putting the “Hours Studied” data in L1 , the “Test Score” data in L2 , and doing LinReg(ax+b)L1,L2 .
When using LinReg(ax+b) , the explanatory variable must come first and the response variable second.
example: An experiment is conducted on the effects of having convicted criminals provide
restitution to their victims rather than serving time. The following table gives the data for 10
criminals. The monthly salaries (X ) and monthly restitution payments (Y ) were as follows:(a) Find the correlation between X and Y and the regression equation that can be used to predict
monthly restitution payments from monthly salaries.
(b) Draw a scatterplot of the data and put the LSRL on the graph.
(c) Interpret the slope of the regression line in the context of the problem.
(d) How much would a criminal earning $1400 per month be expected to pay in restitution?
solution: Put the monthly salaries (x ) in L1 and the monthly restitution payments (y ) in L2 .
Then enter STAT CALC LinReg(a+bx)L1,L2,Y1 .
(a) r = 0.97, = –56.22 + 0.46 (Salary). (If you answered ŷ = 56.22 + 0.46x , you must
define x and y so that the regression equation can be understood in the context of the problem. An
algebra equation, without a contextual definition of the variables, will not receive full credit.)
(b) (c) The slope of the regression line is 0.46. This tells us that, for each $1 increase in the criminal’s
salary, the amount of restitution is predicted to increase by $0.46. Or you could say that the
average increase is $0.46.
(d) Payment = –56.22 + 0.46 (1400) = $587.78.Calculator Tip: The fastest, and most accurate, way to perform the computation above, assuming you
have stored the LSRL in Y1 (or some “Y=” location), is to do Y1(1400) on the home screen. To paste
Y1 to the home screen, remember that you enter VARS Y-VARS Function Y1 . If you do this, you will
get an answer of $594.64 with the difference caused by rounding due to the more accurate 12-place
accuracy of the TI-83/84.