within-treatments sum of squares in the ANOVA table. 3. Select the correct degrees of freedom for all the sums of squares in the ANOVA table. 4. Select the correct values for the mean square due to A X B interaction, the within treatments mean…arrow_forwardChest Injury Rating Car Driver Passenger Car Driver Passenger 1 42 35 10 36 37 2 42 35 11 36 37 3 34 45 12 43 58 4 34 45 13 40 42 5 45 45 14 43 58 6 40 42 15 37 41 7 42 46 16 37 41 8 43 58 17 44 57 9 45 43 18 42 42 Generate test results (α = .01) using data analysis (EXCEL).arrow_forward
- 9arrow_forwardData in the following example are selected from a larger experiment on the use of drugs in the treatment of leprosy (Snedecor and Cochran 1967). Variables in the study are as follows: Drug two antibiotics (A and D) and a control (F) Pre-Treatment (X) a pretreatment score of leprosy bacilli Post-Treatment (Y) a posttreatment score of leprosy bacilli Ten patients are selected for each drug treatment, and six sites on each patient are measured for leprosy bacilli. The covariate (a pretreatment score) is included in the model for increased precision in determining the effect of drug treatments on the posttreatment count of bacilli. An Control Earrow_forwardplease show diagram of p valuearrow_forward
- Treating Chronic Fatigue Syndrome: Patients with chronic fatigue syndrome were tested, and then retested after being treated with fludrocortisones. Listed below are the changes in fatigue after the treatment (based on data from “The Relationship Between Neurally Mediated Hypotension and the Chronic Fatigue Syndrome” by Bou-Holaigah, Rowe, Kan, and Calkins, Journal of the American Medical Association, 274, No. 12). A standard scale from -7 to +7 was used, with positive values representing improvements. Use a 0.05 significance level to test the claim that the mean change is not zero. Does the treatment appear to be effective? 6 5 0 5 6 7 3 3 2 6 5 5 0 6 3 4 3 7 0 4 4 Hypotheses: __________________________________________________________________________________ P-value: ____________________ Conclusion on null hypothesis: Conclusion on claim and answer to the question:arrow_forwarded weight (in kg) and total serum cholesterol (TSC in mg/dl) data for five men (n = 5). They tested for association between weight and TSC at 5% level of significance. weight, X TSC, Y X*X Y*Y X*Y 37 2.9 1369 8.41 107.3 50 0.7 2500 0.49 35.0 67 3.3 4489 10.89 221.1 70 4.2 4900 17.64 294.0 57 3.8 3249 14.44 216.6 SUM: 281 14.9 16507 51.87 874.0 What are the Null and Alternate hypotheses for this test? Clearly label your answers as HỌ or H1. (Do not use a numerical format. Both of your statements must be in sentence form, and say something about the weightarrow_forwardLemons and Car Crashes Listed below are annual data for various years. The data are weights (metric tons) of lemons imported from Mexico and U.S. car crash fatality rates per 100,000 population [based on data from “The Trouble with QSAR (or How I Learned to Stop Worrying and Embrace Fallacy)” by Stephen Johnson, Journal of Chemical Information and Modeling, Vol. 48, No. 1]. Lemon Imports 230 265 358 480 530 Crash Fatality Rate 15.9 15.7 15.4 15.3 14.9 Question ---- > Find the critical values of r using α = 0.05 and use test statistic r = -0.959 to determine whether there is sufficient evidence to support a claim of a linear correlation between the two variables. Group of answer choices r = ± 0.811, There is enough evidence to Support the Claim r = ± 0.878, There NOT is enough evidence to Support the Claim r = ± 0.811, There NOT is enough evidence to Support the Claim r = ± 0.878, There is enough evidence to Support the Claimarrow_forward
- Lemons and Car Crashes Listed below are annual data for various years. The data are weights (metric tons) of lemons imported from Mexico and U.S. car crash fatality rates per 100,000 population [based on data from “The Trouble with QSAR (or How I Learned to Stop Worrying and Embrace Fallacy)” by Stephen Johnson, Journal of Chemical Information and Modeling, Vol. 48, No. 1]. Lemon Imports 230 265 358 480 530 Crash Fatality Rate 15.9 15.7 15.4 15.3 14.9 Question ---- > Find the critical values of r using α = 0.05 and write Ho and H1 Group of answer choices r = ± 0.878 Ho : ρ = 0 H1: ρ < 0 r = ± 0.811 Ho : ρ = 0 H1: ρ > 0 r = ± 0.878 Ho : ρ = 0 H1: ρ ≠ 0 r = 0.878 Ho : ρ > 0 H1: ρ ≠ 0arrow_forwardUsing the data below: How do we create a bivariate analysis?arrow_forwardThe data show the population (in thousands) for a recent year of a sample of aities in South Carolina. 24 19 19 18 62 18 21 24 20 11 22 33 93 13 21 29 13 12 108 45 20 46 108 39 30 51 Send data to Excel Part 1 of 8 The data value 21 corresponds to the 41" percentile. Part: 1/8 Part 2 of 8 The data value corresponds to the 71" percentile.arrow_forward
- Linear Algebra: A Modern IntroductionAlgebraISBN:9781285463247Author:David PoolePublisher:Cengage LearningGlencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw Hill