Deck 13: Regression

Question 1

Suppose that in a random sample of 30 men, the correlation between hours spent working per week and hours spent playing with their children per week is -.50. The men in the sample spent an average of 50 hours per week working, with a standard deviation of 10. They spent an average of 6 hours per week playing with their children, with a standard deviation of 3. If you knew that one man spent 40 hours per week working, how many hours would you predict that he spent playing with his kids?

Answer 1

b = -.5(3/10) $\rightarrow$ -.5(.3) = -.15. a = 6 - (-.15)(50) $\rightarrow$ 6 + 7.5 = 13.5.
Predicted Y: -.15(40) + 13.5 $\rightarrow$ -6 + 13.5 = 7.5 hours.

Question 2

Please wrap words around the regression coefficient and the intercept that you calculated for the previous question (i.e., Question 1).

Answer 2

For every increase of one hour of work there is a corresponding decrease of .15 hours spent playing with kids, on average. When the number of hours worked is zero, the expected number of hours spent playing with kids is 13.5.

Question 3

Suppose that I have a randomly selected sample of 42 American men. In this sample, the average number of hours worked per week is 45 with a standard deviation of 12. The average level of life satisfaction (on a scale from 1 = "hating life" to 10 = "loving life) is 5.00 with a standard deviation of 3.00. The correlation between hours worked and life satisfaction is .40. Please answer the following two questions based on these data.
a. If you know that someone works 35 hours per week, what would you predict her life satisfaction score to be? (Please provide enough information so that I can tell how you reached your conclusion, including your regression coefficient and your intercept).
b. Please wrap words around the regression coefficient that you calculated for the last problem. What does this tell you?

Answer 3

a. I need to calculate the regression coefficient and the intercept for this. The regression coefficient (b) is .4(3/12) $\rightarrow$ .4(.25) = .10. The intercept (a) is 5.00 - (.10*45) $\rightarrow$ a = 5 - 4.5 $\rightarrow$ .50. So the predicted value of Y when X is 35 is: 35(.10) + .50 $\rightarrow$ 3.5 + .50 = 4.00. If someone is working 35 hours per week I would expect her life satisfaction score to be 4.00.
b. For every increase of one hour worked there is a corresponding increase of .10 units of life satisfaction, on average.