1. 1 ES9 Chapter 24 ~ Linear Correlation & Regression Analysis 100110120130140150160 Weight 22 23 24 25 26 27 28 29 30 Waist Size

2. 2 ES9 Chapter Goals More detailed look at linear correlation and regression analysis Develop a hypothesis test to determine the strength of a linear relationship Consider the line of best fit. Use this to make confidence interval estimations.

3. 3 ES9 Linear Correlation Analysis The coefficient of linear correlation, r, is a measure of the strength of a linear relationship Consider another measure of dependence: covariance Recall: bivariate data - ordered pairs of numerical values

4. 4 ES9 Consider the following set of bivariate data: {(8, 22), (5, 28), (8, 18), (4, 16), (13, 27), (15, 23), (17, 17), (12, 13)} Derivation of the Covariance Goal: a measure of the linear relationship between two variables Consider a graph of the data: 1.The point is the centroid of the data 2.A vertical and horizontal line through the centroid divides the graph into four sections

5. 5 ES9 Graph of the Data with Centriod 468101214161820 x 12 14 16 18 20 22 24 26 28 30 y (10.25, 20.5)

6. 6 ES9 Notes 1.Each point (x, y) lies a certain distance from each of the two lines 2. : the horizontal distance from (x, y) to the vertical line passing through the centroid 3. : the vertical distance from (x, y) to the horizontal line passing through the centroid 4.The distances may be positive, negative, or zero 5.Consider the product: a.If the graph has lots of points to the upper right and lower left of the centroid (positive linear relationship), most products will be positive b.If the graph has lots of points to the upper left and lower right of the centroid (negative linear relationship), most products will be negative

7. 7 ES9 Covariance of x and y The covariance of x and y is defined as the sum of the products of the distances of all values x and y from the centroid divided by n  1: Note:

8. 8 ES9 Calculations for Finding Covar (x, y)

9. 9 ES9 Data & Covariance Positive covariance: 012345678 x 0 1 2 3 4 5 6 7 8 y

10. 10 ES9 Data & Covariance Negative covariance: 012345678 x 0 1 2 3 4 5 6 7 8 9 y

11. 11 ES9 Data & Covariance Covariance near 0: 0123456789 x 0 1 2 3 4 5 6 7 8 9 y

12. 12 ES9 Problems 1.The covariance does not have a standardized unit of measure 2.Suppose we multiply each data point in the example in this section by 15 The covariance of the new data set is -514.29 3.The amount of the dependency between x and y seems stronger but the relationship is really the same 4.We must find a way to eliminate the effect of the spread of the data when we measure the strength of a linear relationship

13. 13 ES9 Solution 1.Standardize x and y: 2.Compute the covariance of x and y 3.This covariance is not affected by the spread of the data 4.This is exactly what is accomplished by the coefficient of linear correlation:

14. 14 ES9 Notes 1.The coefficient of linear correlation standardizes the measure of dependency and allows us to compare the relative strengths of dependency of different sets of data 2.Also commonly called Pearson’s product moment, r Calculation of r (for the data presented in this section):

15. 15 ES9 Alternative (Computational) Formula for r 1.This formula avoids the separate calculations of the means, standard deviations, and the deviations from the means 2.This formula is easier and more accurate: minimizes round-off error Alternative (Computational) Formula for r:

16. 16 ES9 Inferences About the Linear Correlation Coefficient Use the calculated value of the coefficient of linear correlation, r*, to make an inference about the population correlation coefficient,  Consider a confidence interval for  and a hypothesis test concerning 

17. 17 ES9 Assumptions... Assumptions for inferences about linear correlation coefficient: The set of (x, y) ordered pairs forms a random sample and the y-values at each x have a normal distribution. Inferences use the t-distribution with n  2 degrees of freedom. Caution: The inferences about the linear correlation coefficient are about the pattern of behavior of the two variables involved and the usefulness of one variable in predicting the other. Significance of the linear correlation coefficient does not mean there is a direct cause-and- effect relationship.

18. 18 ES9 Confidence Interval Procedure 1.A confidence interval may be used to estimate the value of the population correlation coefficient,  2.Use a table showing confidence belts 3.Table 10, Appendix B: confidence belts for 95% confidence intervals 4.Table 10 utilizes n, the sample size

19. 19 ES9 Example:A random sample of 25 ordered pairs of data have a calculated value of r =  0.45. Find a 95% confidence interval for , the population linear correlation coefficient. Example Solution: 1.Population Parameter of Concern The linear correlation coefficient for the population,  2.The Confidence Interval Criteria a.Assumptions: The ordered pairs form a random sample, and for each x, the y-values have a mounded distribution b.Test statistic: The calculated value of r c.Confidence level: 1   = 0.95 3.Sample Evidence n = 25 and r =  0.45

