Data 606 Homework Chapter 7 - Introduction to Linear Regression

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.

Answer: The relationship appear to be moderately strong and positive.

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

Answer: amount of carbohydrates is the response and calories is the explanatory variable.

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

Answer: We might want to fit a linear regression because there is a linear relationship between the two variables and we can make prediction from the linear model.

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

Answer: Yes these data meet the conditions required for fitting a least squares 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.

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

Answer: \(\hat{height}_{i} = 126.1503 + 0.4196798 * girth_{i}\)

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

Answer: On average height of a person is 126.1503 when shoulder girth is not taken into account. For 1 unit increase in shoulder girth height increases by 0.4196798 unit.

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

0.67^2

## [1] 0.4489

\(R_{2}\) = 0.4489 this means that 44.89% of the variability in height can be explained by shoulder girth.

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

Answer:

# slope
slope <- 0.67*(107.2/171.14)
estimated100 <- 126.1503 + slope*100
estimated100

## [1] 168.1183

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

Answer:

160 - 168.1183

## [1] -8.1183

Negative residual here means the model overestimated for this student.

6. 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: It won’t be appropriate to use linear model to predict the height of this child.

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. (Intercept) body wt Std. Error 0.692 -0.515 Estimate -0.357 4.034 t value 0.250 16.119 Pr(>|t|) 20 0.607 0.000 R2 = 64.66% R2 = 64.41% adj 15 10 5 s = 1.452 (a) Write out the linear model.

Answer: \(\hat{heartwt}_{i} = -0.357 + 4.034 * bodywt_{i}\)

2. Interpret the intercept.

Answer: Holding other things constant average heart weight will be -0.357 grams when bodywt is 0.

3. Interpret the slope.

Answer: For 1 unit increase in weight heart weight increase by 4.034 grams on average.

4. Interpret R2. Answer: bodywt explains 64.66% variability in heart weight.

5. Calculate the correlation coefficient.

Answer:

sqrt(0.6466)

## [1] 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 ap- pearance 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.

Answer: The slope is 0.1325028.

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

Answer: Yes. Since the p value of the slope intercept is significant at 5% level of significant.

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

Answer: Linearity: There we don’t see any clear linear relationship. Residual Normality: Residuals are normally distributed. Constant variability: The residuals satisfy constant variability assumption.