Question

Number 1 (15 pts) Say that you want to use simple linear regression to model the relationship between the explanatory variable x= the number of alcoholic drinks consumed per week, and the response variable y= number of hours of exercise per week. You collect data on n=52 people, and find the regression line ŷ=10-0.3x.

(a) (2 pts) What is the theoretical model underlying your regression analysis? (b) (4 pts) Say that Lorenzo has six alcoholic drinks a week. Predict how many hours he will exercise a week. (Only do a pointwise prediction; there’s no need to find a margin of error.) (c) (3 pts) In fact, you learn that Lorenzo exercises twelve hours a week. What is the residual of this observation? (d) (3 pts) Interpret the slope of the regression line. (e) (3 pts) Say that you learn that the standard error for ?1 is ???1 =0.2. Find a 95% confidence interval for the slope.

Answer 1

Result:

Number 1 (15 pts) Say that you want to use simple linear regression to model the relationship between the explanatory variable x= the number of alcoholic drinks consumed per week, and the response variable y= number of hours of exercise per week. You collect data on n=52 people, and find the regression line ŷ=10-0.3x.

(a) (2 pts) What is the theoretical model underlying your regression analysis?

(b) (4 pts) Say that Lorenzo has six alcoholic drinks a week. Predict how many hours he will exercise a week. (Only do a pointwise prediction; there’s no need to find a margin of error.)

When x=6,

predicted number of hours of exercise per week = 10-0.3*6

=8.2 hours

(c) (3 pts) In fact, you learn that Lorenzo exercises twelve hours a week. What is the residual of this observation?

Residual = 12-8.2 =3.8

(d) (3 pts) Interpret the slope of the regression line.

When the number of alcoholic drinks consumed per week increases by 1, the number of hours of exercise per week decreases by 0.3 hours.

(e) (3 pts) Say that you learn that the standard error for ?1 is ???1 =0.2. Find a 95% confidence interval for the slope.

Critical t with 50 Df at 0.05 level of significance= 2.009

CI = b1 ± t* ???1

Lower limit = -0.3-2.009*0.2 = -0.7018

upper limit = -0.3+2.009*0.2 = 0.1018

95% confidence interval for the slope.

95% confidence interval for the slope = (-0.7018, 0.1018).