Data606

1.Nutrition at Starbucks, Part I:

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

Ans) The explanatory variable is Calories and the response variable is carbohydrates.

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

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

Ans) Linearity: The data does seem to linear.

Nearly normal residuals: Residuals are normal.

Constant variability: There isn’t constant variabilty, The residual chart shows that the residuals are scattered more as calories increase.

The data meets the conditions required for fitting a least squares line

2.Body measurements, Part I:

(a) Describe the relationship between shoulder girth and height.

Ans) It seems to be linear and positive relationship exist between shoulder girth and height.

(b) How would the relationship change if shoulder girth was measured in inches while the units of height remained in centimeters?

Ans)The relationship would become more linear and strong.Because residuals become more closer to each other if shoulder girth was measured in inches while the units of height remained in centimeters

3. Body measurements, Part III:

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

Ans)

# calculate slope
(B1 <-  0.67 * (9.41/10.37))

## [1] 0.6079749

# calculate intercept
(B0 <- 171.14 - B1 * 107.2)

## [1] 105.9651

The equation for the regression line is y^ = 105.97 + 0.61x

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

Ans) The slope shows the increase in height as shoulder girth increases. The incerpet is the height with a shoulder girth of 0.

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

Ans)

(R2 <- .67^2)

## [1] 0.4489

This shows that this linear model explains 44.89% of the variation of the height data.

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

Ans)

(height <- B0 + B1 * 100)

## [1] 166.7626

I would predict the students height to be 166.76 cm based on the model.

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

Ans)

(residual <- 160 - height)

## [1] -6.762581

The residual is -6.76. A negative residual means the student’s height is -6.76 cm less than the expected height based on this model.

(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?

Ans) we see that the lowest shoulder girth was 85cm. We wouldn’t use our regression model to estimate this child’s height since he or she is out of our range.

4. Cats, Part I:

(a) Write out the linear model.

Ans) Heart_wt=−0.357+4.034(Body_wt)

(b) Interpret the intercept.

Ans) The intercept is nonsensical because a cat can’t have 0 weight, and negative weight is also not possible. The intercept only serves to adjust the position of the line.

(c) Interpret the slope.

Ans) For every additional 1 kg in body weight, the heart will weigh an additional 4.034 grams on average

(d) Interpret R2

Ans) 65 percent of the variance in heart weight is explained by body weight.

(e) Calculate the correlation coefficient

Ans)

r <- sqrt(64.66/100)

The correlation coefficient is 0.804.

5.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.

Ans)

# calculate intercept
y <- 3.9983
x <- -.0883
beta_0 <- 4.01
# calculate slope
(beta_1 <- (y-beta_0)/x)

## [1] 0.1325028

The slope is 0.132.

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

Ans) These data does show that the slope of the relationship between teaching evaluation and beauty is positive. With the p-value being close to 0. we can conclude the slope to be positive.

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

Ans) Linearity: There is no pattern in the residuals or in the data. This is confirmed

Normality of residuals: The residuals are appropriatly scattered and the histogram and qq plots are approximately normal. This is confirmed

Constant Variance of residuals: There appears to be no difference in variance at the lower levels of professor rating compared with the upper. This is confirmed

Indepednance of observations: There is no pattern in the residuals based on order of data collection. This is confirmed

Data606_HW8

Vijaya Cherukuri

11/1/2019

1.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.

Ans)It seems to be weak positive correlation between number of calories and amount of carbohydrates.Because residuals are not that much closer to the line.

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

Ans) The explanatory variable is Calories and the response variable is carbohydrates.

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

Ans)People on low carb diets may be interested in knowing how many of the calories in a menu item.Even though the regression line is not a perfect fit to the data, it is still a decent predictor

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

Ans) Linearity: The data does seem to linear.

Nearly normal residuals: Residuals are normal.

Constant variability: There isn’t constant variabilty, The residual chart shows that the residuals are scattered more as calories increase.

Independence: Each item on the menu should be independent of each other.

The data meets the conditions required for fitting a least squares line

2.Body measurements, Part I:

(a) Describe the relationship between shoulder girth and height.

Ans) It seems to be linear and positive relationship exist between shoulder girth and height.

(b) How would the relationship change if shoulder girth was measured in inches while the units of height remained in centimeters?

Ans)The relationship would become more linear and strong.Because residuals become more closer to each other if shoulder girth was measured in inches while the units of height remained in centimeters

3. Body measurements, Part III:

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

Ans)

The equation for the regression line is y^ = 105.97 + 0.61x

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

Ans) The slope shows the increase in height as shoulder girth increases. The incerpet is the height with a shoulder girth of 0.

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

Ans)

This shows that this linear model explains 44.89% of the variation of the height data.

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

Ans)

I would predict the students height to be 166.76 cm based on the model.

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

Ans)

The residual is -6.76. A negative residual means the student’s height is -6.76 cm less than the expected height based on this model.

(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?

Ans) we see that the lowest shoulder girth was 85cm. We wouldn’t use our regression model to estimate this child’s height since he or she is out of our range.

4. Cats, Part I:

(a) Write out the linear model.

Ans) Heart_wt=−0.357+4.034(Body_wt)

(b) Interpret the intercept.

Ans) The intercept is nonsensical because a cat can’t have 0 weight, and negative weight is also not possible. The intercept only serves to adjust the position of the line.

(c) Interpret the slope.

Ans) For every additional 1 kg in body weight, the heart will weigh an additional 4.034 grams on average

(d) Interpret R2

Ans) 65 percent of the variance in heart weight is explained by body weight.

(e) Calculate the correlation coefficient

Ans)

The correlation coefficient is 0.804.

5.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.

Ans)

The slope is 0.132.

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

Ans) These data does show that the slope of the relationship between teaching evaluation and beauty is positive. With the p-value being close to 0. we can conclude the slope to be positive.

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

Ans) Linearity: There is no pattern in the residuals or in the data. This is confirmed

Normality of residuals: The residuals are appropriatly scattered and the histogram and qq plots are approximately normal. This is confirmed

Constant Variance of residuals: There appears to be no difference in variance at the lower levels of professor rating compared with the upper. This is confirmed

Indepednance of observations: There is no pattern in the residuals based on order of data collection. This is confirmed