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 contain.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.

Ans: A positive linear relationship.

  1. In this scenario, what are the explanatory and response variables?

Ans: Calories is explanatory, and Carb is response variables.

  1. Why might we want to fit a regression line to these data?

Ans: We want to know how much carb. may produce calories.

  1. Do these data meet the conditions required for fitting a least squares line?

Ans: No, since the residual plot actually shows a nonline relationship. There is not a contradiction: residual plots can show divergences from linearity that can be difficult to see in a scatterplot.

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.

  1. Write the equation of the regression line for predicting height.

Ans: b1=0.67 x 10.37/9.41 = 0.738 , y=b0+0.738x –> b0= 92.29 y=92.29 + 0.738x

  1. Interpret the slope and the intercept in this context.

Ans: Slope:For each additonal shouder girth, the model predicts an additional 0.738 cm in height. Intercept: When shoulder girth is 0, the height is expected to be 92.29 cm.

  1. Calculate R2 of the regression line for predicting height from shoulder girth, and interpret it in the context of the application.

Ans: sqrt(R) = sqrt(0.67) = 0.4489

  1. A randomly selected student from your class has a shoulder girth of 100 cm. Predict the height of this student using the model.

Ans: 92.29 + 0.738 x 100 = 166.09cm

  1. The student from part (d) is 160 cm tall. Calculate the residual, and explain what this residual means.

Ans: ???6.09cm(= 160 ??? 166.09 ) means the model may overestimate 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?

Ans: 92.29 + 0.738 ??? 56 = 133.62cm

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 coecients are estimated using a dataset of 144 domestic cats.

  1. Write out the linear model.

Ans: heartWeight = ???0.357 + 4.034 ??? bodyWeight

  1. Interpret the intercept.

Ans: Expected heart weight of cat with no body weight is -0.357 g.

  1. Interpret the slope.

Ans: For every additional kg in body weight, there is 4.034 g additional to heart weight.

  1. Interpret R2.

Ans: Body weight level expalined 64.66% of heart weight.

  1. Calculate the correlation coecient.

Ans: sqrt(0.6466) = 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 evaluations as an indicator of course quality and teaching e???ectiveness 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 370 CHAPTER 7. INTRODUCTION TO LINEAR REGRESSION 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.

Ans: 3.9983 = 4.01 + b1 x (???0.0883) => b1 = 0.1325

  1. Do these data provide convincing evidence that the slope of the relationship between teaching evaluation and beauty is positive? Explain your reasoning.

Ans: No. Teh scatterplot shows nonlinear relation.

  1. List the conditions required for linear regression and check if each one is satisfied for this model based on the following diagnostic plots.

Ans: The following four condiions required for linear regression: linear, nearly normal residuas, Constant variability and independent observation.