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Homework #9 Score____________ / 15 Name ______________ Additional HW Exercise #3.2 Suppose we are interested in predicting a male's right-hand grip strength from age. The following data is recorded for a random sample of males: Age (years) 15 17 19 11 16 22 17 25 12 14 25 23 Grip Strength (lbs.) 50 54 66 46 58 54 64 80 46 70 76 80 (a) Identify the dependent (response) variable and the independent (explanatory) variable for a regression analysis. The dependent (response) variable is Y = “grip strength”, and the independent (explanatory) variable is “age”. (b) Does the data appear to be observational or experimental? Since the ages look random, it appears that the data is observational. (c) Use SPSS to do calculations necessary for simple linear regression. First, enter the data into an SPSS data file, with the name age for the variable consisting of the ages, and with the name grip for the variable consisting of the grip strengths. Save this SPSS data file with the name grip_strength . Additional HW Exercise #3.2.-continued (d) Use the SPSS output to find each of the following: n = 12 x = 18 y = 62 r = + 0.770 SSxx = (n – 1)sx2 = (11)(4.824)2 = 255.98 SSyy = (n – 1)sy2 = (11)(12.534)2 = 1728.11 SSxy = r SSxx SSyy = (0.770) (255.98)(1728.11) = 512.13 (e) Use the SPSS output to find the equation of the least squares line. ^1 = 2 The least squares line can be written ^0 = ^y = 26 + 2x . 26 (f) Write a one-sentence interpretation of the slope in the least squares line. Grip strength appears to increase on average by about 2 pounds with each increase of one year in age. (g) Find the coefficient of determination, and write a one-sentence interpretation. From the SPSS output, we find r2 = 0.593 About 59.3% of the variation in grip strength is explained by age. (h) Find the estimated standard error of the regression. From the SPSS output, we find s = 8.390 which can also be calculated from parts (d)&(e) using SSyy –^ 1SSxy n–2 Additional HW Exercise #3.3 Suppose we are interested in predicting a subject’s reaction time to a particular stimulus from dosage of a certain drug.. The following data is recorded for a random sample of observations on the subject: Drug Dosage (grams) 4 4 6 6 8 8 10 10 Reaction Time (seconds) 7.5 6.8 4.0 4.4 3.9 3.1 1.4 1.7 (a) Identify the dependent (response) variable and the independent (explanatory) variable for a regression analysis. The dependent (response) variable is Y = “reaction time”, and the independent (explanatory) variable is “drug dosage”. (b) Does the data appear to be observational or experimental? Since the drug dosages do not look random, it appears that the data is experimental. (c) Use SPSS to do calculations necessary for simple linear regression. First, enter the data into an SPSS data file, with the name dosage for the variable consisting of the drug dosages, and with the name time for the variable consisting of the reaction times. Save this SPSS data file with the name reaction . Additional HW Exercise #3.3.-continued (d) Use the SPSS output to find each of the following: n= 8 x= 7 y = 4.1 r = – 0.963 SSxx = (n – 1)sx2 = (7)(2.390)2 = 39.98 SSyy = (n – 1)sy2 = (7)(2.1726)2 = 33.04 SSxy = r SSxx SSyy = (– 0.963) (39.98)(33.04) = – 35 (e) Use the SPSS output to find the equation of the least squares line. ^1 = – 0.875 The least squares line can be written ^0 = ^y = 10.225 – 0.875x . 10.225 (f) Write a one-sentence interpretation of the slope in the least squares line. Reaction time appears to decrease on average by about 0.875 seconds with each increase of one gram in drug dosage. (g) Find the coefficient of determination, and write a one-sentence interpretation. From the SPSS output, we find r2 = 0.927 which is the square of r from part (d). About 92.7% of the variation in reaction time is explained by drug dosage. (h) Find the estimated standard error of the regression. From the SPSS output, we find s = 0.6344 which can also be calculated from parts (d)&(e) using SSyy –^ 1SSxy n–2