A regression analyses was conducted to predict 28 female runners\' time to compl
ID: 3160399 • Letter: A
Question
A regression analyses was conducted to predict 28 female runners' time to complete a 10-mile race (in minutes) based on the number of hours they trained in the month before the race. Below I have provided the Output from the regression analyses. Answer the specified questions.
Predictor
Coefficient
SE of Coeff.
T
P
Constant
129.701
7.414
17.49
0.000
Training time
-0.1492
0.01690
-8.83
0.000
Interpret the slope in terms of a 15-hour increase in training time in the month before the race.
Does the y-intercept have a meaningful interpretation in this context? Why or why not?
Compute the predicted race time for a woman who trained for 25 hours the month before the race.
Find the residual for a woman who trained for 25 hours and completed the race in 124.5 minutes.
Predictor
Coefficient
SE of Coeff.
T
P
Constant
129.701
7.414
17.49
0.000
Training time
-0.1492
0.01690
-8.83
0.000
Explanation / Answer
let Y be the variable denoting the time to complete the race and X be the variable denoting the number of hours trained before the race.
a regression analysis is done how Y is based on X
hence a regression equation of Y on X is done.
from the table we find constant=129.701 and coefficient of training time=-0.1492
hence the regression equation is Y=129.701-0.1492*X
slope=-0.1492 and y intercept is 129.701
now slope denotes the change in Y for unit change in X
let Yn=129.701-0.1492*(X+1) here X is increases by one unit
so change in Y is Yn-Y=129.701-0.1492*(X+1) -129.701+0.1492*X=-0.1492=slope
hence for 15 hours increase in training time the runners time will decrease by slope*15 =0.1492*15 minutes=2.238 minutes [answer]
no the y intercept does not have a meaningful interpretation in this context.
because it gives the value of Y when X=0 as Y=129.701-0.1492*X
so X=0 implies Y=129.701-0.1492*0=129.701
hence it has no relation with the increase or decrease of X.
the predicted race time for a woman who trained for 25 hours the month before the race is
Y=129.701-0.1492*25=125.971 minutes [answer]
hence the residual for a woman who trained for 25 hours and completed the race in 124.5 minutes is
(observed value-predicted value)=(124,.5-125.971)=-1.471 minutes [answer]
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