Chapter 13.2, Problem 17E

The following data on mass rate of burning x and flame length y is representative of that which appeared in the article “Some Burning Characteristics of Filter Paper” (Combustion Science and Technology, 1971: 103–120):

x	1.7	2.2	2.3	2.6	2.7	3.0	3.2
y	1.3	1.8	1.6	2.0	2.1	2.2	3.0
x	3.3	4.1	4.3	4.6	5.7	6.1
y	2.6	4.1	3.7	5.0	5.8	5.3

a. Estimate the parameters of a power function model.

b. Construct diagnostic plots to check whether a power function is an appropriate model choice. c. Test H0: β = 4/3 versus H_a: β < 4/3, using a level .05 test.

d. Test the null hypothesis that states that the median flame length when burning rate is 5.0 is twice the median flame length when burning rate is 2.5 against the alternative that this is not the case.

To determine

Estimate the parameters of power model.

Answer to Problem 17E

The estimate the parameters of power model are 0.626 and 1.254x.

Explanation of Solution

Given info:

The data shows the mass rate of burning x and the length of flame y.

Calculation:

The power model is given below:

$y=\alpha x^{\beta }$

Where, y is transformed into ln(y), x is transformed into ln(x).

The linear function is $\ln \left(y\right)=\ln \left(\alpha \right)+\beta \left(\ln \left(x\right)\right)$

The estimates of the parameters $\alpha ,\beta$ are ${\widehat{\beta }}_0,{\widehat{\beta }}_1$ which is calculated by using the formula:

${\widehat{\beta }}_1=\frac{{\displaystyle \sum {x^{\prime }}_i{y^{\prime }}_i}-\frac{{\displaystyle \sum {x^{\prime }}_i{y^{\prime }}_i}}{n}}{{\displaystyle \sum {\left({x^{\prime }}_i\right)}^2-\frac{{\left({\displaystyle \sum {x^{\prime }}_i}\right)}^2}{n}}}$

$\widehat{{\beta }_0}=\frac{{\displaystyle \sum {y^{\prime }}_i}-\widehat{{\beta }_1}{\displaystyle \sum {x^{\prime }}_i}}{n}$

Where,

${\displaystyle \sum {x^{\prime }}_i}=\ln \left(x\right)$ and ${\displaystyle \sum {y^{\prime }}_i}=\ln \left(y\right)$

The table below shows the calculation of estimating the parameters:

S. No	ln(y)	ln(x)	${\displaystyle \sum {x^{\prime }}_i{y^{\prime }}_i}$	${\displaystyle \sum {\left(y_i^{\text{'}}\right)}^2}$	${\displaystyle \sum {\left(x_i^{\text{'}}\right)}^2}$
1	0.2624	0.5307	0.139256	0.068854	0.281642
2	0.5878	0.7885	0.46348	0.345509	0.621732
3	0.4701	0.833	0.391593	0.220994	0.693889
4	0.6932	0.9556	0.662422	0.480526	0.913171
5	0.742	0.9933	0.737029	0.550564	0.986645
6	0.7885	1.0987	0.866325	0.621732	1.207142
7	1.0987	1.1632	1.278008	1.207142	1.353034
8	0.9556	1.194	1.140986	0.913171	1.425636
9	1.411	1.411	1.990921	1.990921	1.990921
10	1.3084	1.4587	1.908563	1.711911	2.127806
11	1.6095	1.5261	2.456258	2.59049	2.328981
12	1.7579	1.7405	3.059625	3.090212	3.02934
13	1.6678	1.8083	3.015883	2.781557	3.269949
Total	13.3529	15.5016	18.11035	16.57358	20.22989

$\begin{array}{c}\widehat{{\beta }_1}=\frac{\left(18.11035\right)-\frac{\left(13.3529\right)\left(15.5016\right)}{13}}{20.2298-\frac{{\left(15.501\right)}^2}{13}}\\ =\frac{18.11035-15.92241}{20.2298-\frac{240.28}{13}}\\ =\frac{2.18795}{20.2298-18.483}\\ =\frac{2.18795}{1.7468}\end{array}$

        =1.253

$\begin{array}{c}{\widehat{\beta }}_0=\frac{{\displaystyle \sum {y^{\prime }}_i}-\widehat{{\beta }_1}{\displaystyle \sum {x^{\prime }}_i}}{n}\\ =\frac{13.3529-\left(1.253\right)\left(15.5016\right)}{13}\\ =\frac{13.3529-19.424}{13}\\ =\frac{-6.0711}{13}\end{array}$

     = –0.467

Thus, the estimates of the parameters are given below:

$\begin{array}{c}\beta =\widehat{{\beta }_1}\\ =1.253\end{array}$

Similarly,

$\begin{array}{c}\widehat{\alpha }=e^{-0.467}\\ =e^{-0.467}\\ =0.627\end{array}$

To determine

Construct a diagnostic plot for checking the appropriate of power model.

