Question

Growth hormones are often used to increase the weight gain of chickens. In an experiment using $15$chickens, five different doses of growth hormone ($0$, $0.2$, $0.4$, $0.8$, and $1.0$milligrams) were injected into chickens ($3$chickens were randomly assigned to each dose), and the subsequent weight gain (in ounces) was recorded. A researcher plots the data and finds that a linear relationship appears to hold. Computer output from a least-squares regression analysis for these data is shown below.

role="math" localid="1652870715985" $\begin{array}{lcccc}\text{Predictor}& \text{Coef}& \text{SE Coef}& T& \text{P}\\ \text{Constant}& 4.5459& 0.6166& 7.37& <0.0001\\ \text{Dose}& 4.8323& 1.0164& 4.75& 0.0004\\ S=3.135& \mathrm{R}-\mathrm{Sq}=38.4\%\mathrm{R}-\mathrm{Sq}\left(\text{adj}\right)=& 37.7\%& & \end{array}$

(a) What is the equation of the least-squares regression line for these data? Define any variables you use.

(b) Interpret each of the following in context:

(i) The slope

(ii) The $y$intercept

(iii) $s$

(iv) The standard error of the slope

(v) $r^2$

(c) Assume that the conditions for performing inference about the slope B of the true regression line are met. Do the data provide convincing evidence of a linear relationship between dose and weight gain? Carry out a significance test at the A $\alpha =0.05$level.

(d) Construct and interpret a $95\%$confidence interval for the slope parameter.

Short answer

(a) The equation of the least-squares regression line is$\hat{y}=4.5459+4.8323x$.

(b) The values are

(i) $b=4$

(ii) $a=4.5450$

(iii) $s=3.135$

(iv) $S{E}_b=1.0164$

(v)role="math" localid="1652871593895" $r^2=38.4\%$

(c) There's sufficient convincing evidence to justify the argument.

(d) There is a $95\%$chance that the weight gain will be between $2.636876$and $7.027724$. While the does, the ounces climb by $1$milligram.

Step-by-step solution

Part(a) Step 1: Given Information

$\begin{array}{lcccc}\text{Predictor}& \text{Coef}& \text{SE Coef}& T& \text{P}\\ \text{Constant}& 4.5459& 0.6166& 7.37& <0.0001\\ \text{Dose}& 4.8323& 1.0164& 4.75& 0.0004\\ S=3.135& \mathrm{R}-\mathrm{Sq}=38.4\%\mathrm{R}-\mathrm{Sq}\left(\text{adj}\right)=& 37.7\%& & \end{array}$

Part(a) Step 2: Explanation

$a=4.5459$

$b=4.8323$

The general regression line equation

$\hat{y}=a+bx$

By inserting values the regression line becomes:

role="math" localid="1652870965008" $\hat{y}=4.5459+4.8323x$

With $x$Dose and $y$weight gain.

Part(b) Step 1: Given Information

$\begin{array}{lcccc}\text{Predictor}& \text{Coef}& \text{SE Coef}& T& \text{P}\\ \text{Constant}& 4.5459& 0.6166& 7.37& <0.0001\\ \text{Dose}& 4.8323& 1.0164& 4.75& 0.0004\\ S=3.135& \mathrm{R}-\mathrm{Sq}=38.4\%\mathrm{R}-\mathrm{Sq}\left(\text{adj}\right)=& 37.7\%& & \end{array}$

Part(b) Step 2: Explanation

(i) $b=4$is the first output. The weight per milligram might increase by $4.8323$ounces.

(ii). In the result $a=4.5450$, the $y$-intercept is mentioned.

This means that the weight will be $4.5459$ounces if the dosage is $0$milligrams.

(iii) $s=3.135$is the output. This means that the average prediction error is $3.1350$ounces.

(iv) $S{E}_b=1.0164$This suggests that the population's true slope is $1.016-1$on average over all feasible samples.

(v)$r^2$is written as$r^2=R-S_q=38.4\%$. This means that the least-square regression line explains $38.4\%$of the variance in the variables.

Part(c) Step 1: Given Information

$\begin{array}{lcccc}\text{Predictor}& \text{Coef}& \text{SE Coef}& T& \text{P}\\ \text{Constant}& 4.5459& 0.6166& 7.37& <0.0001\\ \text{Dose}& 4.8323& 1.0164& 4.75& 0.0004\\ S=3.135& \mathrm{R}-\mathrm{Sq}=38.4\%\mathrm{R}-\mathrm{Sq}\left(\text{adj}\right)=& 37.7\%& & \end{array}$

Part(c) Step 2: Explanation

Define hypothesis

$H_0:\beta =0$

$H_1:\beta \ne 0$

The test statistic is

$$\begin{gathered} t=\frac{b-{\beta }_0}{S{E}_b} \\ =\frac{4.8323-0}{1.0164} \\ =4.754 \end{gathered}$$

The P-value is the probability of having the test numbers' value or a more dramatic value. The $t$-value in the rowrole="math" localid="1652872078633" $$\begin{gathered} df=n-2 \\ =15-2 \\ =13 \end{gathered}$$ is represented by a number (or interval) in Table B column title:

$P<0.0005$

The null hypothesis is rejected if the $p$-value is less than or equal to the degree of significance.

$P<0.05\Rightarrow \text{Reject}{H}_0$

Part(d) Step 1: Given Information

$\begin{array}{lcccc}\text{Predictor}& \text{Coef}& \text{SE Coef}& T& \text{P}\\ \text{Constant}& 4.5459& 0.6166& 7.37& <0.0001\\ \text{Dose}& 4.8323& 1.0164& 4.75& 0.0004\\ S=3.135& \mathrm{R}-\mathrm{Sq}=38.4\%\mathrm{R}-\mathrm{Sq}\left(\text{adj}\right)=& 37.7\%& & \end{array}$

Part(d) Step 2: Explanation

Degrees of freedom

$$\begin{gathered} df=n-2 \\ =15-2 \\ =13 \end{gathered}$$

Table B, in the row of $df=13$and the column of $\mathrm{c}=95\%$, has the important t-value.

The essential t-value may be found in table B in the $\mathrm{c}=95\%$column and in the $df=13$row.

$t^{*}=2.160$

The boundaries are

role="math" localid="1652872255479" $$\begin{gathered} b-t^{*}\times S{E}_b=4.8323-2.160\times 1.0164 \\ =2.636876 \end{gathered}$$

$$\begin{gathered} b+t^{*}\times S{E}_b=4.8323+2.160\times 1.0164 \\ =7.027724 \end{gathered}$$