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Q. 11

Expert-verifiedFound in: Page 797

Book edition
4th

Author(s)
David Moore,Daren Starnes,Dan Yates

Pages
809 pages

ISBN
9781319113339

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" $\begin{array}{lcccc}\text{Predictor}& \text{Coef}& \text{SE Coef}& T& \text{P}\\ \text{Constant}& 4.5459& 0.6166& 7.37& <0.0001\\ \text{Dose}& 4.8323& 1.0164& 4.75& 0.0004\\ S=3.135& \mathrm{R}-\mathrm{Sq}=38.4\%\mathrm{R}-\mathrm{Sq}\left(\text{adj}\right)=& 37.7\%& \phantom{\rule{0ex}{0ex}}& \phantom{\rule{0ex}{0ex}}\end{array}$

(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.

(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.

$\begin{array}{lcccc}\text{Predictor}& \text{Coef}& \text{SE Coef}& T& \text{P}\\ \text{Constant}& 4.5459& 0.6166& 7.37& <0.0001\\ \text{Dose}& 4.8323& 1.0164& 4.75& 0.0004\\ S=3.135& \mathrm{R}-\mathrm{Sq}=38.4\%\mathrm{R}-\mathrm{Sq}\left(\text{adj}\right)=& 37.7\%& \phantom{\rule{0ex}{0ex}}& \phantom{\rule{0ex}{0ex}}\end{array}$

$a=4.5459$

$b=4.8323$

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.

$\begin{array}{lcccc}\text{Predictor}& \text{Coef}& \text{SE Coef}& T& \text{P}\\ \text{Constant}& 4.5459& 0.6166& 7.37& <0.0001\\ \text{Dose}& 4.8323& 1.0164& 4.75& 0.0004\\ S=3.135& \mathrm{R}-\mathrm{Sq}=38.4\%\mathrm{R}-\mathrm{Sq}\left(\text{adj}\right)=& 37.7\%& \phantom{\rule{0ex}{0ex}}& \phantom{\rule{0ex}{0ex}}\end{array}$

(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.

$\begin{array}{lcccc}\text{Predictor}& \text{Coef}& \text{SE Coef}& T& \text{P}\\ \text{Constant}& 4.5459& 0.6166& 7.37& <0.0001\\ \text{Dose}& 4.8323& 1.0164& 4.75& 0.0004\\ S=3.135& \mathrm{R}-\mathrm{Sq}=38.4\%\mathrm{R}-\mathrm{Sq}\left(\text{adj}\right)=& 37.7\%& \phantom{\rule{0ex}{0ex}}& \phantom{\rule{0ex}{0ex}}\end{array}$

Define hypothesis

${H}_{0}:\beta =0$

${H}_{1}:\beta \ne 0$

The test statistic is

$t=\frac{b-{\beta}_{0}}{S{E}_{b}}\phantom{\rule{0ex}{0ex}}=\frac{4.8323-0}{1.0164}\phantom{\rule{0ex}{0ex}}=4.754$

The P-value is the probability of having the test numbers' value or a more dramatic value. The $t$-value in the row role="math" localid="1652872078633" $df=n-2\phantom{\rule{0ex}{0ex}}=15-2\phantom{\rule{0ex}{0ex}}=13$ 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.

$P<0.05\Rightarrow \text{Reject}{H}_{0}$

Degrees of freedom

$df=n-2\phantom{\rule{0ex}{0ex}}=15-2\phantom{\rule{0ex}{0ex}}=13$

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" $b-{t}^{*}\times S{E}_{b}=4.8323-2.160\times 1.0164\phantom{\rule{0ex}{0ex}}=2.636876$

$b+{t}^{*}\times S{E}_{b}=4.8323+2.160\times 1.0164\phantom{\rule{0ex}{0ex}}=7.027724$

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