Question

Less mess? Kerry and Danielle wanted to investigate whether tapping on a can of soda would reduce the amount of soda expelled after the can had been shaken. For their experiment, they vigorously shook 40 cans of soda and randomly assigned each can to be tapped for 0 seconds, 4 seconds, 8 seconds, or 12 seconds. After waiting for the fizzing to stop, they measured the amount expelled (in milliliters) by subtracting the amount remaining from the original amount in the can.30 Here is some computer output from a

regression of y = amount expelled on x = tapping time:

a. Is a line an appropriate model to use for these data? Explain how you know.

b. Find the correlation.

c. What is the equation of the least-squares regression line? Define any variables that you use.

d. Interpret the values of s and r2.

Short answer

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

Part (b) $r=-0.9240$

Part (c)$\stackrel{⏜}{y}=106.360-2.635x$

Part (d) The least square regression line utilizing the tapping duration as an explanatory variable can explain $85.38$ percent of the variation in the amount ejected.

Step-by-step solution

Part (a) Step 1: Given information

Part (a) Step 2: Explanation

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.

Part (b) Step 1: Calculation

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,

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

Part (c) Step 1: Calculation

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:

$\stackrel{⏜}{y}=b_0+b_{1}x$

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:

$$\begin{gathered} \stackrel{⏜}{y}=b_0+b_{1}x \\ \Rightarrow \stackrel{⏜}{y}=106.360-2.635x \end{gathered}$$

Part (d) Step 1: Calculation

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.