Exercise 106 Page 286

a Using a graphing calculator, we can calculate the least squares regression line.

y = m x + b

We can also include an interpretation of the

r

-value, the

R

-squared value, and the upper and lower boundary of the line. To make sure the linear association is appropriate, let's include a residual plot too.

b Let's start by making a scatterplot.

As we can see, there is a negative linear association between the length of an organelle and the diameter of the cell. To find a linear regression between the length of the organelle and the diameter of the cell, we first have to enter the values into lists. Push $STAT$ , choose Edit, and then enter the values in the first two columns.

To view the linear regression analysis of the dataset push $STAT$ , scroll right to view the CALC options, and then choose the fourth option in the list, LinReg.

The linear regression is $y = - 1.6 + 49.5$ This tells us that as the length of the organelle increases by $1,$ the diameter of the cell decreases by $1.6 .$

From the regression, we see that the $r$ -value is $- 0.92 .$ This means we have a strong negative correlation. Also, the $R$ -squared is $0.86,$ which means $86 %$ of the variability in the diameter of the cell is explained by the length of the organelle. Let's add the linear regression to the scatterplot.

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 = - 1.6 x + b_{u} Lower boundary : y = - 1.6 x + b_{ℓ}

To determine the

y -

intercepts

b_{u}

and

b_{ℓ},

we have to find the observation that is farthest 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 want to find all of the predicted values.