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Correlation and Regression Example # 1 For the following exercise, complete these steps. a. Draw the scatter plot for the variables. b. Compute the value of the correlation coefficient. c. State the hypotheses. d. Test the significance of the correlation coefficient at = 0.05, using Table I. e. Give a brief explanation of the type of relationship. A researcher wishes to determine if a person’s age is related to the number of hours he or she exercises per week. The data for the sample are shown below. a. Draw the scatter plot for the variables. Age x 18 26 32 38 52 59 Hours y 10 5 2 3 1.5 1 10 8 Hours 6 4 2 Age 0 10 20 40 30 50 60 70 b. Compute the value of the correlation coefficient. Age x 18 26 32 38 52 59 Hours y 10 5 2 3 1.5 1 x = 225 y = 22.5 x 2 = 9653 y 2 = 141.25 xy = 625 n=6 x = 225 x 2 = 9653 xy = 625 y = 22.5 y 2 = 141.2 n = 6 n xy – x y r= n x 2 – x 2 n y 2 y 2 x = 225 x 2 = 9653 xy = 625 y = 22.5 y 2 = 141.2 n = 6 6 625 – 225 22.5 r= 6 9653 – 225 2 6 141.25 – 22.5 2 r = – 0.832 c. State the hypotheses. H0 : = 0 and H1 : 0 Age x 18 26 32 38 52 59 Hours y 10 5 2 3 1.5 1 d. Test the significance of the correlation coefficient at = 0.05, using Table I. Age x 18 26 32 38 52 59 Hours y 10 5 2 3 1.5 1 H0 : = 0 and H1 : 0 n=6 d.f . = 4 r = – 0.832 C.V. = ± 0.811 Decision: Reject H0 . e. Give a brief explanation of the type of relationship. Age x 18 26 32 38 52 59 Hours y 10 5 2 3 1.5 1 H0 : = 0 and H1 : 0 n=6 d.f . = 4 r = – 0.832 There is a significant linear relationship between a person’s age and the number of hours he or she exercises per week. Decision: Reject H0 . Correlation and Regression Example # 2 For the following exercise, complete these steps. a. Draw the scatter plot for the variables. b. Compute the value of the correlation coefficient. c. State the hypotheses. d. Test the significance of the correlation coefficient at = 0.05, using Table I. e. Give a brief explanation of the type of relationship. The director of an alumni association for a small college wants to determine whether there is any type of relationship between the amount of an alumnus’s contribution (in dollars) and the years the alumnus has been out of school. The data are shown here. Years x 1 5 3 10 7 6 Contribution y 500 100 300 50 75 80 a. Draw the scatter plot for the variables. Years x 1 5 3 10 7 6 Contribution y 500 100 300 50 75 80 500 Contribution 400 300 200 100 0 2 4 6 8 10 20 30 Years b. Compute the value of the correlation coefficient. Years x 1 5 3 10 7 6 Contribution y 500 100 300 50 75 80 x = 32 y = 1105 x 2 = 220 y 2 = 364,525 xy = 3405 n=6 b. Compute the value of the correlation coefficient. x = 32 x 2 = 220 xy = 3405 y = 1105 y 2 = 364,52 n = 6 n xy – x y r= n x 2 – x 2 n y 2 y 2 b. Compute the value of the correlation coefficient. x = 32 x 2 = 220 xy = 3405 y = 1105 y 2 = 364,52 n = 6 6 3405 – 32 1105 r= 2 2 6 220 – 32 6 364,525 – 1105 r = – 0.883 c. State the hypotheses. Years x 1 5 3 10 7 6 Contribution y 500 100 300 50 75 80 H0 : = 0 and H1 : 0 d. Test the significance of the correlation coefficient at = 0.05, using Table I. Years x 1 5 3 10 7 6 Contribution y 500 100 300 50 75 80 H0 : = 0 and H1 : 0 n=6 d.f . = 4 r = – 0.883 C.V. = ± 0.811 Decision: Reject H0 . e. Give a brief explanation of the type of relationship. Years x 1 5 3 10 7 6 Contribution y 500 100 300 50 75 80 H0 : = 0 and H1 : 0 n=6 d.f . = 4 r = – 0.883 There is a significant linear