d The upper and lower boundary have the same slope as and are equidistant from the line of best fit.
A
a See solution.
B
bLargest Residual: 17
Point: (69.8, 89.6)
C
c 60^(∘) F
D
d See solution.
Practice makes perfect
a Let's first discuss the direction and strength of the line of best fit, then we can analyze the slope.
Direction and strength
Examining the line of best fit, we notice that it has a positive slope and a correlation coefficient that is close to 1. This tells us the association is strong. Also, an r-value of 0.928 corresponds to an R-squared of (0.928)^2≈ 0.86. This means about 86 % of the high temperature the next day is explained by the high temperature on the previous day.
Slope
The slope is 0.85, which means the highest temperature increases on average by 0.85^(∘) F for every 1^(∘) F higher the temperature is on the previous day.
b Examining the residual plot and observations, we notice that when the random day high temperature is 69.8^(∘) F we have the greatest residual. This must correspond to point (69.8,89.6).
c To estimate tomorrow's high temperature based on today's high temperature, we have to substitute x=55 into the line of best fit and simplify.
According to the model, the temperature the next day should be 60^(∘) F.
d To create functions for the upper and lower boundary, we recognize that they will have the same slope as and be equidistant from the line of best fit.
Upper boundary: y=0.85x+b_u
Lower boundary: y=0.85x+b_l
To determine the y-intercepts b_u and b_l, we have to find the observation that is furthest away from the line of best fit. In other words, we have to find the largest residual. From Part B we know that this is (69.8,89.6). Therefore, the upper boundary will be a straight line through this observation.
y=0.85x+b_uLet's substitute the known point in the equation and solve for b_u
The upper boundary is y=0.85x+30.27. Since the upper and lower boundary are equidistant from the line of best fit, we can find the lower boundary's y-intercept by subtracting the difference between the y-intercept of the line of best fit and the upper boundary from 13.17.
13.17-(30.27-13.17)=- 3.93
Now we can write the lower boundary.
y=0.85x-3.93
Let's include the upper and lower boundary in our scatter plot.
If he can replace the complex model with his own is a matter of debate. It would depend on what the meteorologist's models looks like.
