# Reading Computer Printouts

```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 &gt; |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&gt;F If bigger than .05, all the coefficients are probably zero so your explanatory variables are independent of
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) &amp; 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=&plusmn; 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 &frac12; sY above its mean; since r&lt;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 &amp; 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&gt;.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 = &frac12; 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.
```