relationship between a person’s age and his or her contribution. Decision: Reject H0 . Correlation and Regression Example # 3 For the following exercise, complete these steps. a. Draw the scatter plot for the variables. b. Compute the value of the correlation coefficient. c. State the hypotheses. d. Test the significance of the correlation coefficient at = 0.05, using Table I. e. Give a brief explanation of the type of relationship. A criminology student wishes to see if there is a relationship between the number of larceny crimes and the number of vandalism crimes on college campuses in Southwestern Pennsylvania. The data are shown. Is there a relationship between the two types of crimes? Number of larceny 24 6 16 64 10 25 35 crimes, x Number of 21 3 6 15 21 61 20 vandalism crimes y a. Draw the scatter plot for the variables. Number of larceny 24 6 16 64 10 25 35 crimes, x Number of 21 3 6 15 21 61 20 vandalism crimes y vandalism crimes 80 60 40 20 0 10 20 30 40 50 60 70 80 larceny crimes b. Compute the value of the correlation coefficient. Number of larceny 24 6 16 64 10 25 35 crimes, x Number of 21 3 6 15 21 61 20 vandalism crimes y x = 180 y = 147 x 2 = 6914 y 2 = 5273 xy = 4013 n=7 x = 180 x 2 = 6914 xy = 4013 y = 147 y 2 = 527 n = 7 n xy – x y r= n x 2 – x 2 n y 2 y 2 x = 180 x 2 = 6914 xy = 4013 y = 147 y 2 = 527 n = 7 6 4013 – 180 147 r= 2 2 6 6914 – 180 6 5273 – 147 r = 0.104 c. State the hypotheses. Number of larceny 24 6 16 64 10 25 35 crimes, x Number of 21 3 6 15 21 61 20 vandalism crimes y H0 : = 0 and H1 : 0 d. Test the significance of the correlation coefficient at = 0.05, using Table I. Number of larceny 24 6 16 64 10 25 35 crimes, x Number of 21 3 6 15 21 61 20 vandalism crimes y n = 7 d.f .= 5 r = 0.104 C.V. = ± 0.754 Decision: Do not reject H0 . e. Give a brief explanation of the type of relationship. Number of larceny 24 6 16 64 10 25 35 crimes, x Number of 21 3 6 15 21 61 20 vandalism crimes y n = 7 d.f .= 5 r = 0.104 There is not a significant linear relationship between the number of larceny crimes and the number of vandalism crimes. Decision: Do not reject H0 . Correlation and Regression Example # 4 For the following exercise, complete these steps. a. Draw the scatter plot for the variables. b. Compute the value of the correlation coefficient. c. State the hypotheses. d. Test the significance of the correlation coefficient at = 0.05, using Table I. e. Give a brief explanation of the type of relationship. The average daily temperature (in degrees Fahrenheit) and the corresponding average monthly precipitation (in inches) for the month of June are shown here for seven randomly selected cities in the United States. Determine if there is a relationship between the two variables. Average daily 86 81 83 89 80 74 64 temperature, x Average monthly 3.4 1.8 3.5 3.6 3.7 1.5 0.2 precipitation, y a. Draw the scatter plot for the variables. Average daily 86 81 83 89 80 74 64 temperature, x Average monthly 3.4 1.8 3.5 3.6 3.7 1.5 0.2 precipitation, y 5 4 Precipitation 3 2 1 0 60 70 80 90 100 Temperature b. Compute the value of the correlation coefficient. Average daily 86 81 83 89 80 74 64 temperature, x Average monthly 3.4 1.8 3.5 3.6 3.7 1.5 0.2 precipitation, y x = 557 y = 17.7 x 2 = 44,739 y 2 = 55.99 xy = 1468.9 n=7 x = 557 x 2 = 44,739 xy = 1468.9 y = 17.7 y 2 = 55.99 n = 7 n xy – x y r= n x 2 – x 2 n y 2 y 2 x = 557 x 2 = 44,739 xy = 1468.9 y = 17.7 y 2 = 55.99 n = 7 7(1468.9) – ( 557)(17.7) r= ( 