Questions

7.24

  • a.) There appears to be a weak positive relationship between the number of calories and amount of carbohydrates as evidenced by the points in the scatter plot not residing close to the linear model
  • b.) The explanatory variable is Calories and the response variable is carbohydrates
  • c.) A regression line is able to help predict the amount of carbohydrates in a food item for which we know the calories. Even though the regression line is not a perfect fit to the data, it is still a decent predictor as opposed to no predictor
  • d.) Linearity: the data does appear to follow a linearly increased trend despite being dispersed around the regression line Normality: the distribution appears nearly normal Constant Variability: Based on the middle chart, the points seem to hover around zero but as calories increase, so do the residual spread. Conclusion: The data fails to meet the conditions for a least squares line

7.26

  • a.)
xh <- 107.2
yh <- 171.14
sx <- 10.37
sy <- 9.41
r <- .67

# b1
b1 <- (sy/sx) * r

# b0
b0 <- yh - b1 * xh

Equation: y = 105.9650878 + 0.6079749 * x
* b.) Slope: represents the number of centimeters increase (0.6079749) in height for each increase in shoulder girth
Intercept: represents the height in centimeters at girth of 0cm (105.9650878)
* c.) The R2 is 0.4489 which means that 44.89% of the variation is accounted for
* d.)

height <- function(girth) b0 + b1 * girth
height(100)
## [1] 166.7626
  • e.)
yi <- 160
ei <- (yi - height(100)) %>%
  print
## [1] -6.762581

Since the residual is negative, this means that the actual data point is below the result of the linear regression model and that the model is overstating the value * f.) It would not be appropriate. The original dataset had values within 85 and 135 cm. Extrapolation would be necessary here

7.30

  • a.) \(\bar{y} = -.357 + 4.034x\)
  • b.) The -.357 intercept means that at 0kg of body weight, the heart would weight -.357g - an impossible scenario
  • c.) The slope of 4.034 means that the heart’s weight will increase by 4.034g for each 1kg of body weight
  • d.) An R^2 of 64.66% means that 64.66% of the variation is accounted for by the linear model
  • e.) The correlation coefficient is 0.8041144

7.40

  • a.)
b0 <- 4.010

x <- -.0883
y <- 3.9983

b1 <- ((y - b0)/x) %>%
  print
## [1] 0.1325028
  • b.) B1 > 0. For B1 to be >0, Sy and Sx must also be positive, therefore we can conclude the relationshpi between teaching evaluation and beauty will always be positive
  • c.) Linearity - there is no R^2 or correlation coefficient but there does appear to be a slightly linear positive relationship based on the scatterplot
    Constant variance - the residuals look randomly scattered around 0 and meet the condition
    Normal distribution - the data is slightly skewed left but otherwise very close to normal
    Independence - is assumed based on the lack of evidence pointing either way. 463 is also much smaller than the 10% of population of professors nationwide