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 residuals 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.

Exercise

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