91Ó°ÊÓ

To interpret the slope, the $y$ intercept, the standard deviation of the residuals ans the standard error of the slope.

To increase the weight of the chicken, growth hormones are applied. As a result, the experiment was carried out, and the weight gain was recorded. When a researcher plots the data, notices that it looks to be linear relationship. The question includes a computer output assuming that the prerequisites for making inferences about the slope of the regression line are met. As a result,
(i) The slope: The slope $b_1$ of the supplied computer output is presented in the row "Dose" and the column "Coef," i.e.,
$b_1=4.8323$
The slope shows how much y has increased or decreased per unit of x.
Weight gain increases by $4.8323$ ounces per mg of growth hormone on average.
(ii) In the row "Constant" and the column "Coef" of the given computer output, the $y$-intercept of $b_0$ is presented as,
$b_0=4.5459$
When $x$is zero, the $y$-intercept indicates the average $y$-value.
When the growth hormone dose is $0$mg, the weight gain is localid="1654258733317" $4.5459$ounces on average.

(iii) The standard error of the estimate $s$ is presented as:
$s=3.135$
The standard error of predictions, or the average difference between actual $y$-values and expected $y$-values, is represented by the standard error of the estimate s. As a result, the expected weight gain differs by $3.135$ ounces on average from the actual weight gain.
In the row "Dose" and the column "SE Coef" of the given computer output, the standard error of the slope $SE{b}_1$ is presented as:
localid="1654258767678" $S{E}_{b_1}=1.0164$.
(iv) The standard error of the slope indicates the mean deviation of the slope of the sample regression line from the slope of the population regression line. As a result, the slope of the sample regression line differs from the true population regression line by an average of $1.0164$.

To determine the data provide convincing evidence of a linear relationship between dose and weight gain at a significance test $\mathrm{Î\pm }=0.05$ level.

To increase the weight of the chicken, growth hormones are applied.
As a result, the experiment was carried out, and the weight gain was recorded. When a researcher plots the data, notices that it looks to be linear relationship. The question includes a computer output assuming that the prerequisites for making inferences about the slope of the regression line are met. As a result,
$b=4.8323$
$S{E}_b=1.0164$
$n=15$
The hypotheses are now specified as follows:
$H_0:\mathrm{β}=0$
$H_1:\mathrm{β}≠0$
The value of test statistics is determined as follows:
$t=\frac{b-{\mathrm{β}}_0}{S{E}_b}$
$=\frac{4.833-0}{1.0164}$
$≈4.754$

Determine the degrees of freedom is determined as follows:
$df=n-2$
$=15-2$
$=13$
Since, the P-value is the probability of getting the test statistic's value or a number that is more severe.
As a result, the P-value is as follows:
$P<0.0005$
The null hypothesis is rejected if the P-value is less than or equal to the significance level.
Accordingly,
$P<0.05⇒Reject{H}_0$
As a result, there is a sufficient evidence to support the claim.

To construct and interpret a $95\%$ confidence interval for the slope parameter.

To increase the weight of the chicken, growth hormones are applied.
As a result, the experiment was carried out, and the weight gain was recorded. When a researcher plots the data, notices that it looks to be linear relationship. The question includes a computer output assuming that the prerequisites for making inferences about the slope of the regression line are met. As a result,

$b=4.8323$
$S{E}_b=1.0164$
$n=15$
Determine the degrees of freedom as follows:
$df=n-2$
$=15-2$
$=13$
In the table $B$, the crucial value may be found in the row of $df=13$and the column of $c=95\%$, such that
$t^{*}=2.160$
Determine the confidence interval as follows:
$b-t^{*}×S{E}_b$
$=4.8323-2.160×1.0164$
$=2.636876$
$b+t^{*}×S{E}_b$
$=4.8323+2.160×1.0164$
$=7.027724$
As a result, conclude that $95\%$confidence the dose is increased by one milligrams, the weight gain increases between $2.636876$and $7.027724$ounces.