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.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.

Answer: The relation b/w number of calories and amount of carbohyrates is linear.

  1. In this scenario, what are the explanatory and response variables? Answer: In this scenario, the explanatory variable is Calories and response variable is Carbohydrates

  2. Why might we want to fit a regression line to these data? Answer: We want to fit a regression line to these data to find a better least squares line which has minimal residuals

  3. Do these data meet the conditions required for fitting a least squares line? Answer: Not all the conditions are met for fitting a least squares line. A simple linear model is inadequate for modeling these data, since the residuals are not normal.

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.
sx <- 10.37
sy <- 9.41
R <- .67
b1 <- R * sy/sx

xmean <- 107.2
ymean <- 171.14
b0 <- ymean - b1 * xmean
  1. Interpret the slope and the intercept in this context.

Answer: b0 (105.965) represents the intercept and b1(0.608) represents the slope.

  1. Calculate R2 of the regression line for predicting height from shoulder girth, and interpret it in the context of the application.
RSq <- R^2
  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 <- 100
y <- b0 + b1*x
y
## [1] 166.7626
  1. The student from part (d) is 160 cm tall. Calculate the residual, and explain what this residual means.
y <- 160
R <-  y - ymean
R
## [1] -11.14
  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?

Answer: No, since the girth range doesn’t fall under the earlier given data range, we cannot use the linear model.

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.

\[ \hat{heartWt } = -.357 + 4.034 * bodywt \]

  1. Interpret the intercept.

    Answer: The intercept is -.357, which is not meaningful since a negative weight doesn’t make sense.

  2. Interpret the slope.

Answer: The slope is 4.034, which says that for every percentage increase in body weight there is 4.034 times increase in the heart weight in gms.

  1. Interpret R2.

Answer: The R2 value is given as 64.66%, which indicates the variability in heart weight measures.

  1. Calculate the correlation coefficient. Answer: \[ \sqrt{R2 } = \sqrt{.6466} = 0.8041144 \]

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 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.
teachingMean <- 3.9983
beautyMean <- -.0883
intercept <- 4.010

#teachingMean <- intercept + Slope*beautyMean
slope <- (teachingMean - intercept) / beautyMean
slope
## [1] 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.

Answer: Since the slope 0.1325028 is positive, it tends to be a positive relation b/w teaching evaluation and beauty. But at the same time, the slope is not extremely high, so we cannot say that beauty drastically impacts the teaching evaluation.

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

Answer: The below conditions have to be satisfied
1. Normal residuals: Yes, the residuals seems to be near normal.
2. Linearity: Yes, the data doesn’t seem to follow a linear pattern.
3. Constant Variabiity: The variability of the points from least squares lines can be assumed(from the scatter plot) to be near equal.
4. Independent Observations: Yes, the observations seem to be independent.