15: Linear Regression

1 / 16

# 15: Linear Regression - PowerPoint PPT Presentation

15: Linear Regression. Expected change in Y per unit X. Introduction (p. 15.1). X = independent (explanatory) variable Y = dependent (response) variable Use instead of correlation when distribution of X is fixed by researcher (i.e., set number at each level of X )

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about '15: Linear Regression' - ziva

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### 15: Linear Regression

Expected change in Y per unit X

Introduction (p. 15.1)
• X = independent (explanatory) variable
• Y = dependent (response) variable

when distribution of X is fixed by researcher (i.e., set number at each level of X)

studying functional dependency between X and Y

Illustrative data (bicycle.sav) (p. 15.1)
• Same as prior chapter
• X = percent receiving reduce or free meal (RFM)
• Y = percent using helmets (HELM)
• n = 12 (outlier removed to study linear relation)
How formulas determine best line(p. 15.2)
• Distance of points from line = residuals (dotted)
• Minimizes sum of square residuals
• Least squares regression line
Interpretation of Slope (b)(p. 15.3)
• b = expected change in Y per unit X
• Keep track of units!
• Y = helmet users per 100
• X = % receiving free lunch
• e.g., b of –0.54 predicts decrease of 0.54 units of Y for each unit X
Predicting Average Y
• ŷ = a + bx

Predicted Y = intercept + (slope)(x)

HELM = 47.49 + (–0.54)(RFM)

• What is predicted HELM when RFM = 50?

ŷ = 47.49 + (–0.54)(50) = 20.5

Average HELM predicted to be 20.5 in neighborhood where 50% of children receive reduced or free meal

• What is average Y when x = 20?

ŷ = 47.49 +(–0.54)(20) = 36.7

Confidence Interval for Slope Parameter(p. 15.4)

95% confidence Interval for ß =

where

b = point estimate for slope

tn-2,.975 = 97.5th percentile (from t table or StaTable)

seb = standard error of slope estimate (formula 5)

standard error of regression

Illustrative Example (bicycle.sav)
• 95% confidence interval for 

= –0.54 ± (t10,.975)(0.1058)

= –0.54 ± (2.23)(0.1058)

= –0.54 ± 0.24

= (–0.78, –0.30)

Interpret 95% confidence interval

Model:

Point estimate for slope (b) = –0.54

Standard error of slope (seb) = 0.24

95% confidence interval for  = (–0.78, –0.30)

Interpretation:

slope estimate = –0.54 ± 0.24

We are 95% confident the slope parameter falls between –0.78 and –0.30

Significance Test(p. 15.5)
• H0: ß = 0
• tstat (formula 7) with df = n – 2
• Convert tstat to p value

df =12 – 2 = 10

p = 2×area beyond tstat on t10

Use t table and StaTable

Regression ANOVA(not in Reader & NR)
• SPSS also does an analysis of variance on regression model
• Sum of squares of fitted values around grand mean = (ŷi – ÿ)²
• Sum of squares of residuals around line = (yi– ŷi )²
• Fstat provides same p value as tstat
Distributional Assumptions(p. 15.5)
• Linearity
• Independence
• Normality
• Equal variance
Validity Assumptions(p. 15.6)
• Data = farr1852.sav

X = mean elevation above sea level

Y = cholera mortality per 10,000

• Scatterplot (right) shows negative correlation
• Correlation and regression computations reveal:

r = -0.88

ŷ = 129.9 + (-1.33)x

p = .009

• Farr used these results to support miasma theory and refute contagion theory
• But data not valid (confounded by “polluted water source”)