(positive, not very strong) (perfectly negatively correlated)

Transcription

1 HOMEWORK PROBLEMS Part II. Chapter 8 -- Review Exercises Statistics Dr. McGahagan Review Exercises (p ) Problem 3. Men always marry women exactly 10 percent shorter. Let HH = height of husband and HW = height of wife Therefore, HW = a HH will be the equation of the line enabling us to predict the height of the wife, and every point representing a pair of husbands and wives will be exactly on that line -- a man 60 inches tall will have a wife 54 inches tall, a man 70 inches tall will have a wife 63 inches tall, a man 500 inches tall will have a wife 450 inches tall, and so on. The correlation coefficient will be + 1. IMPORTANT LESSON: the correlation coefficient does not give the slope of the line, but is determined by the presence or absence of scatter about the line. Problem 5. Guess the correlation. Choices: -0.5, 0.0, 0.3, 0.6, 0.95 (a) Between freshman and sophomore GPA. Positive but certainly not perfect. Many students who have initial trouble improve in second year. Best guess: 0.6 [incidentally, this is on the order of the correlation between SAT scores and freshman grades] (b) Between freshman and senior grades: Still positive, but with more time comes more variation. Best guess: 0.30 (c) Between weight and length of 2x4 pieces of pine. Best guess: 0.95 There is still likely to be variation if moisture conditions or number of knotholes differ. Problem 7. Associate scatter diagrams with correlation coefficients. Grid of correlation coefficients corresponding with the plots should be: (positive, not very strong) (perfectly negatively correlated) (negative, pretty strong) (positive and very strong) (no real pattern, and not (weak, but pretty definitely negative; even very clear that the compare with the plot to the left relation is positive) to decide which of the final two is which)

2 Problem 8. Human Growth Study. Plot of heights at age 18 against height at age 4 shows fairly strong but not perfect correlation. Average height at age 4 seems about 41 inches -- at about the point where the two lines cross. SD of height at age 4 = about 1.5 inches, because 41 +/- 2 (1.5) will cover almost all points (put an index card vertically at 44 and 38 inches to see this) Average height at age 18 seems about 71 inches (again the point at which the two lines cross). SD of height at age 18 = about 2.5 inches, because 71 +/- 2 (2.5) will cover almost all points (put an index card horizontally at 66 and 76 inches to see this). The correlation coefficient is about 0.75 to high but not perfect correlation. The SD line will be the solid line; we will soon meet the dashed line as the regression line. If you xerox the page (on higher magnification) and draw the horizontal and vertical lines as above, you will see that the SD box corners should pretty much fall on the solid line. Problem 9. Correlation coefficient calculation exercise. Mean SD Corr. (a) X Y 3 2 (b) X Y 2 1 (c) X (note that Y = 2 X) Y 4 2 Computer notes: (bind x (list )) and X will stay the same for all three problems. (bind y (list...)) will change with each problem. (plot y x) and use the PLOT menu to adjust the axis range so X and Y start with zero. You can jitter the plots with the capital J key, distinguishing overlapping points. Problem 11. Quiz in statistics With 10 questions on a quiz, and an average number of 6.4 right, and a SD of 2.0 for the right answers, we note that, letting R = number of right answers and W = number of wrong answers, W = 10 - R There cannot possibly be any point off the SD line, so the correlation coefficient will be minus 1. Simulation: (bind right (rnd 20 10)) Create 50 uniform random integers from 0 to 10, using the Mersenne Twister (bind wrong (- 10 right)) (stats right wrong) (corr right wrong) (plot right wrong) (abline 10-1)

