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When Mentos are dropped into a newly opened bottle of Diet Coke, carbon dioxide is released from the Diet Coke very rapidly, causing the Diet Coke to be expelled from the bottle. To see if using more Mentos causes more Diet Coke to be expelled, Brittany and Allie used twenty-four $2$-cup bottles of Diet Coke and randomly assigned each bottle to receive either $2,3,4,\mathrm{or}5$Mentos. After waiting for the fizzing to stop, they measured the amount expelled (in cups) by subtracting the amount remaining from the original amount in the bottle. Here are their data:

Here is the computer output from a least-squares regression analysis of these data. Construct and interpret a $95\%$confidence interval for the slope of the true regression line.

$$\begin{gathered} \mathrm{Predictor}\mathrm{Coef}\mathrm{SE}\mathrm{Coef}\mathrm{T}\mathrm{P} \\ \mathrm{Constant}1.00210.045122.2150.000 \\ \mathrm{Mentos}0.07080.01235.7700.000 \\ \mathrm{S}=0.06724\mathrm{R}-\mathrm{Sq}=60.2\%\mathrm{R}-\mathrm{Sq}\left(\mathrm{adj}\right)=58.4\% \end{gathered}$$

Step by step solution

01

Given Information

We need to construct and interpret a $95\%$confidence interval for the slope of the true regression line.

02

Simplify

Consider:

$$\begin{gathered} \mathrm{n}=\mathrm{Sample}\mathrm{size}=24 \\ \mathrm{c}=\mathrm{Confidence}\mathrm{level}=95\%=0.95 \end{gathered}$$

The estimate of the slope ${\mathrm{b}}_1$is considered in the row "Mentos" and in the column "Coef" of the given computer output:

${\mathrm{b}}_1=0.0708$

The estimated standard deviation of the slope${\mathrm{SE}}_{\mathrm{b}1}$ is considered in the row "Mentos" and in the column "SE Coef" of the given computer output:

${\mathrm{SE}}_{\mathrm{b}1}=0.0123$

The critical $\mathrm{t}$-value can be found in the Student's $\mathrm{T}$distribution table in the appendix in the row of $\mathrm{df}=\mathrm{n}-2=24-2=22$and in the column of$\mathrm{c}=95\%$

$\mathrm{t}\text{'}=2.074$

The boundaries of the confidence interval then become:

$\mathrm{b}-\mathrm{t}\text{'}脳{\mathrm{SE}}_{\mathrm{b}}=0.0708-2.074脳0.0123=0.0452898$

$b+t\text{'}脳S{E}_b=0.0708+2.074脳0.0123=0.0963102$

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