Chapter 12, Problem T12.11SPT

(a)

To determine

To interpret each of the following: the slope, the y-intercept, the standard deviation of the residual and the standard error of the slope.

Explanation of Solution

Growth hormones are used to increase the weight of the chicken. So the experiment was conducted and the subsequent weight gain was recorded. A researcher plots the data and finds that there is a linear relationship appears to hold. Assuming that the conditions for performing inferences about the slope of the regression line are met, there is an computer output given in the question. Thus, we have,

1. The slope: The slope $b_1$ is given in the row “Dose” and in the column “Coef” of the given computer output, i.e.,

  $b_1=4.8323$

The slope represents the average increase or decrease of y per unit of x. on average, the weight gain increases $4.8323$ ounces per mg of growth hormone.

2. The y-intercept of $b_0$is given in the row “Constant” and in the column “Coef” of the given computer output as,

  $b_0=4.5459$

The y-intercept represents the average y-value when x is zero. Thus, on average, the weight gain is $4.5459$ ounces, when the dose of the growth hormone is zero mg.

3. The standard error of the estimate s is given after “S=” in the given computer output as:

  $s=3.135$

The standard error of the estimate s represents the average error of predictions thus, the average deviation between actual y-values and the predicted y-values. Thus, the predicted weight gain deviates on average by $3.135$ ounces from the actual weight gain.

The standard error of the slope $S{E}_{b_1}$ is given in the row “Dose” and in the column “SE Coef” of the given computer output as:

  $S{E}_{b_1}=1.0164$

4. The standard error of the slope represents the average deviation of the slope of the sample regression line from the slope of the population regression line. Thus, the slope of the sample regression line varies on average by $1.0164$ from the slope of the true population regression line.

(b)

To determine

To explain do the data provide convincing evidence of a linear relationship between dose and weight gain and carry out the significance test at $\alpha =0.05$ level.

Answer to Problem T12.11SPT

There is a convincing evidence of a linear relationship between dose and weight gain.

Explanation of Solution

Growth hormones are used to increase the weight of the chicken. So the experiment was conducted and the subsequent weight gain was recorded. A researcher plots the data and finds that there is a linear relationship appears to hold. Assuming that the conditions for performing inferences about the slope of the regression line are met, there is an computer output given in the question. Thus, we have,

  $\begin{array}{l}b=4.8323\\ S{E}_b=1.0164\\ n=15\end{array}$

Now, the hypotheses is defined by:

  $\begin{array}{l}{H}_0:\beta =0\\ H_1:\beta \ne 0\end{array}$

The value of test statistics is as:

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

The degrees of freedom is as:

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

The P-value is the probability of obtaining the value of the test statistic or a value more extreme. Thus, the P-value is as:

  $P<0.0005$

If the P-value is less than or equal to the significance level, then the null hypothesis is rejected. Thus,

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

Thus, there is a sufficient evidence to support the claim.

(c)

To determine

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

Answer to Problem T12.11SPT

We are $95\%$ confidence that the weight gain is increases between $2.636876\text{ and }7.027724$ ounces when the dose increases by one milligram.

Explanation of Solution

Growth hormones are used to increase the weight of the chicken. So the experiment was conducted and the subsequent weight gain was recorded. A researcher plots the data and finds that there is a linear relationship appears to hold. Assuming that the conditions for performing inferences about the slope of the regression line are met, there is an computer output given in the question.Thus, we have,

  $\begin{array}{l}b=4.8323\\ S{E}_b=1.0164\\ n=15\end{array}$

The degrees of freedom is as:

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

The critical value can be found in the table B in the row of $df=13$ and in the column of $c=95\%$ , i.e.

  $t^{*}=2.160$

The confidence interval be:

  $\begin{array}{l}b-t^{*}\times S{E}_b\\ =4.8323-2.160\times 1.0164\\ =2.636876\\ b+t^{*}\times S{E}_b\\ =4.8323+2.160\times 1.0164\\ =7.027724\end{array}$

Thus, we conclude that we are $95\%$ confidence that the weight gain is increases between $2.636876\text{ and }7.027724$ ounces when the dose increases by one milligram.