Question

Cardiorespiratory fitness is widely recognized as a major component of overall physical well-being. Direct measurement of maximal oxygen uptake (VO2max) is the single best measure of such fitness, but direct measurement is time-consuming and expensive. It is therefore desirable to have a prediction equation for VO2max in terms of easily obtained quantities. A sample is taken and variables measured are age (years), time necessary to walk 1 mile (mins), and heart rate at the end of the walk (bpm) in addition to the VO2 max uptake. The equation from a multiple regression is (V02) = 0.017*(age) - 0.028*(HR) + 0.017*(time) + 3.483. If a person is 29 years old, blood pressure of 130 bpm, and a walking time of 13 minutes, what is his/her expected maximum oxygen uptake?

part A options:

	1) 0.557
	2) -2.926
	3) -6.409
	4) We do not know the observations in the data set, so we cannot answer that question.
	5) 5.102

Suppose that a researcher studying the weight of female college athletes wants to predict the weights based on height, measured in inches, and the percentage of body fat of an athlete. The researcher calculates the regression equation as (weight) = 3.979*(height) + 0.85*(percent body fat) - 87.814. If a female athlete is 67 inches tall and has a 20 percentage of body fat, what is her expected weigh

Part B options:

	1) 371.407
	2) 195.779
	3) 283.593
	4) 48.716
	5) We do not know the observations in the data set, so we cannot answer that question.

A trucking company considered a multiple regression model for relating the dependent variable of total daily travel time for one of its drivers (hours) to the predictors distance traveled (miles) and the number of deliveries of made. After taking a random sample, a multiple regression was performed and the equation is (time) = 0.076*(distance) + 0.579*(deliveries) - 0.466. Suppose for a given driver's day, he is scheduled to drive 58.908 miles and make 10.97 stops. Suppose it took him 14.302 hours to complete the trip. What is the residual based on the regression model?

Part C options:

	1) We do not know the observations in the data set, so we cannot answer that question.
	2) 3.9394
	3) -3.4734
	4) -48.5454
	5) -3.9394

Suppose that a researcher studying the weight of female college athletes wants to predict the weights based on height, measured in inches, the percentage of body fat of an athlete, and age. The researcher calculates the regression equation as (weight) = 4.08*(height) + 1.103*(percent body fat) - 0.995*(age) - 82.074. If a female athlete is 68.575 inches tall, has a 23.901 percentage of body fat, is 22.696 years old, and has a weight of 167.763, the residual is -33.7293. Choose the correct interpretation of the residual.

Part D options:

	1) The height of the athlete is 33.7293 inches larger than what we would expect.
	2) The weight of the athlete is 33.7293 pounds greater than what we would expect.
	3) The weight of the athlete is 33.7293 pounds less than what we would expect.
	4) The weight of the athlete is 167.763 pounds less than what we would expect.
	5) The height of the athlete is 33.7293 inches less than what we would expect.

Answer 1

Part A

We are given

(VO2) = 0.017*age – 0.028*HR + 0.017*time + 3.483

Age = 29

HR = 130

Time = 13

(VO2) = 0.017*29 - 0.028*130 + 0.017*13 + 3.483

(VO2) = 0.557

Correct Answer: 1) 0.557

Part B

We are given

(Weight) = 3.979*(height) + 0.85*(percent body fat) - 87.814

Height = 67

Percent body fat = 20%

(Weight) = 3.979*67 + 0.85*20 - 87.814

(Weight) = 195.779

Correct Answer: 2) 195.779

Part C

Regression equation is given as below:

(time) = 0.076*(distance) + 0.579*(deliveries) - 0.466

We are given

Distance = 58.908

Deliveries = 10.97

Observed time = 14.302 hours

Predicted time = 0.076*(distance) + 0.579*(deliveries) - 0.466

Predicted time = 0.076*58.908 + 0.579*10.97 - 0.466

Predicted time = 10.36264

Residual = Observed value – Predicted value

Residual = 14.302 - 10.36264 = 3.93936

Residual = 3.9394

Correct Answer: 2) 3.9394

Part D

Correct Answer: 3) The weight of the athlete is 33.7293 pounds less than what we would expect.

Explanation: We know that the residual is the difference between observed values and predicted or expected value. A negative value of residual indicates a less observed value than what we would expect.