Chapter 4.2, Problem 4.38E

(a)

To determine

To predict the student’s BAC if a student drinks five beers.

Answer to Problem 4.38E

  $0.077$

Explanation of Solution

From previous exercise we have,

  $\widehat{y}=0.018x-0.013$

The predicted value is found by evaluating the least squares regression at $x=5$ :

  $\begin{array}{l}\widehat{y}=0.018x-0.013\\ =0.018(5)-0.013\\ =0.077\end{array}$

(b)

To determine

To explain will the regression method be as accurate for predicting BAC for a person who drinks fifteen beers.

Explanation of Solution

No it will not be accurate because the given range of x values beers is from $1\text{ to }9$ .Since fifteen does not lie in that range, we would be extrapolating if we use the least square regression equation to make a prediction and then the prediction would be inaccurate.

(c)

To determine

To construct the residual plot by using calculator and explain how well the regression line fit the data.

Explanation of Solution

Calculator commands used:

Assuming that you entered the x -values in list L₁ of your calculator and the y -values in list L₂ .

Use the command “ $2\text{nd}$ ”and “Y=” to open the STAT PLOT menu.

The residual plot is as follows:

  

The regression line appears to fit the data well because the residuals appear to be randomly scatter about zero and there is no obvious pattern in the residuals.

(d)

To determine

To interpret the value of r² in this setting.

Answer to Problem 4.38E

  $79.98\%$

Explanation of Solution

Calculator commands used:

Assuming that you entered the x -values in list L₁ of your calculator and the y -values in list L₂ .

Use the commands STAT, CALC and 8:LinReg( $ax+b$ )

Next we determine the least square regression line by using LinReg( $a+bx$ ) L₁ ,L₂ , Y₁ .

The result should then contain r and r² .

The result will be ${\text{r}}^2=0.7998=79.98\%$ .

  $79.98\%$ of the observed variation in the BAC is explained by the least squared regression line.