91Ó°ÊÓ

In the question, the relationship between the amount expelled and the tapping time in seconds is stated. For the same, a scatterplot and a residual plot are provided. Because there is no substantial curvature in the scatterplot and no strong curvature in the residual plot, it appears that a line is appropriate for these data.

In the question, the relationship between the amount expelled and the tapping time in seconds is stated. For the same, a scatterplot and a residual plot are provided. As a result, the coefficient of determination appears after "R-Sq" in the computer output as:

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

The positive or negative square root of the coefficient of determination $r^2$yields the linear correlation coefficient $r$As a result, we can see that the scatterplot pattern slopes downwards, indicating a negative relationship between the variables and, as a result, a negative linear correlation coefficient $r$This indicates that,

$$\begin{gathered} r=−\sqrt{r^2} \\ =−\sqrt{0.8538} \\ =−0.9240 \end{gathered}$$

In the question, the relationship between the amount expelled and the tapping time in seconds is stated. For the same, a scatterplot and a residual plot are provided. The least square regression line's general equation is:

$\stackrel{Áåœ}{y}=b_0+b_{1}x$

As a result, the estimate of the constant $b_0$is given in the computer output's row "Constant" and column "Coef" as:

$b_0=106.360$

In the computer output, the estimate of the slope $b_1$is presented in the row "Tapping time" and the column "Coef" as:

$b_1=−2.635$

As a result, when we plug the values into the general equation, we get:

$$\begin{gathered} \stackrel{Áåœ}{y}=b_0+b_{1}x \\ ⇒\stackrel{Áåœ}{y}=106.360−2.635x \end{gathered}$$

In the question, the relationship between the amount expelled and the tapping time in seconds is stated. For the same, a scatterplot and a residual plot are provided. As a result, after $"S="$ in the computer output, the standard error of the estimations is given as:

$s=5.00347$

The standard error of the estimations, as we all know, is the average error of forecasts, and thus the average difference between actual and predicted values. As a result, the predicted amount evacuated using the least square regression line differed by $5.00347ml$ on average from the actual amount expelled.

In the computer output, the coefficient of determination is now given after "R-Sq" as:

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

The coefficient of determination, as we know, is a measurement of how much variation in the answers $y$ variable is explained by the least square regression model with the explanatory variable. As a result, the least square regression line utilizing the tapping duration as an explanatory variable can explain $85.38$the percent of the variation in the amount ejected.