Chapter 7 Homework Problems: 7.24, 7.26, 7.30, 7.40

7.24: Nutrition at Starbucks, Part 1

  1. As the amount of calories increases in a Starbucks drink, the number of carbohydrates most certainly increases as well, but the spread of the values is significantly wider as the values increase. We don’t seem to have constant variability here.

  2. Calories is the explanatory variable, carbohydrates is the response variable.

  3. We want to fit a regression line to this data to get a picture of how the values are trending when placed against one another.

  4. No, constant variability is not met.

7.26 Body Measurements, Part III

  1. \(b_1 = R * s_y/s_x\)

\(b_1 = 0.67 * (10.37/9.41) = 0.7383528\)

\(\bar{x},\bar{y}: \bar{y} = b_0 + b_1 * \bar{x}\)

Plug in x bar and y bar and b1, solve for b0

\(b_0 = 91.98857984\)

\(\hat{y} = 91.98857984 = 0.7383528 * shoulder girth\)

  1. For each additional cm in shoulder girth, the model predicts an additional 0.7383528 cm in height. When the shoulder girth is 0, the height is expected to be 91.98857984 cm. We know it doesn’t make sense for shoulder girth to be 0 in any context, so this is essentially meaningless by itself.

  2. \(R^2 = 0.67^2 = 0.4489\) About 44.89% of the variability in height is explained by the shoulder girth.

  3. 91.98857984 + 0.7383528 * 100 = 165.82385984 We would predict that this student is 165.82385984 cm tall.

  4. The residual is 5.82385984. This means our model was off by this amount. It is the difference between the observed value and the predicted value.

  5. It would not be appropriate to use this linear model to predict the height of this child. If we did move forward with the calculation, we would predict that the one year old is 133 cm tall, which is impossible at one year old.

7.30 Cats, Part 1

  1. \(\hat{y} = -0.357 + 4.034 * heart weight\)

  2. When body weight is 0, height is expected to be -0.357. A heart cannot weight negative pounds, so this value makes no sense out of context.

  3. For each additional g in heart weight, the model predicts an additional 4.034 kg in body weight.

  4. With an \(R^2 = 0.6466\), we can say that our model can accurately predict heart weight 64.66% of the time. It is the movement in the independent variable that can be explained by the dependent variable.

  5. The correlation coefficient is 0.8041144.

7.40 Rate my professor

  1. \(b_1 = \frac{(3.9983 - 4.010)}{-0.0883} = 0.1325\)

  2. Yes, there is strong, convincing evidence that the slope of the relationship between teaching evaluation and beauty is positive. In context this would mean that teaching evaluations increase as beauty increases. The same can be said for decreasing.

  3. Conditions for linear regression:
    Linearity: Yes
    Nearly Normal Residuals: Yes
    Constant Variability: Yes

0.67 * (10.37 / 9.41)
## [1] 0.7383528
sqrt(0.6466)
## [1] 0.8041144