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## Welcome to CUNY DATA606 Statistics and Probability for Data Analytics 
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library(knitr)

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 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.

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

The relationship between number of calories and amount of carbs is not a strong linear relationship.

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

Explanatory variable is calories. Response variable is carb.

(c) Why might we want to fit a regression line to these data?

To predict values of response variable from the explanatory variable. Here the response variable is Amount of Carbs and number of calories is the explanatory variable.

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

Linear: Yes but there is a weak linear relationship.

Residual: yes but the histogram is not completely symmetrical and does not seems to be nearly normal.

Constant variability: No, Plot suggests there is no constant variability.

Independent observations: Yes, Observations are independent

Overall “NO” the data doesnt 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.

(a) Write the equation of the regression line for predicting height.

Answer:

x = (sdh/sdshgirth) * CR

x = (9.41/10.37) * .67

x = 0.6079749

H = x0 + 0.6079749*shoulder_girth

171.14 = x0 + 0.6079749(107.20)

x0 = 105.96509072

H = 105.96509072 + 0.6079749*shoulder_girth

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

Slope: Every additional centimeter of shoulder girth the average height increases by 0.6079749 centimeters. Intercept: For a shoulder girth of 0 centimeters, the average height is 105.96509072 centimeters

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

R^2 = .67^2 = 0.4489

44.9% (0.4489) of the variability in height is explained by shoulder 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.

shoulder_girth = 100 cm

Height = 105.96509072 + 0.6079749*shoulder_girth

Height = 166.93999072 cm

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

Residual = 160 - 166.93999072 = -6.93999072 cm

The residual is negative, so model over estimates height based on shoulder girth.

(f) 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?

It would not be good to use this model to estimate the height of a one year old with a shoulder girth of 56 cm because that would require extrapolation.

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.

a) Write out the linear model.

\({\beta}_0 = -0.357\), \({\beta}_1 = 4.034\), Linear model: \(\hat{y}_h = {\beta}_0 + {{\beta}_1}x = \hat{y}_h = -0.357 + 4.034x\)

(b) Interpret the intercept.

\({\beta}_0 = -0.357\), translates to cat with body weight of zero kg will have heart weight of -0.357 g. This cannot be true as weight cannot exist in negative numbers. As this observational data, caution must be observed while using model in the places it has not been analyzed.

(c) Interpret the slope.

Every kg of increase in body weight is associated with a heart weight increase of 4.034 g.

(d) Interpret R2.

64.66% of the variability in heart weight is explained by body weight.

(e) Calculate the correlation coefficient.

\(R^2 = 64.66\%\), it means 64.66 percent of observed data can be explained using liner model: \(\hat{y}_h = -0.357 + 4.034x\).

Appendix

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