20. 20 ES9 Solution Continued 4.The Confidence Interval The confidence interval is read from Table 10, Appendix B Find r =  0.45 at the bottom of Table 10 Visualize a vertical line through that point Find the two points where the belts marked for the correct sample size cross the vertical line Draw a horizontal line through each point to the vertical scale on the left and read the confidence interval The values are  0.68 and  0.12 5.The Results  0.68 to  0.12 is the 95% confidence interval for 

21. 21 ES9 Table 10 -0.12 -0.68 Scale of p (population correlation coefficient) -0.45 Scale of r (sample correlation) The numbers on the curves are sample sizes:

22. 22 ES9 Hypothesis Testing Solution 1.Null hypothesis: the two variables are linearly unrelated,  = 0 3.Test statistic: calculated value of r 4.Probability bounds or critical values for r: Table 11, Appendix B 5.Number of degrees of freedom for the r-statistic: n  2 2.Alternative hypothesis: one- or two-tailed, usually  0

23. 23 ES9 Example: In a study of 32 randomly selected ordered pairs, r = 0.421. Is there any evidence to suggest the linear correlation coefficient is different from 0 at the 0.05 level of significance? Example Solution: 1.The Set-up a.Population parameter of concern: The linear correlation coefficient for the population,  b.The null and the alternative hypothesis: H o :  =0 H a :  0

24. 24 ES9 Solution Continued 2.The Hypothesis Test Criteria a.Assumptions: The ordered pairs form a random sample and we will assume that the y-values at each x have a mounded distribution b.Test statistic: r* (calculated value of r) with df = 32  2 = 30 c.Level of significance:  = 0.05 3.The Sample Evidence n = 32 and r* = r = 0.421

25. 25 ES9 Solution Continued 4.The Probability Distribution (p-Value Approach) a.The p-value: Use Table 11: 0.01 < P < 0.02 b.The p-value is smaller than the level of significance,  5.The Results a.Decision: Reject H o b.Conclusion:At the 0.05 level of significance, there is evidence to suggest x and y are correlated 4.The Probability Distribution (Classical Approach) a.Critical Value: The critical value is found at the intersection of the df = 30 row and the two-tailed 0.05 column of Table 11: 0.349 b.r* is in the critical region ~ or ~

26. 26 ES9 Line of best fit results from an analysis of two (or more) related variables Linear Regression Analysis Try to predict the value of the dependent, or output, variable The variable we control is the independent, or input, variable

27. 27 ES9 Method of Least Squares: Method of Least Squares Notes: 1.A scatter diagram may suggest curvilinear regression 2.If two or more input variables are used: multiple regression The line of best fit: The slope: The y-intercept:

28. 28 ES9  0 :The y-intercept, estimated by b 0 Linear Model The Linear Model: This equation represents the linear relationship between the two variables in a population  1 :The slope, estimated by b 1  :Experimental error, estimated by The random variable e is called the residual e is the difference between the observed value of y and the predicted value of y at a given x The sum of the residuals is exactly zero Mean value of experimental error is zero:   = 0 Variance of experimental error:

29. 29 ES9 Estimating the Variance of the Experimental Error Estimating the Variance of the Experimental Error: Assumption: The distribution of y’s is approximately normal and the variances of the distributions of y at all values of x are the same (The standard deviation of the distribution of y about y is the same for all values of x) Consider the sample variance: 1.The variance of y involves an additional complication: there is a different mean for y at each value of x 2.Each “mean” is actually the predicted value, y 3.Variance of the error e estimated by: Degrees of freedom: n  2

30. 30 ES9 Rewriting : Alternative (Computational) Formula for Variance of Experimental Error SSE = sum of squares for error

31. 31 ES9 Example: A recent study was conducted to determine the relation between advertising expenditures and sales of statistics texts (for the first year in print). The data is given below (in thousands). Find the line of best fit and the variance of y about the line of best fit. Example

32. 32 ES9 Solution

33. 33 ES9 Solution Continued The equation for the line of best fit: The variance of y about the regression line: Note:Extra decimal places are often needed for this type of calculation

34. 34 ES9 Scatter diagram, regression line, and random errors as line segments: Illustration Sales Advertising Costs 35404550556065 250 275 300 325 350 375 400 425 450 475 500

35. 35 ES9 Regression Analysis The regression equation is C2 = 39.8 + 7.19 C1 Predictor Coef StDev T P Constant 39.81 69.11 0.58 0.580 C1 7.191 1.484 4.84 0.001 S = 36.63 R-Sq = 74.6% R-Sq(adj) = 71.4% Analysis of Variance Source DF SS MS F P Regression 1 31491 31491 23.47 0.001 Residual Error 8 10734 1342 Total 9 42225 Minitab Output

