HW7

Answer

There are a moderate to weak positive relationship between the number of calories and the amount of carbohydrates. It appears from regression line and as weel from residual plot that datapoints are moving away from the regression line as Calories increasing and making the error higher for the model.

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

Answer

carbohydrates is response variable (Y axis), and the explanatory variable is Calories on the x axis.

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

Answer

The regression line helps to predict the amount of carbohydrate (in grams) a starbucks menu food items contains by looking at the number of calories in the item.

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

Answer

Linearity: The data should show a linear trend. With the help of the scatterplot we can see that data shows a moderate to weak linear relationship.

Nearly normal residual: From the histogram for residual, the residuals in this case have a slightly left skewed.

Constant variability: The residual plot suggests that there is no constant variability for residuals. The data fit the linear model much better for lower number of calories than for higher as shown by much larger residuals value.

Independent observation: Its being the item from Menu , asuamed to be independent.

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.

The general equation for the regression line: y = B0 + B1*x where B0 is yintercept, and B1 is the slope, represent.

# Shoulder girth is the explanatory variable
shoulder.mean <- 107.20 # in cm
shoulder.SD <- 10.37 # in cm


# Height is the response
height.mean <- 171.14 # in cn
height.SD <- 9.41 # in cm

# R for correlation
R <- 0.67

# Calculate the slope (or otherwise known as B1)
B1 <- R * (height.SD/shoulder.SD)

# Now to calculate for B0, we will use the values (x,y) = (107.20, 171.14). They are also the mean values, and they lie along the regression line.
# Now to rearrange the equation to solve for B0. B0 = y - B1*x
B0 <- 171.14 - B1 * 107.20

B1;B0

## [1] 0.6079749

## [1] 105.9651

height= 105.9651+0.6079749 * shoulder_girth

2. Interpret the slop and the intercept in this context b1.

Answer

The intercept (105.9651) is the height in centimeters at girth of 0 cm. The slope (0.6079749) is the number of centimeters increase in height for each increase in shoulder girth. Not all values make sense when we plug them into this linear regresion equation. For example, a person with a shoulder width of 0 cm (which can’t happen) would indicate a height of 105.965 cm.

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

Answer

R.squared <- R^2
paste("R squared: ", round(R.squared,3))

## [1] "R squared:  0.449"

This means that only 0.449 of the variation found in this data is explained by the linear model. Rsqurare is max when its 1 (best fit) , 0 (worst fit). This r^2 doesn’t give good indication of the related variables in model.

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

Yhat = bo + bi X

st.shoulder <- 100 # in cm
st.height <- B0 + B1 * st.shoulder
paste("Estimated height of a student with a shoulder girth of 100 cm is: ", round(st.height,3), "cm.")

## [1] "Estimated height of a student with a shoulder girth of 100 cm is:  166.763 cm."

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

Answer

actaul_value <- 160
predicated <- st.height

residual <- actaul_value - predicated

paste("The residual is: ", round(residual,3))

## [1] "The residual is:  -6.763"

This residual is the difference between the actaul value and predicated value.

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

x= 1 lets find Yhat i.e height

# Yhat = B0 + B1 X
X <- 1
yhat1 <- B0 + B1 * X

yhat1

## [1] 106.5731

yhat56 <- 56
x56 <- (yhat56 - B0)/B1
x56

## [1] -82.18281

residual <- yhat56 - yhat1
residual

## [1] -50.57306

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. We can also see the residual is neagtive, and negative value for a residual it means the actual value was LESS than the predicted value.

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.

1. Write out the linear model.

Answer

yhat = B0 + B1 * x, B0 <- yintercept B1 <- Slope Heart Weight (g) = -0.357 + 4.034 * `Body Weight (kg)

2. Interpret the intercept.

Answer

negative intercept of -0.357 means that for weight of 1 kg , hert weight will be -0.357 + 4.034 = 3.67 grams. But for ZERO Value of weight , Hear weight = -0.357

3. Interpret the slope.

Answer

For an increment of 1 kg in body weight in a cat, the heart weight would increase by 4.034 g.

4. Interpret R2.

Answer

R2 = 64.66%, it means that linear model would be able to explain ~65% variability in the heart weight (in g) of the cat.

5. Calculate the correlation coefficient.

Answer

The correlation coefficient is R, we know r2, so R = sqrt(R) = 8.0411442

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 evaluations as an indicator of course quality and teaching e???ectiveness 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.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

From Table we know Yintercept (Bo)= 4.010, We knwo response varible (average teaching evaluation score) Yhat = 3.9983 and explanatory value (average standardized beauty score ) X= -0.0883

3.9983 = 4.010 + B1 * (-0.0883)

B1 <- (3.9983 - 4.010)/(-0.0883)
paste("B1 or the slope: ", round(B1, 3))

## [1] "B1 or the slope:  0.133"

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

Answer

Scatterplot doesn’t show any correlation between beauty and teaching evaluation.But the summary table for P value shows ~ 0, this is clear indication of that there is a relation between these variables.

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 : From the scatterplot of the data, there may be a weak positive linear relationship. Normal Q-Q Plot shows a linear relation.

Nearly normal residuals : The residuals do appear to be nearly normal, by looking at its distribution on the histogram.

Constant variability : By looking at the residual plots, there appears to be constant variability throughout th plot.

Independent observations : We do not have much information on how the data sample was collected beyond the fact that it was collected for 463 professors. We might assume independence of observations.