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A sample of 45 students was randomly selected from the population of current Smith College students. Based on this sample, a linear regression was calculated to predict the percent body fat of a student based on the number of hours the student exercised per week. The sample regression line is: y=26-0.7x

a. What is the estimate for the intercept?

b. What is the estimate for the slope of the regression line?

c. What is the variable for x?

d. Find the predicted percent body fat for a student in the population who excerises 6 hours per week.

e. Find the residual for a student in a population who exercises 6 hours per week and has a 15% body fat. 

 

** would be helpful if you explain how you got the answer. Thank you so much

 Jan 14, 2015

Best Answer 

 #2
avatar+33665 
+10

For part e.  The regression line gives y = 26 - 0.7*6 = 21.8% body fat.  

The difference between this and the actual body fat for this student is 21.8% - 15% = 6.8%

.

 Jan 14, 2015
 #1
avatar+118724 
+5

x is the number of hours of exercise.

y is the predicted % body fat.

y=26-0.7x

 

a. What is the estimate for the intercept?       

When there is no exercise  -0.7x=-0.7*0=0     SO the y intercept is  26   (26 % body fat)

b. What is the estimate for the slope of the regression line?        That is the number in front of the x    -0.7

c. What is the variable for x?       Hours of exercise

d. Find the predicted percent body fat for a student in the population who excerises 6 hours per week.

Substitute in x=6 and find y

e. Find the residual for a student in a population who exercises 6 hours per week and has a 15% body fat.

Don't know what this means.  

 Jan 14, 2015
 #2
avatar+33665 
+10
Best Answer

For part e.  The regression line gives y = 26 - 0.7*6 = 21.8% body fat.  

The difference between this and the actual body fat for this student is 21.8% - 15% = 6.8%

.

Alan Jan 14, 2015
 #3
avatar+118724 
0

Thanks Alan 

 Jan 15, 2015

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