Class 28 Get Ready….

Height and Weight Is CM or Inches the better predictor of KG? Whichever has the lower standard error Will also have a variety of better stats NOT whichever has the bigger coefficient A multiple regression lets you test H0: all b’s = 0 (nothing in the model matters) H0: b1=0 given all the other b’s When using both CM and INCHES We reject H0 b1=b2=0 We fail to reject H0 b1=0 given b2 We fail to reject H0 b2=0 given b1 You need either CM or INCHES but not both Because they are highly correlated Regressions ALWAYS go thru the sample averages

Things I expect you will know How to interpret a regression using p-1 dummy variables The p possible forecasts will equal the sample average Y for each of the p groups The intercept is the average of the left-out group The coefficients are differences in group averages. The p-value/significance F will match that from ANOVA single factor

Things I expect you will know How to interpret a residual (error) It is Y - 𝑌 It is the distance each Y is from the line. Positive means above the line. They measure the difference between actual Y and expected Y (based on the X’s) The most over-weight girl (for her height) is the girl with the largest positive residual. Check the box to get residuals.

Things I expect you will know How to interpret a coefficient in a multiple regression. It measures the change in expected Y for a unit change in that X keeping all other Xs constant. If I keep miles and stops constant and change from williams to spencer, expect 0.97 hours less. If I change from Williams to Spencer, expect 0.33 hours more. It is the easy way to answer some questions. If the previous rating goes from 17.5 to 20, how will the expected ratings change? (by 0.18571 per point)

Things I expect you will know How to use a regression model to calculate a point forecast. Plug and chug. I use SUMPRODUCT You must know what Xs to plug in. It is a package deal….you must know and plug in ALL the Xs.

Things I expect you will know How to use a regression model to calculate a probability. The question gives you the Y. You Plug and chug to get the 𝑌 . You calculate t = (Y - 𝑌 )/ standard error Use t.dist.rt( t , dof) Dof is n – total number of regression terms. Requires the FOUR assumptions.

Things I expect you will know If the coefficient of X1 changes when X2 is included in the model….. You know X1 and X2 are correlated. You can use the two regression results to tell whether X1 and X2 are positively or negatively correlated. Ds was positively correlated with Miles Fact was negatively correlated with Stars Nobel was positively correlated with Yanks Speed was positively correlated with Dcorporate Exam 1 was negatively correlated with Exam 2.

Oh…Fact Movies had fewer Stars! UNDERSTANDING Coefficient Regression Table Constant 13.24615 Fact 1.40107 Coefficient Regression Table Constant 12.568 Fact 1.799 Stars 1.259 Oh…Fact Movies had fewer Stars!

Oh…Fact Movies had fewer Stars! Secret Formula Coefficient Regression Table Constant 13.24615 Fact 1.40107 Coefficient Regression Table Constant 12.568 Fact 1.799 Stars 1.259 Regress Y on X1 𝑐 = 𝑏 − 𝑏 1 𝑏 2 Regress Y on X1 and X2 Oh…Fact Movies had fewer Stars! Regress Y on X1 and X2 Regress X2 on X1

Secret Formula 𝑐 = 1.40−1.80 1.26 𝑐 =−0.32 Regress Y on X1 Coefficient Regression Table Constant 13.24615 Fact 1.40107 Coefficient Regression Table Constant 12.568 Fact 1.799 Stars 1.259 Regress Y on X1 Regress Y on X1 and X2 𝑐 = 1.40−1.80 1.26 Regress Y on X1 and X2 𝑐 =−0.32 Regress X2 on X1

Oh…Fact Movies had fewer Stars! UNDDERSTANDING Coefficient Regression Table Constant 13.24615 Fact 1.40107 Coefficient Regression Table Constant 12.568 Fact 1.799 Stars 1.259 Oh…Fact Movies had fewer Stars!

Fact Movies averaged 0.32 fewer Stars! UNDERSTANDING Secret Formula Coefficient Regression Table Constant 13.24615 Fact 1.40107 Coefficient Regression Table Constant 12.568 Fact 1.799 Stars 1.259 Fact Movies averaged 0.32 fewer Stars!

Regression is the line through a cloud of points Scatter-plot the cloud It is up to YOU to interpret the results. Don’t assume X causes Y Y might be causing X Both might be caused by Z Don’t assume better fitting lines are better at forecasting They usually are not…..too good a fit means too complicated a model…..means poorer performance.