44,739) – ( 557) 2 7( 55.99) – (17.7) 2 7 r = 0.883 c. State the hypotheses. Average daily 86 81 83 89 80 74 64 temperature, x Average monthly 3.4 1.8 3.5 3.6 3.7 1.5 0.2 precipitation, y H0 : = 0 and H1 : 0 d. Test the significance of the correlation coefficient at = 0.05, using Table I. Average daily 86 81 83 89 80 74 64 temperature, x Average monthly 3.4 1.8 3.5 3.6 3.7 1.5 0.2 precipitation, y H0 : = 0 and H1 : 0 n =7 d.f . = 5 r = 0.883 C.V. = ± 0.754 Decision: Reject H0 . e. Give a brief explanation of the type of relationship. Average daily 86 81 83 89 80 74 64 temperature, x Average monthly 3.4 1.8 3.5 3.6 3.7 1.5 0.2 precipitation, y H0 : = 0 and H1 : 0 n =7 d.f . = 5 r = 0.883 There is a significant linear relationship between temperature and precipitation. Decision: Reject H0 . Correlation and Regression Example # 5 Find the equation of the regression line and find the y value for the specified x value. Remember that no regression should be done when r is not significant. Ages and Exercise Age x 18 26 32 38 52 59 Hours y 10 5 2 3 1.5 1 Find y when x = 35 years. Ages and Exercise Age x 18 26 32 38 52 59 Hours y 10 5 2 3 1.5 1 a= y x xy x2 – n x2 – x 2 22.5 9653 – 225 625 a= 2 6 9653 – 225 Find y when x = 35 years. Ages and Exercise Age x 18 26 32 38 52 59 Hours y 10 5 2 3 1.5 1 a= x xy y x2 – n x2 – x 2 a = 10.499 Find y when x = 35 years. Ages and Exercise Age x 18 26 32 38 52 59 Hours y 10 5 2 3 1.5 1 n xy – x y b= n x2 – x 2 6 625 – 225 22.5 b= 2 6 9653 – 225 Find y when x = 35 years. Ages and Exercise Age x 18 26 32 38 52 59 Hours y 10 5 2 3 1.5 1 n xy – x y b= n x2 – x 2 b = – 0.18 Find y when x = 35 years. Ages and Exercise Age x 18 26 32 38 52 59 Hours y 10 5 2 3 1.5 1 a = 10.499 b = – 0.18 y = a + bx y = 10.499 – 0.18x y = 10.499 – 0.18(35) y = 4.199 hours Correlation and Regression Example # 6 Find the equation of the regression line and find the y value for the specified x value. Remember that no regression should be done when r is not significant. Years and Contribution Years x 1 5 3 10 7 6 Contribution y, $ 500 100 300 50 75 80 Find y when x = 4 years. Years and Contribution Years x 1 5 3 10 7 6 Contribution y, $ 500 100 300 50 75 80 a= y x xy x2 – n x2 – x 2 1105 220 – 32 3405 a= 2 6 220 – 32 1105 220 – 32 3405 a= 2 6 220 – 32 243,100 – 108,960 a= 1320 – 1024 134,140 a= 296 a = 453.176 Find y when x = 4 years. Years and Contribution Years x 1 5 3 10 7 6 Contribution y, $ 500 100 300 50 75 80 n xy – x y b= n x2 – x 2 6(3405) – (32)(1105) b= 2 6( 220) – (32) 6(3405) – (32)(1105) b= 6( 220) – (32) 2 20,430 – 35,360 b= 296 – 14,930 b= 296 b = – 50.439 Find y when x = 4 years. Years and Contribution Years x 1 5 3 10 7 6 Contribution y, $ 500 100 300 50 75 80 a = 453.176 b = – 50.439 y = a + bx y = 453.176 – 50.439x y = 453.176 – 50.439(4) y = $251.42 Correlation and Regression Example # 7 Find the equation of the regression line and find the y value when x = 70 ºF. Remember that no regression should be done when r is not significant. Temperatures ( in. F ) and precipitation (in.) Avg. daily temp. x 86 81 83 89 80 74 64 Avg. mo. Precip. y 3.4 1.8 3.5 3.6 3.7 1.5 0.2 x = 557 y = 17.7 x 2 = 44,739 xy = 1468.9 x = 557 y = 17.7 x 2 = 44,739 xy = 1468.9 a= x xy y x2 – n x 2 – x 2 (17.7)(44,739) – (557)(1468.9) a= 7(44,739) – (557)2 a = – 8.994 x = 557 y = 17.7 x 2 = 44,739 xy = 1468.9 n xy – x y b= n x 2 – x 2 b= 7(1468.9) – (557)(17.7) 7(44,739) – (557)2 b = 0.1448 x = 557 y = 17.7 a = – 