Does how long young children remain at the lunch table help predict how much they eat? Here are data on a random sample of $20$toddlers observed over several months. “Time” is the average number of minutes a child spent at the table when lunch was served. “Calories” is the average number of calories the child consumed during lunch, calculated from careful observation of what the child ate each day.

(a) A scatterplot of the data with the least-squares line added is shown below. Describe what this graph tells you about the relationship between these two variables.

Minitab output from a linear regression on these data is shown below.

$\begin{array}{lcccc}\text{Predictor}& \text{Coef}& \text{SE Coef}& T& \text{P}\\ \text{Constant}& 560.65& 29.37& 19.09& 0.000\\ \text{Time}& -3.0771& 0.8498& -3.62& 0.002\\ \mathrm{S}=23.3980& \mathrm{R}-\mathrm{Sq}=& 42.1\%\mathrm{R}-\mathrm{Sq}\left(\mathrm{adj}\right)=& 38.9\%& \end{array}$

(b) What is the equation of the least-squares regression line for predicting calories consumed from the time at the table? Define any variables you use.

(c) Interpret the slope of the regression line in context. Does it make sense to interpret the intercept in this case? Why or why not?

(d) Do these data provide convincing evidence of a negative linear relationship between time at the table and calories consumed in the population of toddlers? Carry out an appropriate test at the A $\alpha =0.01$level to help answer this question.

The study wanted to see if how long young toddlers stay at the lunch table influences how much they eat. As a result, the scatterplot for the variables utilized is also included in the question. As a result, we can deduce from the scatterplot that

Direction: Because the scatterplot slopes downhill, the direction is negative.

Form: Because the points appear to nearly lie along a line, the form is linear.

Strength: Moderate, as the points are not too far away but also not too close together.

Minitab output from a linear regression on these data is shown below.

$\begin{array}{lcccc}\text{Predictor}& \text{Coef}& \text{SE Coef}& T& \text{P}\\ \text{Constant}& 560.65& 29.37& 19.09& 0.000\\ \text{Time}& -3.0771& 0.8498& -3.62& 0.002\\ \mathrm{S}=23.3980& \mathrm{R}-\mathrm{Sq}=& 42.1\%\mathrm{R}-\mathrm{Sq}\left(\mathrm{adj}\right)=& 38.9\%& \end{array}$

Now, the study looked at whether the length of time young children spend at the lunch table helps predict how much they eat. and the data's computer output is provided. As we all know, the general regression equation is as follows:

$\hat{y}=a+bx$

In the "Coef" column of the computer output, the estimates $a$and $b$are given:

$$\begin{gathered} \hat{y}=a+bx \\ =560.65-3.0771x \end{gathered}$$

$\hat{y}$represents the expected calories, and $x$represents the time spent at the table.

Minitab output from a linear regression on these data is shown below

$\begin{array}{lcccc}\text{Predictor}& \text{Coef}& \text{SE Coef}& T& \text{P}\\ \text{Constant}& 560.65& 29.37& 19.09& 0.000\\ \text{Time}& -3.0771& 0.8498& -3.62& 0.002\\ \mathrm{S}=23.3980& \mathrm{R}-\mathrm{Sq}=& 42.1\%\mathrm{R}-\mathrm{Sq}\left(\mathrm{adj}\right)=& 38.9\%& \end{array}$

Now, the study looked at whether the length of time young children spend at the lunch table helps predict how much they eat. and the data's computer output is provided. As we all know, the general regression equation is as follows:

$\hat{y}=a+bx$

In the "Coef" column of the computer output, the estimates $a$and $b$are given:

role="math" localid="1652797079072" $$\begin{gathered} \hat{y}=a+bx \\ =560.65-3.0771x \end{gathered}$$

$\hat{y}$represents the expected calories, and $x$represents the time spent at the table.

And the $y$-intercept $a$ is the regression equation's constant, resulting in $560.65$. This indicates that after zero minutes, the calories should be $560.65$.

Minitab output from a linear regression on these data is shown below

$\begin{array}{lcccc}\text{Predictor}& \text{Coef}& \text{SE Coef}& T& \text{P}\\ \text{Constant}& 560.65& 29.37& 19.09& 0.000\\ \text{Time}& -3.0771& 0.8498& -3.62& 0.002\\ \mathrm{S}=23.3980& \mathrm{R}-\mathrm{Sq}=& 42.1\%\mathrm{R}-\mathrm{Sq}\left(\mathrm{adj}\right)=& 38.9\%& \end{array}$

The following is taken from the computer output:

$$\begin{gathered} n=20 \\ b=-3.0771 \\ {\mathrm{SE}}_b=0.8498 \end{gathered}$$

As a result, we define the hypothesis as follows:

$H_0:\beta =0$

$H_a:\beta <0$

As a result, the test statistics have the following value:

$$\begin{gathered} t=\frac{b-\beta }{{\mathrm{SE}}_b} \\ =\frac{-3.0771-0}{0.8498} \\ =-3.621 \end{gathered}$$

The degrees of freedom are as follows:

$$\begin{gathered} df=n-2 \\ =20-2 \\ =18 \end{gathered}$$

As a result, the $p$-value is:

$0.005<P<0.01$

The null hypothesis is rejected if the $p$-value is less than or equal to the significance level, as follows:

$P<0.01\Rightarrow Reject{H}_0$