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1. Calculate the value and understand the meaning of the slope and the $y$ intercept of the regression line.
2. Predict values of $y$ for given values of $x$, and calculate the prediction error for a given prediction.

Note: The “hat” over the y (pronounced “y-hat”) indicates that this is an estimate of y and not necessarily an actual value of y.

$\hat{y}=b_{1}x+b_0$

$\begin{array}{cc}{b}_1=r\cdot {\displaystyle \frac{s_y}{s_x}}& b_0=\overline{y}-\left(b_1\cdot \overline{x}\right)\end{array}$

1. Find the value of the regression coefficients $b_0$ and $b_1$ for the temperature data in Table 3.
2. Write out the equation of the regression line for the temperature data.
3. Clearly explain the meaning of the regression equation.

1. We will outline the steps used in calculating the value of $b_1$ using the temperature data.
   • Step 1 Calculate the respective sample means $\overline{x}$ and $\overline{y}$. We have already done this in Example 4 (page 193): $\overline{x}=40$ and $\overline{y}=64$.
   • Step 2 Calculate the respective sample standard deviations $s_x$ and $s_y$. We have already done this in Example 4: $s_x\approx 7.90569415$ and $s_y\approx 11.93733639$.
   • Step 3 Find the correlation coefficient $r$. This was computed in Example 4: $r\approx 0.92717265$.
   • Step 4 Combine the statistics from Steps 2 and 3 to calculate $b_1$:
     

     $b_1=r\cdot {\displaystyle \frac{s_y}{s_x}}=0.92717265\cdot {\displaystyle \frac{11.93733639}{7.90569415}}=1.4$

   • Step 5 Use the statistics from Steps 1–4 to calculate $b_0$:
     

     $b_0=\overline{y}-\left(b_1\cdot \overline{x}\right)=64-\left(1.4\right)\left(40\right)=8$

2. Thus, the equation of the regression line for the temperature data is
   

   $\hat{y}=1.4x+8$

3. Because $y$ and $x$ represent high and low temperatures, respectively, this regression equation is read as follows: “The estimated high temperature for an American city is 1.4 times the low temperature for that city plus 8 degrees Fahrenheit.”

YOUR TURN#6

1. The $y$ intercept $b_0=8$
2. The slope $b_1=1.4$

1. In statistics, we interpret the slope of the regression line as the estimated change in y per unit increase in x. In our temperature example, the units are degrees Fahrenheit, so we interpret our value $b_1=1.4$ as follows:

   “For each increase of 1°F in low temperature, the estimated high temperature increases by 1.4°F.”

2. The $y$ intercept is interpreted as the estimated value of y when x equals zero. Here, we interpret our value $b_0=8$ as follows:

   “When the low temperature is 0°F, the estimated high temperature is 8°F.”

YOUR TURN#7

This relationship holds because $b_1=r\cdot {\displaystyle \frac{s_y}{s_x}}$ and neither $s_y$ nor $s_x$ can be negative.

• $b_1$ is positive if and only if $r$ is positive.
• $b_1$ is negative if and only if $r$ is negative.

$\hat{y}=1.4\left(low\right)+8=1.4\left(40\right)+8=64$

YOUR TURN#8

Developing Your Statistical Sense

• Positive prediction error: The data value lies above the regression line, so the observed value of $y$ is greater than predicted for the given value of $x$.
• Negative prediction error: The data value lies below the regression line, so the observed value of $y$ is lower than predicted for the given value of $x$.
• Prediction error equal to zero: The data value lies directly on the regression line, so the observed value of $y$ is exactly equal to what is predicted for the given value of $x$.

Prediction errors are also called estimation errors or residuals.

All values of $\hat{y}$ (the predicted values of $y$) lie on the regression line.

1. Philadelphia: $\text{low}=40$, $\text{high}=70$
2. Washington, DC: $\text{low}=45$, $\text{high}=65$

1. The actual data point for Philadelphia is shown in the scatterplot in Figure 27 (denoted as “Phil”). In Example 9, we calculated the predicted high temperature for Philadelphia to be $\hat{y}=64°\mathrm{F}$. In Figure 27, $\hat{y}=64°\mathrm{F}$ represents the $y$-value of the point on the regression line where it intersects $x=40°\mathrm{F}$. That is, the actual high temperature $y=70°\mathrm{F}$ lies directly above predicted temperature $\hat{y}=64°\mathrm{F}$ for low temperature $x=40°\mathrm{F}$.
2. The actual high temperature in Washington that day was $y=65$. Using the regression equation, the predicted high temperature is $\hat{y}=1.4\left(45\right)+8=71$. So the prediction error is $y-\hat{y}=65-71=-6°\mathrm{F}$. The data point lies below the regression line, so that its actual high temperature of 65°F is lower than predicted given its low temperature of 45°F.

Philadelphia, Pennsylvania

FIGURE 27 Prediction error for Philadelphia and Washington, DC.

YOUR TURN#9

The further outside the range of $x$, the more unreliable the prediction becomes.

1. 50°F
2. 60°F

1. $\hat{y}=1.4\left(50\right)+8=78°\mathrm{F}$. Because $x=50°\mathrm{F}$ lies between 30°F and 50°F, inclusive, this estimate does not represent extrapolation.
2. $\hat{y}=1.4\left(60\right)+8=92°\mathrm{F}$. Because $x=60°\mathrm{F}$ does not lie between 30°F and 50°F, this estimate represents extrapolation.

YOUR TURN#10