Practice Questions 9.3

If a point gives the middle of a column of data values, what gives the middle of a scatterplot?

A) A line going through the middle of the scatterplot.
B) A set of points clustered around the middle of the scatterplot (x̄, ȳ).
C) A plane fitting over the scatterplot.
D) The crossing point of a line connecting the rightmost and leftmost points with a line connecting the highest and lowest points.

What does the statistical method of linear regression do?

A) It finds the endpoints of the middle line.
B) It finds the coordinates of points (x,y) on the middle line.
C) It finds the values of the coordinates of the point (x̄,ȳ).
D)  It finds the equation (slope and y-intercept) of the middle line.

How many of each variable below is there in simple linear regression?

A. Response variable. 
B. Predictor variable.

From a statistics standpoint, why are the values of the slope and y-intercept needed?

A) They are needed to define a unique line.
B) Two statistics are needed when there are two columns of data values.
C) They make use of readily available mathematics.
D) They summarize the relationship information in the data set.

What new information does linear regression give over linear correlation?

A)  It gives the magnitude (size) of the effect of the predictor variable on the response variable.
B) It can detect and measure almost any type of variable relationship.
C) It can quantify the linear component of a data set.
D) It can give the strength and direction of a non-linear relationship.

Below is the equation of a regression line, match each symbol with its meaning.
ŷ = b0 + b1x

A. (x)
B. (b0)
C. (b1)
D. (ŷ)

__ The predicted value of the y-variable.
__ The y-intercept of the regression line.
__ The slope of the regression line.
__ The value of the x-variable.

What two assumptions need to be met to do linear regression?

A) All data values are independent of each other.
B) That a linear relationship does exist for these two variables.
C) That the scatterplot does not show any major exceptions.
D) That the data values are normally distributed (vertically) around the regression line.

Using the regression equation below, how much would you expect Weight (lbs) to change on average if Height (ins) went up one inch?

A) Weight would go down 12.3 lbs.
B) Weight would go down 3.3 lbs.
C)  Weight would go up 9 lbs.
D) Weight would not change.

Using the regression equation below, how much would you expect Weight (lbs) to change on average if Height (ins) went down ten inches?

Weight = -12.3 + 9 × Height

A)  Weight would go down 90 lbs.
B) Weight would go down 102.3 lbs.
C) Weight would go up 90 lbs.
D) Weight would go down 12.3 lbs.

Using the regression equation below, how much would you expect Weight (lbs) to change on average if Height (ins) went up two inches?

Weight = -12.3 + (-9) × Height

A) Weight would go down 21.3 lbs.
B) Weight would go down 30.3 lbs.
C)  Weight would go down 18 lbs.
D) Weight would go down 12.3 lbs.

What are the two uses of linear regression?

A) To know where to draw the middle line in a scatterplot.
B) To explain the change in the response variable from a change of the predictor variable.
C) To predict the value of the response variable from the value of the predictor variable.
D) To coordinate the movement of the response variable with the movement of the predictor variable.

If car prices followed the regression equation below, would you wait to buy a car or would you buy a car now?

Car Price = 50,000 + (-100) × Months

If car prices followed the regression equation below, would you wait to buy a car or would you buy a car now?

 Car Price = 50,000 + (-100) × Months 

A) Wait to buy as car prices are on average dropping $1,100 per month.
B) Wait to buy as car prices are on average dropping $100 per month.
C) Buy now when the car price is known for certain.
D) Buy now as car prices are on average rising $100 per month.

If car prices followed the regression equation below, what would you expect the car price to be (on average) a year from now?

Car Price = 50,000 + (-100) × Months

A) $50,000.
B) $49,900.
C) $48,800.
D) $51,200.

Match each use of linear regression below with the parts of the regression equation needed for that use.

A. Explanation of the effect of the predictor variable on the response variable.
B. Prediction of the value of the response variable from the value of the predictor variable.

__ Slope of the regression equation (b(1))
__ y-Intercept and slope of the regression equation ((b(0),b(1)).

What is NOT an appropriate comment because of the Scope of the Model?

A) Regression does not well describe a non-linear relationship inside the range of the predictor variable.
B)  Regression is good for extrapolation outside of the range of the predictor variable.
C) Regression accurately describes the linear relationship inside the range of the predictor variable.
D) Regression is good for interpolation inside the range of the predictor variable.

In concept, what is a regression residual?

A)  The vertical distance a data point (x,y) is from the regression line.
B) The distance a data point (x,y) is from the center point (x̄,ȳ) of the data set.
C) The horizontal distance a data point (x,y) is from the regression line.
D) The perpendicular distance a data point (x,y) is from the regression line.

Using the regression equation below, what would be the residual for a point with the coordinates (10,80)?

Weight = -12.3 + 9 × Height

A) Residual = -2.3
B) Residual = +22.3
C) Residual = +10.0
D)  Residual = +2.3

Using the regression equation below, what would be the residual for a point with the coordinates (24,45000)

Car Price = 50,000 + (-100) × Months

A) Residual = +2,600
B) Residual = +2,400
C) Residual = -2,400
D)  Residual = -2,600

Which of the points below would be closest to the regression line?

Point A: with residual = +150
Point B: with residual = -120
Point C: with residual = -10
Point D: with residual = +12

A)  Point C, as it has the smallest magnitude of the residuals (-10).
B) Point B, as it has the most negative of the residuals (-120).
C) Point A, as it has the largest magnitude of the residuals (+150).
D) Point D, as it has the smallest positive value of the residuals (+12).

In concept, how does linear regression find the regression line?

A) By finding the midpoint of the response values at every predictor value.
B) By averaging several lines of good fit to get the line of best fit.
C) By finding the midpoint of the response values at the midpoint of the predictor variable.
D)  By finding the line where the residuals of the points are the smallest.