aSlope shows how the function's value changes when you move 1 unit in the positive horizontal direction.
B
b The residual is the actual value minus the predicted value.
C
c Calculate the predicted value and then add the residual.
D
d Substitute t=80 in the upper and lower boundary.
E
e Given the residual plot, does a linear model make sense?
A
a See solution.
B
b Too low.
C
c About 16 950 people
D
dUpper boundary: 25 800 people
Lower boundary: 11 800 people
E
e No, see solution.
Practice makes perfect
a The slope shows how the function changes when t increases by 1. Therefore, a slope of 0.41 tells us that for every degree Fahrenheit the temperature rises, the number of people increases on average by 410 people.
b Residuals are calculated by subtracting the predicted values from the actual values. Residual=Actual value-Predicted value
Examining the residual plot, we see that when temperatures were in the 80's the residuals are positive. Therefore, if the residuals were positive, the model must have understated the attendance.
c By substituting t=95 in the function we can calculate the predicted attendance on that day.
The predicted attendance was 24 950 people. Examining the residual plot, we notice that at t=95 the residual was - 8000 people. Therefore, by adding the residual to the predicted value we get the actual attendance.
24 950+(- 8000)=16 950 people
d By substituting t= 80 in the equation for the upper and lower boundary line, we can find the attendance for this temperature.
Lower boundary:& a = - 7+0.41( 80)=25.8
Upper boundary:& a = - 21+0.41( 80)=11.8
Therefore, the lower boundary is 11 800 people and the upper boundary is 25 800 people.
e The residual is calculated by subtracting the predicted value from the actual value. Since the residuals at the lower and higher temperatures are below 0 and above 0, the model is either understating or overstating the attendance. Therefore, a linear model is not appropriate — there should be a random scatter around the x-axis.
