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READING COMPUTER PRINTOUTS Variable price (Y) size (X) N 93 (total sample size) 93 (“) Mean 99.533 (Y-bar) 1.650 (X-bar) Std. Dev. 44.184 (sY) .525 (sX) Source Model DF 1 (# of explanatory variables) Sum of Squares 145097 (TSS-SSE; numerator of r2) Mean Square 145097 Error (cond. distrib) Total (marginal dist.) 91 (n-(k+1)) 92 (model + error df) 34508 (SSE) 179605 (TSS) 379 Variable DF Parameter Estimator Standard Error Intercept (α or a) 1 -25 (α) 6.7 ( σ ) a Size (β or b) 1 75 (β) 3.9 ( σ ) b Root MSE (σ-hat) 19.473 ( ˆ = estimated standard deviation, estimated variability of Y values) σ T for Ho: Parameter = 0 R Square .8 (r2=coefficient of determination, because .8 is close to 1, Y-hat is a good predictor of correlation, strong + association) Prob > |T| -3.8 .0003 (very low, reject the null of Ho: β = 0 .0001 (“ ) 19.6 (t-statistic) ***F Value: tells you if the model is doing any good; Prob>F If bigger than .05, all the coefficients are probably zero so your explanatory variables are independent of your response variable. E(Y)= α + β X “regression function”; Y = α + βX “linear function”; Y-hat = a + b1X1+ b2X2 “prediction equation” or “least squares line”; the best equation according to the least squares criterion for estimating a linear regression because it has the smallest SSE. note: a and b, α and β are “regression coefficients” a=(Y-bar)=b(X-bar) & b=the sum of all {(X-(X-bar)(Y-(Y-bar))} divided by the sum of all (X-(X-bar)2, also note: Y-hat = a + b1X1+ b2X2+e = prediction model Interpreting regression coefficients: Every 1 unit increase in the explanatory variable has an increase in b (units). ex. equation Y-hat = -498.7 + 32.6(X1)+9.1(X2); Y = crime (deaths/100,000); X1=poverty rate; Interp: every 1% in poverty rate has an increase in 32 deaths per100,000 people. SSE = ∑ (Y − Yˆ ) 2 = “sum of squared errors” or “residual sum of squares” = sum of all errors from the prediction equation E(Y) = a+bX; the variation of the observed points around the prediction line (Y-hat = a +bX); note: a residual is the vertical distance between the observed point and the prediction line = Y – Y-hat; note that the sum of all residuals (Y – Y-hat) = 0 TSS = ∑ (Y − Y ) 2 = “total sum of squares” = sum of all errors from the mean Model Sum of Squares=(TSS-SSE) = “regression sum of squares” or “explained sum of squares” = the numerator of r2, and represents the amount of the total variation in TSS in Y that is explained by X using the least squares line. (Y=a+bX) MSE = σˆ = SSE = “mean square error” or “root square error” or “estimated conditional standard deviation” where df for the (Y − Yˆ ) 2 = n −2 n −2 estimate =(n-2) ; Measures the variability of Y-values for all fixed Xs. note: with p-unknown parameters, df = n-p) σˆ σb = ∑(X − X ) 2 sY= standard deviation of the marginal distribution of Y (σ Y if a population); sY is usually greater than the conditionsal distribution because the errors of margeral are greater than the errors of conditional w/ correlation. r= sx sy when sx = ∑ (X − X ) 2 and n −1 sx = ∑ (Y − Y ) n −1 2 = TSS “pearsons correlation coefficient” or “standardized regression coefficient” =standardized slope to test n −1 strengths of association with a linear equation; r is always between -1 and +1; at r=0, b=0 and there is independence between the two variables being tested; at r=± 1 it is called a “perfect positive/negative linear association” because all points are on the prediction line; The prediction equation using Y to predict X is equal to that of X to predict Y, r ≠ unit dependent r2 = TSS − SSE ∑ (Y − Y ) 2 − ∑ (Y − Yˆ ) 2 - “the coefficient of determination”—r2 is always between 0 and 1; if SSE = 0, r2=0; if r2=1, all points fall on the prediction line; TSS = ∑ (Y − Y ) 2 no prediction error if SSE=TSS, because that is when r2=0; closer r2 is to 1, the better Y-hat (Y=hat=a+bX) as a prediction compared with Y-bar (mean); r2 is not unit dependent. Regression toward the mean: a 1 s.d. in X = an r s.d. in Y, where r = pearson’s correlation coefficient. If correlation = r=.5, an x-value being sx above the mean is predicted to have a Y-value ½ sY above its mean; since r<1, X’s will be farther from their means than their Y’s are from theirs. MULTIVARIATE REGRESSIONS Regression Sum of Squares: RSS = TSS-SSE SSE R2= proportion of the total variation in Y explained by the multiple regression model now called “coefficient of mult. determination.”; 2 TSS − SSE R = Ra2= 1-n-1/n-(k+1)* n-1/n-(k+1) multicollinearity: when explanatory variables are highly correlated and only slightly boost R2. r = “multiple correlation coefficient” = square root of R squared TSS = 1− TSS Q5: There are 5 kids who are patients, ages 6, 8, 10, 12 & 14, 2 kids picked at random. Find the sampling distribution of the mean age of the two children: (6+8)/2=7 (8+10)/2=9 (10+12)/2=11 (12+14)/2=13 .1 if m = 7, 8, 12 or 13 (6+10)/2=8 (8+12)/2=10 (10+14)/2=12 sampling distrib: p(m) = .2 if m = 9, 10, or 11 (6+12)/2=9 (8+14)/2=11 0 otherwise (6+14)/2=10 sample mean: E[X-bar]=[(7+8+12+13)*1/10]+[(9+10+11)*2/10)=10 standard error of sample mean: Var[X-bar] = .1{(7-10)2+(8-10)2+(12-10)2+(13-10)2}+ .2*{(9-10)2+(10-10)2+(11-10)2}=3 Q1: Prior opinion was even split between pro and con on abortion. On first day in office, 8 of 10 letters received were anti abortion. If letters = representative, what are chances of getting this? “I am concerned a majority of my constituency does not support my pro-choice stand.” hypothesis test: Ho: π = .5 Ha: π>.5 (proportion of .. is greater than .5). What is the Prob of 8 or worse anti letters given the null hypothesis? Prob(# anti abortion letters ≥ 8| π=.5) = Prob(8) + Prob(9) + Prob(10); n = 10, π = .5; so ((10x9)/2)*(510) + 10(510) + (510) = .055. The latter is the P-value for the test statistic (of 8 observed opposed to abortion letters) so with a significance level of 5%, the assumption of the proportion of anti abortion voters = ½ would not be rejected. sample sizes: 99% sure of public opinion w/in 2% points. $20 per persion interviewed: How much will the survey cost? .02 = 2.576*([(π(1-π))/n]. π = .5 so largest value (conservative) for n = 4,148, with cost estimated at $82,960.