Week 4 Assignment 1. For each correlation coefficient below, calculate what proportion of variance is shared by the two correlated variables: a. b. c. d. 2. r = 0.25 r = 0.33 r = 0.90 r = 0.14 For each coefficient of determination below, calculate the value of the correlation coefficient: r2 = 0.54 r2 = 0.13 r2 = 0.29 r2 = 0.07 a. b. c. d. 3. Answer: (0.25)(0.25) = 0.0625 Answer: (0.33)(.033) = 0.1089 Answer: (0.90)(0.90) = 0.81 Answer: (0.14)(0.14) = 0.0196 Answer: sqrt (0.54) = 0.735 Answer: sqrt (0.13) = 0.361 Answer: sqrt (0.29) = 0.539 Answer: sqrt (0.07) = 0.265 Suppose a researcher regressed surgical patients' length of stay (dependent variable) in the hospital on a scale of functional ability measured 24 hours after surgery. Given the following, solve for the value of the intercept constant and write out the full regression equation: Mean length of stay = 6.5days; mean score on scale = 33; slope = -0.10 Answer: solve for 'a' Formula: *y = a+bx Y=6.5 X= 33 B = -0.10 6.5 = a + (-0.10) (33) = 6.5 6.5 = a - (-3.3) 6.5 + 3.3 = 9.8 Regression: y= 9.8 - (-0.10) (33) * information needed for num. 4 4. Using the regression equation calculated in Exercise 3, compute the predicted value of Y (length of hospital stay) for patients with the following functional ability scores: a. b. c. d. 5. X = 42 X = 68 X = 23 X = 10 Answer: Y = 9.8 - (-0.10)(42) = 5.6 Answer: Y = 9.8 - (-0.10)(68) = 3 Answer: Y = 9.8 - (-0.10)(23) = 7.5 Answer: Y = 9.8 - (-0.10)(10) = 7.8 Use the regression equation below for predicting graduate GPA for the three presented cases. Y = -1.636 + 0.793(undergrad GPA) + 0.004(GREverbal) - 0.0009(GREquant) +0.009(Motivation) Subject 1 2 3 undergrad GPA 2.9 3.2 3.4 GREverbal GREquant Motivation 540 55 590 65 550 70 560 550 600 Answers: Subject 1 Y = -1.636 + 0.793(2.9) + 0.004(560) - 0.0009(540) + 0.009(55) = 2.9127 Subject 2 Y = -1.636 + 0.793(3.2) + 0.004(550) - 0.0009(590) + 0.009(65) = 3.1556 Subject 3 Y = -1.636 + 0.793(3.4) + 0.004(600) - 0.0009(550) + 0.009(70) = 3.5952 6. Using the following information for R2, k, and N, calculate the value of the F statistic for testing the overall regression equation and determine whether F is statistically significant at the 0.05 level: a. Answer: F = 0.13(120-1-5)/(1-0.53)5 = 6.31 6.31> 2.29 Sig. We can reject Null b. R2 = 0.53, k = 5, N = 30 Answer: F = 0.53(30-1-5)/(1-0.53)5 = 5.41 5.41> 2.62 Sig. We can reject Null c. R2 = 0.28, k = 4, N = 64 Answer: F = 0.28(64-1-4)/(1- 0.28) 4 = 5.74 5.74> 2.52 Sig. We can reject Null d. 7. R2 = 0.13, k = 5, N = 120 R2 = 0.14, k = 4, N = 64 Answer: F = 0.14(64-1-4)/(1- 0.14) 4 = 2.40 2.40< 2.52 Non sig. We Accept Null According to the University of Chicago, as men age, their cholesterol level goes up. A new drug (XAB) is being tested to determine if it can lower cholesterol in aging males and at what dose. The data for the first test subject is below: Dose (mg) Cholesterol level (mg/dL) 2 310 3 124 5 201 6 110 8 52 10 20 a. Plot the data and include a regression line in StatCrunch. Copy and paste your graph into your Word document for full credit. b. What is the correlation coefficient r and what does it mean in this case? Answer: R (correlation coefficient) = -0.85274561, close to -1 moderately strong negative correlation, which concurs with scatterplot. c. What is the coefficient of determination and what does it mean in this case? Answer: 0.72717508, 73% of one variation is explained by the other variable. d. Is there a statistically significant correlation between dose and cholesterol level in this case? Answer: P-value 0.03 < 0.05 non sig correlations. e. What is the predicted cholesterol level for a person taking a dose of 4 mg? What about if they are not taking the drug at all (0 mg)? Answer: What is the predicted cholesterol level for a person taking a dose of 4 mg? What about if they are not taking the drug at all (0 mg)? *Formula Y = a + bx Y = (rounded 305.75) 306 + (-30) x Y= (rounded 305.75) 306 + (-30) (4 mg) = 186 mg/dl Y= (rounded 305.75) 306 + (-30) (0 mg) = 306 mg/dl