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Answers without the blur. Sign up and see all textbooks for free! Q. 11

Expert-verified Found in: Page 797 ### The Practice of Statistics for AP

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 $\stackrel{^}{y}=4.5459+4.8323x$.

(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.

See the step by step solution

## Part(a) Step 1: Given Information

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

## Part(a) Step 2: Explanation

$a=4.5459$

$b=4.8323$

The general regression line equation

$\stackrel{^}{y}=a+bx$

By inserting values the regression line becomes:

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

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

## Part(b) Step 1: Given Information

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

## 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

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

## Part(c) Step 2: Explanation

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.

$P<0.05⇒\text{Reject}{H}_{0}$

## Part(d) Step 1: Given Information

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

## Part(d) Step 2: Explanation

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

$b+{t}^{*}×S{E}_{b}=4.8323+2.160×1.0164\phantom{\rule{0ex}{0ex}}=7.027724$ ### Want to see more solutions like these? 