Nutrition at Starbucks, Part I. (8.22, p. 326) 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.

Solution: It looks like a linear relationship where the carbohydrates increases as the calories increases.

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

Solution: Carbohydrate is the responses variable while calories is the explanatory variable

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

Solution: To predict the amount of carbs based on the number of calories consumed

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

Solution: Linearity: The data seems to be linear. Normal residuals: Residuals tends to be normal. Constant variability: There isn’t constant variabilty. Independence: Observations should be independent of each other.

Body measurements, Part I. (8.13, p. 316) Researchers studying anthropometry collected body girth measurements and skeletal diameter measurements, as well as age, weight, height and gender for 507 physically active individuals.19 The scatterplot below shows the relationship between height and shoulder girth (over deltoid muscles), both measured in centimeters.

\begin{center} \end{center}

  1. Describe the relationship between shoulder girth and height.

Solution: There is a positive linear relationship between shoulder girth and height.

  1. How would the relationship change if shoulder girth was measured in inches while the units of height remained in centimeters?

Solution: Changing the units for one of the variables will not change the direction or strength of the relationship between the two variables.

Body measurements, Part III. (8.24, p. 326) Exercise above 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.

Solution:

x <- 10.37
y <- 9.41
R <- 0.67

b1 <- (y / x) * R
b1
## [1] 0.6079749
xhat <- 107.2
yhat <- 171.14

b0 <- yhat - b1 * xhat
b0
## [1] 105.9651

The equation: y’ = 105.97 + 0.61x

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

Solution: The slope shows the increase in height as shoulder girth increases.

  1. Calculate \(R^2\) of the regression line for predicting height from shoulder girth, and interpret it in the context of the application.

Solution:

R2 <- .67^2
R2
## [1] 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.

Solution:

h <- function(x) {
y  <- 105.97 + 0.61*x
return(y)
}

h(100)
## [1] 166.97

The predicted student height is 166.76 cm.

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

Solution:

r <- 160 - h(100)
r
## [1] -6.97

The residual is -6.97 which indicates that te student height is less than expected 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?

Solution:

No. It will not be appropriate to estimate this child’s height using the model since the height is out of the range.

Cats, Part I. (8.26, p. 327) 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.

\begin{center} \end{center}

  1. Write out the linear model.

Solution: Hwt=−0.357+4.034(Bwt)

  1. Interpret the intercept.

Solution: Not applicable

  1. Interpret the slope.

Solution: For additional 1 kg in body weight, the heart will weigh an additional 4.034 grams

  1. Interpret \(R^2\).

Solution: 65 percent of the variance in heart weight is explained by body weight.

  1. Calculate the correlation coefficient.

Solution:

r <- sqrt(64.66/100)
r
## [1] 0.8041144

The correlation coefficient is 0.8041

Rate my professor. (8.44, p. 340) 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. The scatterplot below shows the relationship between these variables, and also provided is a regression output for predicting teaching evaluation score from beauty score.

\begin{center}

\end{center}

  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.

Solution:

y <- 3.9983
x <- -0.0883
beta0 <- 4.01

#Slope
beta1 <- (y - beta0)/x
beta1
## [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.

Solution:

The slope beta1 is positive at 0.1325 indicating a positive correlation between beauty and teaching

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

Solution:

Linearity: There is n pattern in the data which shows that the data appear to be linear. Normal Residuals: The residuals appear to be nearly normal. Constant Variability: The obsrvations seem to be evenly spread out indicating constant variability. Independent Observations: Observations seems to be independent of each other.