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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:

$H_0:尾=0$

$H_a:尾<0$

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

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

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:

$P<0.01鈬�Reject{H}_0$