36. 36 ES9 Inferences Concerning the Slope of the Regression Line Confidence Interval for  1 : 1-  confidence interval estimate for the population slope of the line of best fit Hypothesis Test for  1 : Tests the null hypothesis,  1 = 0, the slope of the line of best fit is equal to 0, that is, the line is of no use in predicting y for a given value of x

37. 37 ES9 1.b 1 has a sampling distribution that is approximately normal Assume:Random samples of size n are repeatedly taken from a bivariate population Sampling Distribution of the Slope b 1 2.The mean of b 1 is  1 3.The variance of b 1 is: provided there is no lack of fit

38. 38 ES9 Estimator for : Standard Error of Regression The standard error of regression (slope) is and is estimated by Example (continued): For the advertising costs and sales data:

39. 39 ES9 Assumptions for inferences about the slope parameter  1 : The set of (x, y) ordered pairs forms a random sample and the y- values at each x have a normal distribution. Since the population standard deviation is unknown and replaced with the sample standard deviation, the t-distribution will be used with n  2 degrees of freedom. Inferences About Slope Continued 1 1b stb  )2/,2(n  Confidence Interval Procedure: The 1   confidence interval for  1 is given by

40. 40 ES9 Example: Find the 95% confidence interval for the population slope  1 for the advertising costs and sales example Example Solution: 1.Population parameter of Interest The slope,  1, for the line of best fit for the population 2.The Confidence Interval Criteria a.Assumptions: The ordered pairs form a random sample and we will assume the y-values (sales) at each x (advertising costs) have a mounded distribution b.Test statistic: t with df = 10  2 = 8 c.Confidence level: 1   = 0.95 3.Sample Evidence Sample information:

41. 41 ES9 Solution Continued 4.The Confidence Interval a.Confidence coefficients: t (df,  /2) = t (8, 0.025) = 2.31 b.Interval: 5.The Results The slope of the line of best fit of the population from which the sample was drawn is between 5.707 and 8.676 with 95% confidence )676.8,707.5( 4845.11915.7 2037.2)31.2(1915.7 1 1    b sb t (n-2,  /2)

42. 42 ES9 Hypothesis-Testing Procedure 1.Null hypothesis is always H o :  1 = 0 2.Use the Students t distribution with df = n  2 3.The test statistic:

43. 43 ES9 Example: In the previous example, is the slope for the line of best fit significant enough to show that advertising cost is useful in predicting the first year sales? Use  = 0.05 Example Solution: 1.The Set-up a.Population parameter of concern: The parameter of concern is  1, the slope of the line of best fit for the population b.The null and alternative hypothesis: H o :  1 = 0 (x is of no use in predicting y) H a :  1 > 0 (we expect sales to increase as costs increase)

44. 44 ES9 Solution Continued 2.The Hypothesis Test Criteria a.Assumptions: The ordered pairs form a random sample and we will assume the y-values (sales) at each x (advertising costs) have a mounded distribution b.Test statistic: t* with df = n  2 = 8 c.Level of significance:  = 0.05 3.The Sample Evidence a.Sample information: b.Calculate the value of the test statistic:

45. 45 ES9 Solution Continued 4.The Probability Distribution (p-Value Approach) a.The p-value: P = P(t* > 4.8444, with df = 8) < 0.001 b.The p-value is smaller than the level of significance,  5.The Results a.Decision: Reject H o b.Conclusion:At the 0.05 level of significance, there is evidence to suggest the slope of the line of best fit is greater than zero. The evidence indicates there is a linear relationship and that advertising cost (x) is useful in predicting the first year sales (y). 4.The Probability Distribution (Classical Approach) a.Critical value: t (8, 0.05) = 1.86 b.t* is in the critical region ~ or ~

46. 46 ES9 Confidence Interval Estimates for Regression Use the line of best fit to make predictions Predict the population mean y-value at a given x Predict the individual y-value selected at random that will occur at a given value of x The best point estimate, or prediction, for both is y

47. 47 ES9 Notation & Background Notation: 1.Mean of the population y-values at a given value of x: 2.The individual y-value selected at random for a given value of x: Background: 1.Recall: the development of confidence intervals for the population mean  when the variance was known and when the variance was estimated 2.The confidence interval for and the prediction interval for are constructed in a similar fashion 3.y replaces as the point estimate 4.The sampling distribution of y is normal

48. 48 ES9 Background Continued 5.The standard deviation in both cases is computed by multiplying the square root of the variance of the error by an appropriate correction factor 6.The line of best fit passes through the centroid: Consider a confidence interval for the slope  1 If we draw lines with slopes equal to the extremes of that confidence interval through the centroid, the value for y fluctuates considerably for different values of x (See the Figure on the next slide.) It is reasonable to expect a wider confidence interval as we consider values of x further from We need a correction factor to adjust for the distance between x 0 and This factor must also adjust for the variation of the y-values about y