Answer to Problem 17E

The diagnostic plot is given below:

Explanation of Solution

Calculation:

Software procedure:

Step-by-step procedure to construct a diagnostic plot is given below:

• Choose Stats>Regression> Regression.
• Select Simple and click OK
• Under Response, choose the column containing ln(y).
• Under Predictors, choose the column containing ln(x).
• Click Graphs, select residuals versus fits.
• Click OK.

Output obtained from MINITAB is given below:

The residual plot versus fitted values shows that the errors are randomly distributed with mean 0. This tells that the power model is appropriate to use for the given data.

The R-square value is 96% which tells that ln of mass rate of burning x can explain 96% of the variation in ln of flame length.

Hence, a power model is appropriate.

To determine

Test the hypotheses $H_0:\beta =\frac{4}{3}$ versus $H_a:\beta <\frac{4}{3}$ at 5% level of significance.

Answer to Problem 17E

There is sufficient evidence to conclude that $\beta =\frac{4}{3}$.

Explanation of Solution

Calculation:

$H_0:\beta =\frac{4}{3}$

That is, the slope coefficient equals to $\frac{4}{3}$.

$H_a:\beta <\frac{4}{3}$

That is, the slope coefficient is lesser than $\frac{4}{3}$.

Test statistic:

$\begin{array}{c}t=\frac{\widehat{\beta }-\beta }{SE\left(\widehat{\beta }\right)}\\ =\frac{1.253-\frac{4}{3}}{0.07752}\\ =\frac{1.253-1.333}{0.07752}\\ =\frac{-0.08}{0.07752}\end{array}$

   =–1.032

P-value:

Software procedure:

Step-by-step procedure to find the P-value is given below:

• Choose Graph>Probability distribution Plot>View Probability.
• Select t, enter 11 for degrees of freedom.
• Under Shaded Area tab, select X value and click on Both tails.
• Enter ­1.032 for X value.
• Click OK.

Output obtained from MINITAB is given below:

Conclusion:

The P-value is 0.1621 and the level of significance is 0.05.

The P-value is greater than the level of significance is 0.05.

That is, 0.1621>0.05.

Thus, the null hypothesis is not rejected.

Thus, there is sufficient evidence to conclude that $\beta =\frac{4}{3}$.

To determine

Test the hypothesis that the whether the median flame with 5.0 burning rate is twice the median flame length when the burning rate is 2.5 or not.

Answer to Problem 17E

There is no sufficient evidence to conclude the median flame with 5.0 burning rate is twice the median flame length when the burning rate is 2.5

Explanation of Solution

Calculation:

The median flame with 5.0 burning rate is twice the median flame length when the burning rate is 2.5 can be expressed as,

$\begin{array}{l}{\mu }_{Y.5}=2{\mu }_{Y\left(2.5\right)}\\ \alpha {\left(5\right)}^{\beta }=2\alpha {\left(2.5\right)}^{\beta }\end{array}$

The hypothesis test is given below:

$H_0:\beta =1$

$H_a:\beta \ne 1$

Test statistic:

$\begin{array}{c}t=\frac{\widehat{\beta }-\beta }{SE\left(\widehat{\beta }\right)}\\ =\frac{1.253-1}{0.0775}\\ =\frac{1.253-1}{0.0775}\\ =\frac{0.253}{0.0775}\end{array}$

   =3.265

P-value:

Software procedure:

Step-by-step procedure to find the P-value is given below:

• Choose Graph>Probability distribution Plot>View Probability.
• Select t, enter 11 for degrees of freedom.
• Under Shaded Area tab, select X value and click on Both tails.
• Enter 3.26 for X value.
• Click OK.

Output obtained from MINITAB is  given below:

Thus, the P-value is $2\left(0.004\right)=0.008$

Conclusion:

The P-value is 0.008 and the level of significance is 0.01.

The P-value is lesser than the level of significance.

That is 0.008<0.01.

Thus, the null hypothesis is rejected.

Thus, there is no sufficient evidence to conclude the median flame with 5.0 burning rate is twice the median flame length when the burning rate is 2.5.