3 Chapter 9 -- Review Exercises Statistics Dr. McGahagan Problem 2. True or False? (a) FALSE. If correlation coefficient is -0.8, below-average values of the dependent variable are associated with ABOVE average values of the independent variable. Try (scatter -0.8) and compare to (scatter 0.8) (b) FALSE. If y is always less than x, the correlation coefficient may be positive or negative. Try (bind x (list )), (bind y (/ x 10)) (< y x) = (T T T T T) so y is always less than X. (corr x y) = 1.00 Then (bind y (- (/ x 10))) and repeat: (< y x) = (T T T T T) so y is always less than X. (corr x y) = Problem 5. Can X and Y be perfectly correlated? Given the following sets of numbers, can you fill in the final number to ensure rho = + 1? (a) X = ( ) Y = ( ) All points given lie along a line through the points (1, 1) and (2, 3) The slope of the line is + 2, so the equation of the line would be Y = a + 2 X To fit the other points, we must have: 1 = a + 2 (1) so a = -1 3 = a + 2 (2) so a = -1 The required point will be Y = (4) = 7 You may check your solution by defining the lists: (bind y (list )) (bind x (list )) Then (corr x y) = will show that you have solved the problem (b) X = ( ) Y = ( ) The first two points lie on a line through the points (1, 1) and (2, 3) The equation of that line was Y = X, as found in part (a) The third point (3, 4) does NOT lie on that line: Y = (3) = 5, so an X value of 3 corresponds to a value of 5, not 4. It is therefore impossible for all 4 points to line on the same line.

4 Problem 9. Student evaluations of assistants and exam performance. Calculate the correlations: (bind assistant (list )) (bind course (list )) (bind final (list )) The correlations are: (corr assistant course) = ; not much relation. (b) is true. (corr assistant final ) = definite NEGATIVE correlation. Hence (a) is false. The more students liked the assistant, the worse they did on the final. Hypothesis: the well-liked assistants were entertaining, but not demanding. (corr course final) = Positive relation here; though it is unclear from the correlation whether students put forward effort because they liked the course, or liked the course because they realized they were doing well in it. Problem 12. Education of husbands and wives Correlation of about 0.8, at a guess. (a) Vertical/horizontal stripes occur because few people have, and even fewer report, years of school -- the data are discrete, not continuous. (b) Few points appear because they overlap. If the graph were in EcLS, the capital J key would jitter the points (add some random noise so they did not overlap). (c) Shaded areas on which plot indicate that: (i) Wife completed exactly 16 years of schooling. Plot C (ii) Wife completed more schooling than husband. No plot shows this; given the actual plot, you could draw a 45-degree line (y = x) with the command (abline 0 1). Points above and to the left of that line would indicate wife's education > husband's education. (iii) Husband completed more than 16 years of schooling. Plot B. (iv) Husband completed exactly 12 years, and wife completed fewer than 12 years. Plot A.

5 Review Questions -- Chapter Regression Statistics Dr. McGahagan * Problem 4. Educational levels of husband and wife. Average ed. level = 12 years, SD = 3 years for both; correlation coefficient = 0.5 (a) If husband has 18 years of schooling, he is (18-12) / 3 = 2 SD above average. The wife would be expected to be 0.5 * 2 SD = 1 SD above average, or to have 15 years of education. (b) If wife has 15 years of schooling, she is 1 SD above average. Her husband would be expected to be half a standard deviation above average, that is to have = 13.5 years of education. (c) Isn't this strange? Well-educated men marry above average, but less well educated women, but can it be also true that the women marry still less educated men? While very well educated men or women might prefer to marry someone as well educated as themselves, the marriage pool doesn't include many of them, and chances are they will on average marry someone a bit less highly educated. Note the "on average" -- it will be the somewhat unusual very well educated man or woman who marries someone more educated than themselves, but "somewhat unusual" does not mean "non-existent". * Problem 6. Which line is the regression line? The dotted, roughly 45 degree line is the SD line. Regression lines are always flatter than the SD line, when the response variable (the dependent variable) is on the vertical axis. The dashed line tells you the expected value of y given x; the solid line indicates the expected value of X given Y (note that you have to interchange the axes so that the response variable is on the vertical axis). * Problem 9. Percentile rank and correlations. Correlation of midterm scores and final scores is 0.50 (a) Percentile rank on midterm = 5 percent. SD below mean on midterm: (normal-quant 0.05) = or find 90 in the area column of the normal table (5 percent on either side). The area percent is defined by a z-score of +/ Expected SD below mean on final: 0.5 * = This translates into an area defined by +/ of between and 60.47, let's say a 60 percent central area, so that 20 percent is below and 20 percent above This means the average student at the 5th percentile could be predicted to be at the 20th percentile on the final. In EcLS: (normal-cdf (* rho (normal-quant pctile1)) = (normal-cdf (* 0.5 (normal-quant 0.05)) =.2054 (b) Percentile rank on midterm: Since 20 percent rank higher, we must first look for central area of 60, that is defined by about +/ SDs Score will be expected to be 0.5 * 0.85 = SDs, which gives a central area of between and in the table -- say one third of the total area for simplicity. About a third will be above SDs, so the expected percentile rank is 2/3 = 67th percentile. In EcLS: (normal-cdf (* 0.5 (normal-quant 0.80)) = or the 66th percentile. (c) and (d). Whether we know the student is average at the midterm or simply assume it because we have no further information, we expect the student to be average on the final. * Problem 10. Regression effect. Expect the student at the 40th percentile to improve somewhat. Exactly how much depends on the correlation coefficient. See Exercise set C, problem 2 for a similar problem.