49. 49 ES9 Confidence Interval for Slope 35404550556065 250 275 300 325 350 375 400 425 450 475 500 Slope is 8.676 Slope is 5.707 Sales Advertising Costs

50. 50 ES9 Confidence interval for the mean value of y at a given value of x, Notes: 1.The numerator of the second term under the radical sign is the square of the distance of x 0 from 2.The denominator is closely related to the variance of x and has a standardizing effect on this term x Confidence Interval )(SS )( 1 ˆ )( )( 1 ˆ 2 0 2 2 0 x xx n sty xx xx n sty e e       standard error of y (n-2,  /2)

51. 51 ES9 Example:It is believed that the amount of nitrogen fertilizer used per acre has a direct effect on the amount of wheat produced. The data below shows the amount of nitrogen fertilizer used per test plot and the amount of wheat harvested per test plot. Example a.Find the line of best fit b.Construct a 95% confidence interval for the mean amount of wheat harvested for 45 pounds of fertilizer

52. 52 ES9 Solution Using Minitab, the line of best fit: Confidence Interval: 1.Population Parameter of Interest The mean amount of wheat produced for 45 pounds of fertilizer, 2.The Confidence Interval Criteria a.Assumptions: The ordered pairs form a random sample and the y-values at each x have a mounded distribution b.Test statistic: t with df = 16  2 = 14 c.Confidence level: 1   = 0.95 3.Sample Information:

53. 53 ES9 45| 2 2 0 for interval confidence 95%,74.21 to92.13 91.383.17 )3587.0)(096.5)(14.2(83.17 0662.00625.0)096.5)(14.2(83.17 94.8746 )06.6945( 16 1 )096.5)(14.2(83.17 )(SS )( 1 ˆ         xy e x xx n sty  Solution Continued 4.The Confidence Interval: (n-2,  /2)

54. 54 ES9 Confidence Belts for Confidence interval: green vertical line Confidence interval belt: upper and lower boundaries of all 95% confidence intervals 30405060708090100110120 10 15 20 25 30 35 40 45 Upper boundary for Line of best fit Wheat Fertilizer Lower boundary for

55. 55 ES9 Prediction Interval Prediction interval of the value of a single randomly selected y: Example:Find the 95% prediction interval for the amount of wheat harvested for 45 pounds of fertilizer Solution: 1.Population Parameter of Interest y x=45, the amount of wheat harvested for 45 pounds of fertilizer )(SS )( 1 1 ˆ 2 0 x xx n sty e   (n-2,  /2)

56. 56 ES9 Solution Continued 2.The Confidence Interval Criteria a.Assumptions: The ordered pairs form a random sample and the y-values at each x have a mounded distribution b.Test statistic: t with df = 16  2 = 14 c.Confidence level: 1   = 0.95 3.Sample Information

57. 57 ES9 45 2 2 0 for interval prediction 95%,41.29 to24.6 5859.1183.17 )0624.1)(096.5)(14.2(83.17 1287.1)096.5)(14.2(83.17 0662.00625.01)096.5)(14.2(83.17 94.8746 )06.6945( 16 1 1)096.5)(14.2(83.17 )(SS )( 1 1 ˆ          x e y x xx n sty Solution Continued 4.The Confidence Interval (n-2,  /2)

58. 58 ES9 0 x y Prediction belts for 30405060708090100110120 10 15 20 25 30 35 40 45 Lower boundary for 95% prediction interval on individual y-values at any x Upper boundary on individual y-values Line of best fit x 0 = 45 Wheat Fertilizer

59. 59 ES9 Precautions 1.The regression equation is meaningful only in the domain of the x variable studied. Estimation outside this domain is risky; it assumes the relationship between x and y is the same outside the domain of the sample data. 2.The results of one sample should not be used to make inferences about a population other than the one from which the sample was drawn 3.Correlation (or association) does not imply causation. A significant regression does not imply x causes y to change. Most common problem: missing, or third, variable effect.

60. 60 ES9 13.6 ~ Understanding the Relationship Between Correlation & Regression We have considered correlation and regression analysis When do we use these techniques? Is there any duplication of work?

61. 61 ES9 Remarks 1.The primary use of the linear correlation coefficient is in answering the question “Are these two variables related?” 2.The linear correlation coefficient may be used to indicate the usefulness of x as a predictor of y (if the linear model is appropriate) The test concerning the slope of the regression line (H o :  1 = 0) tests the same basic concept 3.Lack-of-fit test: Is the linear model appropriate? Consider the scatter diagram

62. 62 ES9 Conclusions 1.Linear correlation and regression measure different characteristics. It is possible to have a strong linear correlation and have the wrong model? 2.Regression analysis should be used to answer questions about the relationship between two variables: a.What is the relationship? b.How are the two variables related?