Class 28 Assignment Variable School Graduation Rate % of Classes Under 20 Student/Faculty Ratio Alumni Giving Rate Description The name of the University Percentage of enrollees who graduate Percentage of Classes offered with <= 20 students. Number of students enrolled divided by total number of faculty Percentage of living alumni who gave to the University in 2000 Mean   83.042 55.729 11.542 29.271 Median 83.5 59.5 10.5 29 Mode 92 65 13 Standard Deviation 8.607 13.194 4.851 13.441 Skewness -0.282 -0.501 0.582 0.370 Minimum 66 3 7 Maximum 97 77 23 67 Count 48

Regress Giving Rate on Grad Rate Check if coeff is positive 1. Test the hypothesis that graduation rate and alumni giving rate are (linearly) independent. We expect universities with higher graduation rates to have higher mean giving rates. [15 points] Regress Giving Rate on Grad Rate Check if coeff is positive Divide reported p-value (found in two places) by 2. Reject if less than 0.05.   Coefficients Standard Error t Stat P-value Intercept -68.76 12.58 -5.46 1.82E-06 Graduation Rate 1.18 0.15 7.83 5.24E-10

2. If the graduation rate of school A is 5 percentage points higher than that of school B, how much higher do we expect school A’s giving rate to be? [10 points] Using the above regression (graduation rate is all we know), the expected giving rate will be 1.18*5 = 5.9 percentage points higher for school A.

3. If you learn that A and B above have identical student to faculty ratios, what is your revised answer to question 2? Be certain to explain why it went up (if it went up) or why it went down (if it went down) or why it stayed the same. Direct your response to a university administrator. [15 points]   Coefficients Standard Error t Stat P-value Intercept -19.10631 15.55006 -1.22870 0.22557 Graduation Rate 0.75574 0.16023 4.71669 0.00002 Student/Faculty Ratio -1.24595 0.28430 -4.38250 0.00007 IF we keep SFR constant, expected Giving Rate goes up 0.76 points per point of graduation rate. If we don’t keep SFR constant, expected Giving Rates went up 1.18 points per point. Schools with higher grad rates had LOWER SFR (that makes sense) If we don’t hold SFR constant, increases in grad rate mean decreases in SFR and the combined effect of the two is 1.18. So….if grad rate is higher (but SFR is not), expected 0.76 increase. If grad rate is higher (and SFR is lower as in the data), expect 1.18 increase.

Don’t Use this variable. 4. Provide a point forecast of alumni giving rate for a university with graduation rate of 80, 65 percent of its classes with 20 or fewer students, and a student/faculty ratio of 20. [25 points]   Coefficients Standard Error t Stat P-value Intercept -20.7201 17.5214 -1.1826 0.2433 Graduation Rate 0.7482 0.1660 4.5082 0.0000 % of Classes Under 20 0.0290 0.1393 0.2084 0.8358 Student/Faculty Ratio -1.1920 0.3867 -3.0823 0.0035 Don’t Use this variable.   Coefficients Intercept -19.10631 Graduation Rate 0.75574 Student/Faculty Ratio -1.24595 1 80 20 POINT FORECAST 16.43 Use this model. Plug and Chug. The best model includes Grad Rate and SFR (% classes <20 not needed)

The university with the most negative residual. 5. Of the 48 universities in the data set, which one has the most surprisingly low alumni giving rate? [10 points] The university with the most negative residual. Use the best model, ask for residuals, find the minimum. MICHIGAN!

ANOVA or Regression of SFR on 2 dummies. 6. Bo notices that some of the 48 have “university” in their names, some have “college” and the rest have “institute”. Bo wonders whether these names are predictive of student/faculty ratio? (Formulate and test a relevant hypothesis.) [25 points] Three groups (p=3) ANOVA or Regression of SFR on 2 dummies. SUMMARY OUTPUT ANOVA   df SS MS F Significance F Regression 2 103.7348 51.8674 2.3290 0.1090 Residual 45 1002.1818 22.2707 Total 47 1105.9167 Coefficients Standard Error t Stat P-value Intercept 11.8636 0.7114 16.6754 0.0000 Dcollege -0.3636 3.4120 -0.1066 0.9156 Dinstitute -7.3636 -2.1582 0.0363

Get Ready….. More practice problems (answers) on website. I’ll host Sunday night Office Hours. I am available Monday and Tuesday until 2pm. Email pfeiferp@virginia.edu Check the website to see where I am…you are welcome to join us.