Q: 7.24, 7.26, 7.30, 7.40

7.24

  1. Describe the relationship between number of calories and amount of carbohydrates (in grams) that Starbucks food menu items contain.

Ans

The relationship is a linear, positive , moderately strong relationship between number of calories and amount of carbohydrates

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

Explanatory variable: Calories on the x axis.

Response variable: Carbohydrates on the y axis.

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

So as to predict the amount of carbs a menu item has based on its calorie content.

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

Linearity: Yes the relationship between the calories and carbohydrates appear to be linear

Nearly normal Residuals: The residual plot displays a nearly normal histogram

Constant Variability: this relationship violates constant variabliliy

Therefore we conclude conditions are not met for fitting a least square line

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.

Ans

\[ y = \beta_0 + \beta_1 \cdot x \] \[ \beta_1 = \frac{S_y}{S_x}R \]

Sx <- 10.37
Sy <- 9.41

R <- 0.67

b1 <- (Sy / Sx) * R
b1
## [1] 0.6079749
x0 <- 107.2
y0 <- 171.14

b0 <- y0 - b1 * x0
b0
## [1] 105.9651

The equation for the regression line for predicting height will be:

y = 105.9650878 + 0.6079749 * x

  1. Interpret the slope and the intercept in this context. Explanatory variable - shoulder girth

Response variable - Height

the intercept is the estimated average height if the grith is 0

the slope is the estimated rate of per unit increase of height to a unit increase in shoulder grith

  1. Calculate R2 of the regression line for predicting height from shoulder girth, and interpret it in the context of the application.
R <-  0.67
R2 <- R^2
R2
## [1] 0.4489

44.89% of the variability in the height of perople is explained by the model

  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 = 105.9650878 + 0.6079749 * x
y
## [1] 166.7626

the predicted height of the student is 166.7626

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

the residual is 160 - 166.7626 = -6.7626

  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?

A measure of 56 is outside the sample and we would require extrapolation and would not be appropriate to predict height for children.

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.

\[\bar{y} = - 0.357 + 4.034 \cdot x \]

  1. Interpret the intercept.

The intercept will be at -0.357. This value tells us that this model will predict a negative heart weight when the cat’s body weight is 0.

  1. Interpret the slope

the slope of 4.034 tells us that the heart weight increases by 4.034 grams for each 1kg of body weight increase.

  1. Interpret R 2 .

the R2=64.66% tells us that the linear model describes 64.66% of the variation in the heart weight.

  1. Calculate the correlation coefficient. sqrt(R2) The correlation coefficient is 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

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.
b0 <- 4.010

x <- -0.0883
y <- 3.9983

b1 <- (y - b0) / x

The slope for 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.

since B1 os 0.13250 and looking at the slope is positive

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

Linearity: linearity is satisfied weakly

Nearly normal residuals: As shown in the residuals distribution and Q-Q plot, they are in fact nearly normal.

Constant variability: The scatterplot of the residuals does appear to have constant variability.

Independent observations: Assuming independence 463 professors would likely be < 10% of nationwide professors,and are rated individually