Exercise 56 Page 218

a In the solution to the mentioned exercise, we calculated the line of best fit.

y = - 1.58 x + 5.37

Let's add this to the scatterplot from the mentioned exercise.

b The upper and lower boundary have the same slope as the line of best fit, and they are equidistant from the line of best fit.

Upper boundary : y = - 1.58 x + b_{u} Lower boundary : y = - 1.58 x + 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. Since the residual is the actual value minus the predicted value, we first have to find all of the predicted values.

x 0.5 1 1.5 2 2.5 - 1.58 x + 5.37 - 1.58 (0.5) + 5.37 - 1.58 (1) + 5.37 - 1.58 (1.5) + 5.37 - 1.58 (2) + 5.37 - 1.58 (2.5) + 5.37 Predicted 4.58 3.79 3 2.21 1.42

Now we can calculate the residual and identify the largest one.

x 0.5 2.5 12 0.5 121 1.5 2.5 Actual 4.5 1.6 3 1.8 5 4.2 2.4 3.6 3.3 1.4 Predicted 4.58 1.42 3.79 2.21 4.58 3.79 2.21 3.79 3 1.42 Residual - 0.08 0.18 - 0.79 - 0.41 0.42 0.41 0.19 - 0.19 0.3 - 0.02

As we can see, the greatest residual is

- 0.79 .

Since the residual is negative, we know that the lower boundary will be a straight line through the observation

(1, 3) .

y = - 1.58 x + b_{ℓ}

Let's substitute this point in the equation and solve for

b_{ℓ} .

y = - 1.58 x + b_{ℓ}

SubstituteII

$x = 1$ , $y = 3$

3 = - 1.58 (1) + b_{ℓ}

▼

Solve for

b_{ℓ}

MultByOne

$a \cdot 1 = a$

3 = - 1.58 + b_{ℓ}

AddEqn

$LHS + 1.58 = RHS + 1.58$

4.58 = b_{ℓ}

RearrangeEqn

Rearrange equation

b_{ℓ} = 4.58

Now we can write the function for the lower boundary.

y = - 1.58 x + 4.58

Since the lower and upper boundary are equidistant from the line of best fit, we can determine the upper boundary's

y -

intercept by adding the difference between the

y -

intercepts of the line of best fit and lower boundary to

5.37 .

b_{u} = 5.37 + (5.37 - 4.58) = 6.16

The upper boundary is

y = - 1.58 x + 6.16 .

Now we can graph the upper and lower boundary.

c To predict the number of days a cold lasts after

3

months of taking the supplement, we should substitute

x = 3

into our model.

y = - 1.5 x + 5.37

Substitute

$x = 3$

y = - 1.5 (3) + 5.37

Multiply

y = - 4.5 + 5.37

AddTerms

Add terms

y = 0.87

Since the values only have one decimal, we would predict the cold to last between

0.5

and

1

days.

d If a person has taken no supplements we have

x = 0,

which corresponds to the

y -

intercept. Since the

y -

intercept of our model is

5.37,

the cold should last for about

5.4

days.

e At

6

months the predicted value would be negative. Therefore, we cannot really say anything useful about the number of days the cold will last when you take the supplement for

6

months.

f The residual is calculated by subtracting the predicted value from the actual value.

Residual = Actual value - Predicted value

If the residual is negative, the actual value must be less than the predicted. In the case of a cold this is good news, because the cold lasted less days than predicted.

Subchapter links

4.2.1 Residual Plots p.218-220

4.2.2 Correlation p.225-226

4.2.3 Association is Not Causation p.230-231

4.2.4 Interpreting Correlation in Context p.236-237

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