8.994 x 2 = 44,739 xy = 1468.9 b = 0.1448 y = a+ bx y = – 8.994 + 0.1448x y = – 8.994+ 0.1448(70) y = 1.1 inches Correlation and Regression Coefficient of Determination and Standard Error of the Estimate Find the coefficients of determination and non- determination when r = 0.70 and explain the meaning of: r 2 = 0.49 49% of the variation of y is due to the variation of x. Find the coefficients of determination and non- determination when r = 0.70 and explain the meaning of: 1– r 2 = 0.51 51% of the variation of y is due to chance. Chapter 10 Correlation and Regression Section 10-5 Exercise #15 Correlation and Regression Example # 2 Coefficient of Determination and Standard Error of the Estimate Compute the standard error of the estimate. x = 225 y = 22.5 x = 9653 2 y 2 = 141.25 xy = 625 n=6 a = 10.499 b = – 0.18 2 y – a y – b xy s = est n –2 141.25 – 10.499(22.5) – ( – 0.18)(625) s = est 6– 2 141.25 – 10.499(22.5) – ( – 0.18)(625) sest = 6 2 sest = 4.380625 sest = 2.09 Correlation and Regression Example # 3 Coefficient of Determination and Standard Error of the Estimate Find the 90% prediction interval when x = 20 years. Age x 18 26 32 38 52 59 Hours y 10 5 2 3 1.5 1 x = 225 xy = 625 y = 22.5 n=6 2 a = 10.499 x = 9653 b = – 0.18 y 2 = 141.25 y = 10.499 – 0.18x = 10.499–0.18(20) = 6.899 1 n( x X ) 2 y – t 2 sest 1+ n + y n x 2 ( x )2 < 1 n( x X ) 2 < y + t 2 sest 1+ n + n x 2 – ( x )2 x = 225 xy = 625 y = 6.899 y = 22.5 n=6 2 a = 10.499 x = 9653 b = – 0.18 y 2 = 141.25 6.899 – (2.132)(2.09) 1 + 1 + 6(20 – 37.5) 2 < y 6 6(9653) 2252 2 < 6.899 + (2.132)(2.09) 1+ 1 + 6(20 – 37.5) 6 6(9653) – 2252 6.899– (2.132)(2.09)(1.19)< y < 6.899 + (2.132)(2.09)(1.19) 1.60 < y < 12.20 Correlation and Regression Example # 4 Coefficient of Determination and Standard Error of the Estimate Find the 90% prediction interval when x = 4 years. Years x 1 5 3 10 7 6 Contributions y, $ 500 100 300 50 75 80 x = 35 xy = 3405 y = 1105 n=6 2 a = 453.176 x = 220 b = – 50.439 y 2 = 364,525 y = 453.176 – 50.439x = 453.176 – 50.439(4) = 251.42 1 n( x – X ) 2 y – t 2 sest 1+ n + y n x 2 – ( x )2 < 1 n( x – X )2 < y + t 2 sest 1+ n + n x 2 – ( x )2 x = 35 xy = 3405 y = 1105 n=6 2 a = 453.176 x = 220 b = – 50.439 y 2 = 364,525 y = 251.42 251.42 – (2.132)(94.22) 1 + 1 + 6(4 – 5.33) 2 < y 6 6(220) – 322 < 251.42 + (2.132)(94.22) 1 + 1 + 6(4 – 5.33) 2 6 6(220) – 322 251.42 – (2.132)(94.22)(1.1) < y < 251.42+ (2.132)(94.22)(1.1) $30.46 < y < $472.38 Chapter 10 Correlation and Regression Section 10-6 Multiple Regression Correlation and Regression Multiple Regression Example # 1 A manufacturer found that a significant relationship exists among the number of hours an assembly line employee works per shift x1, the total number of items produced x2, and the number of defective items produced y. The multiple regression equation is y = 9.6 + 2.2x1 – 1.08x 2. Predict the number of defective items produced by an employee who has worked 9 hours and produced 24 items. y = 9.6 + 2.2x1 – 1.08x 2 y = 9.6 + 2.2 9 – 1.08 24 y = 3.48 or 3 items Correlation and Regression Multiple Regression Example # 2 An educator has found a significant relationship among a college graduate’s IQ x1, score on the verbal section of the SAT x2, and income for the first year following graduation from college y. Predict the income of a college graduate whose IQ is 120 and verbal SAT score is 650. The regression equation is y ' = 5000 + 97x1+ 35x 2 . y = 5000+ 97(120) – 35(650) y = $39,390