7.24) a) The relationship between the number of calories and the amount of carbohydrates in grams in the Starbucks menu is positive and linear.
b) The explanatory variable is the number of calories and the response variable is the amount of carbohydrates in grams.
c) We might want to fit a regression line to this data because it would be enable us to predict the amount of carbohydrates in a food item for which we only know the calories. This would be helpful to individuals who are watching their carbohydrate intake and want to keep track of how much carbohydrates they consume.
d) The conditions for fitting a least squares regression line:
Linearity - Yes it fulfils this condition. The relationship between number of calories and amount of carbohydrates is fairly linear.
Nearly normal residuals - It isn’t perfect, but I think we can consider the residuals to be nearly normal.
Constant Variability - There is not constant variability in the data. The data is close to the regression line for low calorie items, but far from the regression line for high calorie items.
Independent Observations - The observations are independent.
The conditions for fitting a least squares regression line are not all met.
7.26) a) b1 = (sd height/sd shoulder_girth) * R
b1 = (9.41/10.37) * .67
b1 = 0.6079749
H = b0 + .608SG
171.14 = b0 + .608(107.20)
bo = 105.9624
H = 105.96 + .608SG
b) The intercept is the value of the height when shoulder girth is zero. You cannot have zero shoulder girth. This value just adjusts the height of the line and has no meaning by itself. The slope tells us that every 1 cm of increase in shoulder girth is associated with a .608 cm increase in height.
c) R^2 = .67^2 = 0.4489
About 45% of the variability in height is explained by shoulder girth.
d) H = 105.96 + 0.608SG
H = 166.76 cm
e) Residual = 160 - 166.76 = -6.76 cm
The residual is negative. This tells us that the model over-estimates height based on shoulder girth.
f) It would not be appropriate to use this model to estimate the height of a one year old with a shoulder girth of 56 cm because that would require extrapolation.
7.30) a) Heart Weight = -.357 + 4.034 * Body Weight
b) Intercept - When body weight is zero, heart weight is -.357 g. This doesn’t make sense. The y-intercept adjusts the height of the line.
c) Slope - Every kg of increase in body weight is associated with a heart weight increase of 4.034 g.
d) About 64.66% of the variability in heart weight is explained by body weight.
e) R = \(\sqrt{.6466}\) = 0.8041144
7.40) a) evaluation = b0 + b1*beauty
3.9983 = 4.010 + b1(-.0883)
b1 = .1325
b) Since the p value is so small (zero), we reject the null hypothesis and conclude that teaching evaluation and beauty are positively correlated.
c) Conditions for linear regression -
Linearity - Yes
Nearly normal residuals - It is slightly left skewed, but since the sample size is large, that is ok.
Constant Variability - Yes
Independent Observations - Yes