Chapter 7 - Introduction to Linear Regression Practice: 7.23, 7.25, 7.29, 7.39 Graded: 7.24, 7.26, 7.30, 7.40

7.24 Nutrition at Starbucks, Part I. The scatterplot below shows the relationship between the number of calories and amount of carbohydrates (in grams) Starbucks food menu items con- tain.21 Since Starbucks only lists the number of calories on the display items, we are interested in predicting the amount of carbs a menu item has based on its calorie content.

  1. Describe the relationship between number of calories and amount of carbohydrates (in grams) that Starbucks food menu items contain. The relationship is a direct poisitive linear correlation

  2. In this scenario, what are the explanatory and response variables? x(Carbs) is the explanatory, y(Calories) is the response

  3. Why might we want to fit a regression line to these data? To find whether or not high calorie foods are high in carbs, or to help suggest menu items for carb sensitive customers.

  4. Do these data meet the conditions required for fitting a least squares line? Yes, because the residual plot reveals increased frequency at higher calories and tighter bound errors at lower calories. Least squares should help us fit this accordingly.

7.26 Body measurements, Part III. 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) Write the equation of the regression line for predicting height. \[ \hat{height} = beta_0 - beta_1*\hat{shoulder girth} \] (b) Interpret the slope and the intercept in this context. \[ beta_1 = R * s_y/s_x \]

R <- 0.67
x_mean <- 107.20
y_mean <- 171.14
beta_1 <- 0.67 * (9.41/10.37)
beta_0 <- y_mean - (beta_1*x_mean)
  1. Calculate R^2 of the regression line for predicting height from shoulder girth, and interpret it in the context of the application.
R_squared <- R * R
  1. A randomly selected student from your class has a shoulder girth of 100 cm. Predict the height of this student using the model.
x_rand <- 100
y_rand <- beta_0 + beta_1*x_rand
cat("the predicted height of the student is:", y_rand, "cm" )
## the predicted height of the student is: 166.7626 cm
  1. The student from part (d) is 160 cm tall. Calculate the residual, and explain what this residual means. \[ e_i = y_i \hat{y}i \]
e_i <- 160 - y_rand
cat("the residual is:", e_i, "cm")
## the residual is: -6.762581 cm

A positive residual means that the model underestimates the height.

  1. A one year old has a shoulder girth of 56 cm. Would it be appropriate to use this linear model to predict the height of this child?

No, this calculation would require extrapolation.

7.30 Cats, Part I. The following regression output is for predicting the heart weight (in g) of cats from their body weight (in kg). The coefficients are estimated using a dataset of 144 domestic cats.

  1. Write out the linear model. -0.357 + 4.034 * body_wt
  2. Interpret the intercept. The negative intercept isn’t helpful as a data point, but tells us how to build our slope. A cat can’t have a negative heart rate.
  3. Interpret the slope. For each kg of body weight added, the heart weight rises by 4.034 grams

  4. Interpret R2. 64.66 % of the time, the linear model describes the variation in heart weight(g) by body weight(kg.)

  5. Calculate the correlation coefficient.

R <- sqrt(.6466)
cat("the correlation coefficient (R) =", R)
## the correlation coefficient (R) = 0.8041144

7.40 Rate my professor. Many college courses conclude by giving students the opportunity to evaluate the course and the instructor anonymously. However, the use of these student evalu- ations as an indicator of course quality and teaching effectiveness is often criticized because these measures may reflect the influence of non-teaching related characteristics, such as the physical appearance of the instructor. Researchers at University of Texas, Austin collected data on teaching evaluation score (higher score means better) and standardized beauty score (a score of 0 means average, negative score means below average, and a positive score means above average) for a sample of 463 professors.24 The scatterplot below shows the relationship between these variables, and also provided is a regression output for predicting teaching evaluation score from beauty score.

  1. Given that the average standardized beauty score is -0.0883 and average teaching evaluation score is 3.9983, calculate the slope. Alternatively, the slope may be computed using just the information provided in the model summary table.
x_mean <- -.0883
y_mean <- 3.9983
beta_0 <- 4.010
beta_1 <- (y_mean - beta_0)/x_mean

cat("The slope is:", beta_1)
## The slope is: 0.1325028
  1. Do these data provide convincing evidence that the slope of the relationship between teaching evaluation and beauty is positive? Explain your reasoning. The relationship is weak; the slop is near 0 and the scatter plot shows that as well. These data show a lack of sufficient evidence that there is a relationship between teaching eval and beauty.
  2. List the conditions required for linear regression and check if each one is satisfied for this model based on the following diagnostic plots. When fitting a least squares line, we generally require
  1. Linearity. The data should show a linear trend. If there is a nonlinear trend (e.g. left panel of Figure 7.13), an advanced regression method from another book or later course should be applied. There is linearity; figure 3 (The normal q_q plot shows normal distribution)
  2. Nearly normal residuals. Generally the residuals must be nearly normal. figure 1, figure 4
  3. Constant variability. The variability of points around the least squares line remains roughly constant.