6 Review Questions -- Chapter RMS Error * Problem 4. Midterm and Final Scores, part I. Average grade on midterm = 50; SD on midterm = 25 Average grade on final = 55; SD on final = 15 Correlation of midterm and final = 0.60 (a) For about one-third of students, prediction of final will be off by more than --- RMSEs. For about 2/3 of the students, the prediction will be off by less than the same unknown value. But we do know that a central area of about 2/3 will be from - 1 to + 1; so the blank should be filled by 1 RMSE.This is given by the problem 1 formula as (sqrt (1-0.36)) * 15 = 0.8 * 15 = 12 points. (b) Given a midterm of 80, the student scored (80-50) / 25 = 30 / 25 = 1.2 SDs above the mean for the midterm. He is therefore expected to score 0.6 * 1.2 = 0.72 SDs above the mean on the final. The standardized score of the final will have been computed as (X - 55) / 15 = 0.72; hence X = * 15 or we would predict a score of In our notation, E [final midterm = 80] = 65.8 (c) This prediction will, if all assumptions are met (normal distribution of residuals, no pattern to residuals), have a 50 percent chance of being within 0.7 RMSEs (see the normal table; look for central area of 51.61). That is, it has an even chance of falling within 0.7 * 12 = 8.4 points of the predicted value. * Problem 5. Midterm and Final Scores, Part II. (a) Assuming normality, the percentage of students scoring above 80 on the final can be found without regression: Z-score of 80 points = (80-55) / 15 = 25 / 15 = 5/3 = 1.67 Corresponding central area (for Z = 1.65, closest value in table) = percent Two tail area = = 9.89 percent, so one tail area (above 80 points) = percent. (b) Given a score of 80 on the midterm, the expected score on the final is 65.8 (see last problem). RMSE is 12 points, so a score of 80 would be ( ) / 12 = 1.18 RMSEs above the mean. The central area for Z = +/- 1.2 (closest to 1.18) is 76.99, so about 23 percent of the points would be outside this area, or half that = 12.5 points above 80. * Problem 7. Correlation and regression (Extended) There is a correlation between math and physics test scores; both have mean of 60 and equal SDs. We would expect students who scored an above average 75 on the math test to score above average on the physics test as well -- but, due to the regression effect, not as much above average. So the expected score will be more than 60 but less than 75. Suppose the SDs of each test are 20, and the correlation coefficient 0.90 Then the score of 75 was (75-60) / 20 = 15 /20 = 0.75 SDs above average on the math test. We would expect a score of 0.9 * 0.75 = SDs above average on the physics test. This Z-score would have been computed as; = (X - 60) / 20 so X - 60 = * 20 = 13.5 and X = 73.5 Note that due to the high correlation, we don't expect the new score to be much less