(a) The relationship between the two variables appears to be positive and moderately strong.
(b) In this scenario, the explanatory variable is calories, and the response variable is carbohydrates.
(c) We might want to fit a regression line to this data to see how well calories can predict carbohydrates at Starbucks.
(d) Yes. The data meets all three conditions required for fitting a least-squares line.
Linearity: There does not appear to be a clear pattern of the residuals, suggesting that it may be due to a linear relationship.
Nearly normal residuals: The residuals approach the normal distribution.
Constant variablity: There appear to be roughly the same degree of residuals above and below the horizonal line, suggesting constant variability.
Exercise 7.15 introduces data on shoulder girth and height of a group of individuals. The mean shoulder girth is 107.20 cm with a standard deviation of 10.37 cm. The mean height is 171.14 cm with a standard deviation of 9.41 cm. The correlation between height and shoulder girth is 0.67.
(a) \(\hat{height} = 0.4196798 \times\)
shoulder girth\(+ 126.1503\)
# Coordinates from mean
x.26 <- 107.2
y.26 <- 171.14
# Correlation coefficient
R.26 <- 0.67
# Slope
b1.26 <- R.26 * (x.26 / y.26)
b1.26## [1] 0.4196798# Intercept
b0.26 <- (b1.26 * x.26 - y.26)*-1
b0.26## [1] 126.1503(b) Interpretations:
Slope: The slope predicts that for every 10 cm of shoulder girth that a person gains, they will gain about 130 cm of height.
Intercept: The intercept is the average height of an individual with 0 shoulder girth. This does not make sense in this context because 0 is not meaningful – a person cannot have no shoulders.
(c) \(R^2 = 0.67^2 = 0.4489\). This means about 45% of the variability in height can be explained by shoulder girth.
(d) For a shoulder girth of 100 cm, the model predicts a height of about 168 cm.
yhat.26 <- b1.26 * 100 + 126.1503
yhat.26## [1] 168.1183(e) The residual is \(160 - 168 = -8 cm\). A negative residual means that the model overestimates the height of the individual.
(f) It would not be appropriate to use this model to calculate the height of a 1 year old. It would be extrapolation, since it is outside of the model’s range.
(a) \(\hat{heartwt} = 4.034 * bodywt - 0.357\)
(b) The intercept means that for a cat with a body weight of 0 grams, their heart weight would be \(-0.357\) grams. This is not meaningful – it is impossible for a cat to have no body weight.
(c) The slope means that for every 1 gram of body weight that a cat gains, they will gain 3.677 grams of heart weight.
(d) The \(R^2\) means that \(64.66\)% of the variability in heart weight is explained by body weight.
(e) Correlation coefficient \(= \sqrt{R^2} = \sqrt{0.6466} \approx 0.8041\)
(a) The slope is 0.1325028.
# Given values
x.40 <- -0.0883
y.40 <- 3.9983
b0.40 <- 4.010
# Slope
b1.40 <- (y.40 - b0.40) / x.40
b1.40## [1] 0.1325028(b) The p-value is so small that we reject the null hypothesis and conclude that beauty and teaching evaluation are not positively correlated.
(c) No, the conditions for linear regression are not all met. It does not meet the condition of nearly-normal residuals.
Linearity: There does not appear to be a clear pattern of the residuals, suggesting that it may be due to a linear relationship.
Nearly normal residuals: The residuals apear to be slightly left-skewed, based on the histogram and normal plot.
Constant variablity: There appear to be roughly the same degree of residuals above and below the horizonal line, suggesting constant variability.