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.
There is a positive relationship between number of calories and the amount of carbohydrates that starbucks food menu items contain
  1. In this scenario, what are the explanatory and response variables?
explanatory: calories

response: carbs
  1. Why might we want to fit a regression line to these data?
In order to predict carbs from calories
  1. Do these data meet the conditions required for fitting a least squares line?
Linearity: yes, the relationship is linear

Variability: the relationship appears to not have constant variability

Normal Residuals: the residuals appear to be nearly normal

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. The scatterplot below shows the relationship between height and shoulder girth (over deltoid muscles), both measured in centimeters.

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  1. Describe the relationship between shoulder girth and height.
There is a linear relationship between shoulder girth and shoulder height with constant variability of the response.
  1. How would the relationship change if shoulder girth was measured in inches while the units of height remained in centimeters?
The slope of the line would increase, since the magnitude of the predictor increases. 1 step to the right would result in a greater increase in the response.

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.
slope = (9.41/10.37)*0.67 = .607

intercept = 171.14 - slope*107.20 = 171.14 - 0.607*107.20 = 106.06

y = 0.607x + 106.06
  1. Interpret the slope and the intercept in this context.
Interpreting the slope and intercept: positive slope suggests positive relationship between girth and height. As girth increases 1, height increases .607.
  1. Calculate \(R^2\) of the regression line for predicting height from shoulder girth, and interpret it in the context of the application.
R2 = R^2 = 0.6079749^2 = .369

About 37% of the variability of the response is accounted for by this 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.
y = 0.607*100 + 106.6 = 167.3
  1. The student from part (d) is 160 cm tall. Calculate the residual, and explain what this residual means.
resid = 160 - 167.3 = -7.3

Residual is negative, meaning the actual datapoint is less than the model prediction
  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?
The original data set had a response variable values between 80 and 140 cm. A measure of 56 is outside the sample and we would require extrapolation and would not be appropriate.

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.

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  1. Write out the linear model.
summary(m_cats_hwt_bwt)
## 
## Call:
## lm(formula = cats$Hwt ~ cats$Bwt)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -3.5694 -0.9634 -0.0921  1.0426  5.1238 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  -0.3567     0.6923  -0.515    0.607    
## cats$Bwt      4.0341     0.2503  16.119   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.452 on 142 degrees of freedom
## Multiple R-squared:  0.6466, Adjusted R-squared:  0.6441 
## F-statistic: 259.8 on 1 and 142 DF,  p-value: < 2.2e-16
y = 4.0341x - 0.35
  1. Interpret the intercept.
Should a cat have a body weight of 0 kg, it would have a heart weight of -0.357 g, which doesn't make much intuitive sense since the value of 0 is outside our training data
  1. Interpret the slope.
As body weight increases by 1, we can expect eh size of the heart to increase by 4.0341
  1. Interpret \(R^2\).
with an R2 value of .6441, we can say that this model accounts for 64% of the variability of the response variable
  1. Calculate the correlation coefficient.
R = sqrt(R2) = .802

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.

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\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.
summary(m_eval_beauty)
## 
## Call:
## lm(formula = eval ~ beauty)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1.80015 -0.36304  0.07254  0.40207  1.10373 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  4.01002    0.02551 157.205  < 2e-16 ***
## beauty       0.13300    0.03218   4.133 4.25e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.5455 on 461 degrees of freedom
## Multiple R-squared:  0.03574,    Adjusted R-squared:  0.03364 
## F-statistic: 17.08 on 1 and 461 DF,  p-value: 4.247e-05
  1. Do these data provide convincing evidence that the slope of the relationship between teaching evaluation and beauty is positive? Explain your reasoning.
With a p-value < .05, we reject the null hypothesis and claim sufficient information to deternmine the relationship
  1. List the conditions required for linear regression and check if each one is satisfied for this model based on the following diagnostic plots.

Independent observations: survery ws conducted randomly and annonymously.

Nearly normal distribution residual. The histogram and the theoritical quantile show that the distribution is nearly normal.

Linearity is not strong

Constant variabilty. Variability is constant from the graphs