7.24 Nutrition at Starbucks, Part I.

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

It seems there is a weak positive relationship between the number of calories and amount of carbohydrates .

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

Calories is explanatory and carbs is response.

c. why might we want to fit a regression line to these data?

Customers would like to kow how many Calories on average are coming from carbs.

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

No. The residuals are not constant as residuals increase as the calories increase.

7.26 Body measurements, Part III.

a.Write the equation of the regression line for predicting height.
avg_shoulder = 107.20 
sd_shoulder = 10.37 
avg_height = 171.14 
sd_height = 9.41 
cor = 0.67 

b1 <- (sd_height/sd_shoulder) * cor
b0 <- avg_height - b1 * avg_shoulder
b1
## [1] 0.6079749
b0
## [1] 105.9651

Equation: y= 0.6079749X + 105.9650878

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

Slope = 0.6079749 means every 1 cm increase in should girth, the height increase by 0.6079749 cm on average

Intercept = 105.9650878 means if a person has 0 cm in shoulder girth, the person should be 105.9561 cm tall on average.

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

R2=0.4489. 44.89% of the variation is accounter for should girth.

d. A randomly selected student from your class has a shoulder girth of 100 cm. Predict the height of this student using the model.
pred = b1*100+b0
pred
## [1] 166.7626
e. The student from part (d) is 160 cm tall. Calculate the residual, and explain what this residual means.
residual = pred - 160 
residual
## [1] 6.762581

Residual is 6.7625805. It means there is a 6.7625805 cm difference between the prediction and the actual observation.

7.30 Cats, Part I.

a. Write out the linear model.
b0 = -0.357
b1 = 4.034

y = 4.034X - 0.357

b. Interpret the intercept.

The intercept here means the off-set between the body mass and heart mass.

c. Interpret the slope

The slope means the heart weight increase 4.034 g over 1 kg increase in body weight.

d. Interpret R2.

64.66% of the heart weight variation is explained by the catโ€™s body mass.

e. Calculate the correlation coefficient.
r2 = 0.6466 
r = sqrt(r2)
r
## [1] 0.8041144

7.40 Rate my professor.

a. 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 <- -.0883
y <- 3.9983
b1 = (y-b0)/x
b1
## [1] 0.1325028
b.Do these data provide convincing evidence that the slope of the relationship between teaching evaluation and beauty is positive? Explain your reasoning.

The p-value=0.0000 mean the model is statistically significant. The slope seems large enough