Chapter 3.2, Problem 60E

(a)

To determine

To explain is a line an appropriate model to use for these data.

Answer to Problem 60E

Yes, a line appears to be appropriate for these data.

Explanation of Solution

The relationship of the amount expelled and the tapping time in seconds is given in the question. The scatterplot and the residual plot for the same is given. Thus, a line appears to be appropriate for these data because there is no strong curvature present in the scatterplot and there is no strong curvature in the residual plot either.

(b)

To determine

To find the correlation.

Answer to Problem 60E

  $r=-0.9240$

Explanation of Solution

The relationship of the amount expelled and the tapping time in seconds is given in the question. The scatterplot and the residual plot for the same is given. Thus, the coefficient of determination $r^2$ is given in the computer output after “R-Sq” as:

  $\begin{array}{l}{r}^2=85.38\%\\ =0.8538\end{array}$

Thus, the linear correlation coefficient r is the positive or negative square root of the coefficient of determination $r^2$ . Thus, we note that the pattern in the scatterplot slopes downwards, which indicates that there is a negative association between the variables and thus the linear correlation coefficient r is negative. This then implies that,

  $\begin{array}{l}r=-\sqrt{r^2}\\ =-\sqrt{0.8538}\\ =-0.9240\end{array}$

(c)

To determine

To find out what is the equation of least square regression line.

Answer to Problem 60E

  $\widehat{y}=106.360-2.635x$ .

Explanation of Solution

The relationship of the amount expelled and the tapping time in seconds is given in the question. The scatterplot and the residual plot for the same is given. The general equation of the least square regression line is:

  $\widehat{y}=b_0+b_{1}x$

Thus, the estimate of the constant $b_0$ is given in the row “Constant” and in the column “Coef” of the computer output as:

  $b_0=106.360$

And the estimate of the slope $b_1$ is given in the row “Tapping time” and in the column “Coef” of the computer output as:

  $b_1=-2.635$

Thus, putting the values in the general equation we will get,

  $\begin{array}{l}\widehat{y}=b_0+b_{1}x\\ \Rightarrow \widehat{y}=106.360-2.635x\end{array}$

(d)

To determine

To interpret the value of s and $r^2$ .

Explanation of Solution

The relationship of the amount expelled and the tapping time in seconds is given in the question. The scatterplot and the residual plot for the same is given. Thus, the standard error of the estimate s is given after “S=” in the computer output as:

  $s=5.00347$

As we know that the standard error of the estimate s represents the average error of predictions thus the average deviation between actual and the predicted values. Thus the predicted amount expelled using the equation of least square regression line deviated on average by $5.00347$ ml from the actual amount expelled.

Now, the coefficient of determination is given after “R-Sq” in the computer output as:

  $\begin{array}{l}{r}^2=85.38\%\\ =0.8538\end{array}$

As we know that the coefficient of determination measures the proportion of variation in the responses y variable that is explained by the least square regression model using the explanatory x variable. Thus, we can say that $85.38\%$ of the variation in the amount expelled is explained by the least square regression line using the tapping time as explanatory variable.