Chapter 12, Problem 11PT

(a)

To determine

To find: the equation on the basis of given of the least-squares regression line and explain the variable.

Answer to Problem 11PT

  $\widehat{y}=4.5459+4.8323x$

Explanation of Solution

Given:

  

The coefficient of a and b

  $\begin{array}{l}a=4.5459\\ b=4.8323\end{array}$

Calculation:

regression line of least squares

  $\widehat{y}=a+bx$

Then regression line of least squares becomes:

  $\widehat{y}=4.5459+4.8323x$

With x Dose and y weight gain.

(b)

To determine

To Explain: the given terms.

Explanation of Solution

Given:

  

1. The slope
2. The standard error of the error
3. They $y$ interpret
4. $r^2$
5. s

1. Output as:

b = 4. The implies the weight could rise by 4.8323 ounces per milligram

2. The y-intercept is mention in the output a=4.5450

This implies that if the dosage is 0 milligrams the weight is projected to be 4.5459 ounces.

3. output as: s = 3.135

This implies that the expected error on predictions is 3.135ounces.

4. the output as: $S{E}_b=1.0164$

This means that the real slope of the population ranges 1.016-1 on average for all possible samples.

5. $r^2$ is mention in the output as: $r^2=R-S_q=38.4\%$

This implies that the least-square regression line describes 38.4 percent of the variance between the variables

(c)

To determine

To find: the significance test for the $\alpha =0.05$ level on the basis of given information.

Answer to Problem 11PT

Yes, there’s sufficient convincing evidence to justify the argument.

Explanation of Solution

Given:

  

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

Formula used:

  $t=\frac{b-{\beta }_0}{S{E}_b}$

Calculation:

The hypotheses testing

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

The test statistic is

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

The P-value is the chance of having the value of the test numbers, or a more drastic value. The P-value is the number (or interval) in Table B column title which contains the t-value in the row $df=n-2=15-2=13$ :

  $P<0.0005$

If the P-value is below or equal to the degree of importance, the null hypothesis is rejected. $P<0.05\Rightarrow \text{Reject }{H}_0$

There's sufficient convincing evidence to justify the argument.

(d)

To determine

To construct: and explain the slope parameter of 95 percent confidence interval.

Answer to Problem 11PT

(2.636876, 7.027724)

There is 95 percent assurance that between 2.636876 and 7.027724 the weight gain will rise

Ounces rise by 1 milligram while the does.

Explanation of Solution

Given:

  

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

Formula used:

For the interval

  $\begin{array}{l}b-t^{\ast }\times S{E}_b\\ b+t^{\ast }\times S{E}_b\end{array}$

Calculation:

The degrees of freedom

  $df=n-2=15-2=13$

The critical t-value can be found in table B in the row of $df=13$ and in the column of c=95%

The critical t-value can be shown in table B in the column of c=95% section and in the row of $df=13$

  $t^{\ast }=2.160$

The boundaries are

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

There is 95 percent assurance that between 2.636876 and 7.027724 the weight gain will rise

Ounces rise by 1 milligram while the does.