Chapter 12, Problem 61CE

a.

To determine

Write the fitted regression equation.

Answer to Problem 61CE

The regression equation is,

$\text{Monthly machine downtime}=1,743.57-1.2163\text{monthly maintenance spending}\text{.}$

Explanation of Solution

Calculation:

An output of a regression is given. The X variable is monthly maintenance spending and Y be the monthly machine downtime. The sample size is 15.

Regression:

Suppose $x_1\mathrm{...}{x}_n$ be n sample values of independent variable and the corresponding dependent variable values are $y_1\mathrm{...}{y}_n$. The slope and the intercept of ordinary least square can be defined as $b_0=\overline{y}-b_1\overline{x}$ and $b_1=\frac{S{S}_{xy}}{S{S}_{xx}}$.

Where, $S{S}_{xx},S{S}_{yy},S_{xy}$ are the sum of squares due to x, y and xy respectively. $\overline{x}\text{and}\overline{y}$ are the sample mean of the independent and dependent variable respectively.

The total sum of squares is denoted as, $SST={\displaystyle \sum _{i=1}^n{\left(y_i-\overline{y}\right)}^2}$.

The regression sum of squares is denoted as, $SSR={\displaystyle \sum _{i=1}^n{\left({\widehat{y}}_i-\overline{y}\right)}^2}$.

The error sum of squares is denoted as, $SSE={\displaystyle \sum _{i=1}^n{\left(y_i-{\widehat{y}}_i\right)}^2}$.

From the regression the fitted line is denoted as, $\widehat{y}=b_0+b_{1}x$ .

From the output, $b_0=1,743.57$$b_1=-1.2163$.

Hence, the regression equation is,

$\text{Monthly machine downtime}=1,743.57-1.2163\text{monthly maintenance spending}\text{.}$

b.

To determine

Find the degrees of freedom for the two tailed-test.

Find the two-tailed critical value of t using Appendix D.

Answer to Problem 61CE

The degree of freedom is 13 for the t-test.

The critical-value using Appendix D is 2.160.

Explanation of Solution

Calculation:

Critical value:

Here from the output, the sample size, $n=15$.

The degrees of freedom is,

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

For two tailed test, the critical value for t-test will be, $t_{\frac{\alpha }{2},\left(n-2\right)}$.

It is assumed that the level of significance, $\alpha =0.05$.

From the Appendix D: STUDENT’S t CRITICAL VALUES:

• • Locate the value 13 in the column of degrees of freedom.
• • Locate the 0.025 in level of significance.
• • The intersecting value that corresponds to the degrees of freedom 13 with level of significance 0.025 is 2.160.

Thus, the critical-value using Appendix D is 2.160.

c.

To determine

Write the conclusion about the slope.

Answer to Problem 61CE

There is an association between monthly machine downtime and monthly maintenance spending.

Explanation of Solution

Calculation:

Let ${\beta }_1$ be the slope parameter.

Hypotheses:

Null hypothesis:

$H_0:{\beta }_1=0$

That is, there is no association between X and Y.

Alternative hypothesis:

$H_0:{\beta }_1\ne 0$

That is, there is an association between X and Y.

Decision rule:

If $p\text{-value}\le \alpha$ , reject the null hypothesis.

If $p\text{-value >}\alpha$ , fail to reject the null hypothesis.

From the output, the p-value for the t-test of slope is 0.0161.

The level of significance is 0.05.

Conclusion:

Here the p-value is less than the level of significance.

That is, $0.00161\left(=p\text{-value}\right)<0.05\left(=\alpha \right)$ .

Hence, by the decision rule the null hypothesis will be rejected.

That is, the slope is significantly differs from zero.

Therefore, it can be concluded that there is an association between monthly machine downtime and monthly maintenance spending.

d.

To determine

Interpret the 95% confidence limits for the slope.

Explanation of Solution

Interpretation:

From the given output, the 95% confidence interval for the slope is (–2.1671, –0.2656).

It can be said that there is 95% confident that the slope lies between –2.1671 and –0.2656. The interval does not contain zero. That is, all values are negative that implies a relationship between monthly maintenance spending and monthly machine downtime.

e.

To determine

Verify $F=t^2$ for the slope.

Explanation of Solution

Calculation:

From the output the F statistic is 7.64.

For the slope the t-statistic is –2.764.

$\begin{array}{c}{t}^2={\left(-2.764\right)}^2\\ =7.64\\ =F\end{array}$

Hence, it can be concluded that $F=t^2$.

f.

To determine

Describe the fit of the regression.

Explanation of Solution

Calculation:

From the output, the R-squared value is 0.37.

$R^2$(R-squared):

The coefficient of determination ($R^2$) is defined as the proportion of variation in the observed values of the response variable that is explained by the regression. The squared correlation gives fraction of variability of response variable (y) accounted for by the linear regression model.

The $R^2$ value is 37%, which means that the percentage of variation in the observed values of monthly machine downtime that is explained by the regression is 37%, which indicates that 37% of the variability in monthly machine downtime is explained by the variability in monthly maintenance spending with a linear relationship.