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Instructor's Solutions Manual for Applied Regression Analysis and Other Multivariable Methods Chapter 5
f The line does not differ from the line plotted in (e). The evidence suggests that there is no significant linear relationship. Determining a well fitting line is difficult given the dispersion of the data.
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400 425 450 WGT (X) (grams)
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- a j9 0 = 19.882 $=-0.
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r=-0.76 P = 0. Since the P-value is > 0.05, we fail to reject HO and conclude there is not a significant linear relationship between wavelength and year.
b The 95% CI for Pi equals: /?, ±^_2A975 • SEp= -0.109 ± (2.101)(0.144) = (-0.41,0.19).
c See graph (below). 540
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490 1870 1880 1890 1900 1910 1920 1930 1940 YEAR (X)
Instructor's Solutions Manual for Applied Regression Analysis and Other Multivariable Methods Chapter 5
- a See graph (below). As X increases the dispersion of Y increases.
b J3G= -2.546 /?,= 0.032 Y = -2.546 + 0.032X
c See graph. This graph implies a better linear relationship between X and Y.
d J3Q= -6.532 $=1.430 Y = -6.532 + 1.43QAT
e The natural log transformation provides the best representation. The natural log plot illustrates the linear relationship better and the dispersion of the data is more similar at each level of toluene exposure. The first plot indicates that there may be a violation of homoscedasticity for the untransformed data.
100 200 300 400 500 600 700 800 900 1000 PPM_TOLU (X) LN_PPMTL (X)
- a See graph for plot. As house size increases (X) house price (Y) increases.
b J3 0 = -17.365 $=5.
c Y =-17.365 + 5. 582X. See graph for estimated line. The line fits the data well.
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10 15 20 25 HOUSE SIZE (X) (100s of sq. ft.)
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Instructor's Solutions Manual for Applied Regression Analysis and Other Multivariable Methods
c See graph. No it does not seem to fit the data very well.
Chapter 5
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T=-5.85 P< 0. Since the P-value is < 0.05, we reject Ho&nd conclude that the slope is not equal to zero. There is a significant linear relationship between marginal cost and output.
e A model containing a curvilinear term would better fit the data.
- a There may be a slight negative linear relationship.
b Y = J3 0 + J3 1 X + E. (^3) Q =76.008, #=-0.01 5. The baseline OWNEROCC = 76% and as OWNCOST increases by $1,000 the percentage of OWNEROCC decreases by ~ 2%.
c F=76.008-0.015X. See graph for line. The line fits the data well. 80
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200 400 600 800 1000 1200 1400 OWNCOST (X)
