Step 1:

Given that, the data display the information of the values of variables ${\rm{N}}{{\rm{o}}_x}$ emission rate(y) measured in $\left( {\frac{{{\rm{MBtu}}}}{{{\rm{hr - f}}{{\rm{t}}^{\rm{2}}}}}} \right)$ and burner area liberation rate (x) measured in ppm. The variable burner area liberation rate is measured in (ppm).

 
Step 2:

a.

Linear regression model:

A linear regression model is $\hat y = {\hat \beta _0} + {\hat \beta _1}x$, where $\hat y$ is the predicted values of response variable and x is the predictor variable. The ${\hat \beta _0}$ denote the estimate of y-intercept of the line and ${\hat \beta _1}$ be the estimate of slope.

The y-intercept is computed as follows,

\[\begin{array}{c} {{\hat \beta }_0} = \bar y - {{\hat \beta }_1}\bar x\\ = \frac{{\sum\limits_i {{y_i}} - {{\hat \beta }_1}\sum\limits_i {{x_i}} }}{n} \end{array}\]

The slope coefficient of linear regression is computed as follows,

\[\begin{array}{c} {{\hat \beta }_1} = \frac{{{S_{xy}}}}{{{S_{xx}}}}\\ = \frac{{\left[ {\sum\limits_i {{x_i}{y_i}} - \frac{{\left( {\sum\limits_i {{x_i}} } \right) \times \left( {\sum\limits_i {{y_i}} } \right)}}{n}} \right]}}{{\sum\limits_i {x_i^2} - \frac{{{{\left( {\sum\limits_i {{x_i}} } \right)}^2}}}{n}}} \end{array}\]

The sample size is $n = 14$.

The below table is used for computation of slope and intercept:

x	y	x^2	y^2	xy
100	150	10000	22500	15000
125	140	15625	19600	17500
125	180	15625	32400	22500
150	210	22500	44100	31500
150	190	22500	36100	28500
200	320	40000	102400	64000
200	280	40000	78400	56000
250	400	62500	160000	100000
250	430	62500	184900	107500
300	440	90000	193600	132000
300	390	90000	152100	117000
350	600	122500	360000	210000
400	610	160000	372100	244000
400	670	160000	448900	268000
Sumx= 3300	Sumy=5010	Sumx^2=913750	Sumy^2=2207100	Sumxy=1413500

Thus, the slope coefficient of linear regression is:

\[\begin{array}{c} {{\hat \beta }_1} = \frac{{\left[ {\sum\limits_i {{x_i}{y_i}} - \frac{{\left( {\sum\limits_i {{x_i}} } \right) \times \left( {\sum\limits_i {{y_i}} } \right)}}{n}} \right]}}{{\sum\limits_i {x_i^2} - \frac{{{{\left( {\sum\limits_i {{x_i}} } \right)}^2}}}{n}}}\\ = \frac{{\left[ {1413500 - \frac{{3300 \times 5010}}{{14}}} \right]}}{{913750 - \frac{{{{\left( {3300} \right)}^2}}}{{14}}}}\\ = \frac{{232571.4286}}{{135892.8571}}\\ = 1.7114 \end{array}\]

Therefore, the point estimate of ${\hat \beta _1}$ is 1.7114.

Thus, the y-intercept is:

\[\begin{array}{c} {{\hat \beta }_0} = \frac{{\sum\limits_i {{y_i}} - {{\hat \beta }_1}\sum\limits_i {{x_i}} }}{n}\\ = \frac{{5010 - \left( {1.7114 \times 3300} \right)}}{{14}}\\ = \frac{{ - 637.62}}{{14}}\\ = - 45.5443 \end{array}\]

Therefore, the y-intercept is $ - 45.5443$ .

Therefore, the regression line for the variable’s emission rate and burner area liberation rate is:

\[\begin{array}{c} \hat y = {{\hat \beta }_0} + {{\hat \beta }_1}x\\ = - 45.5443 + 1.7114x \end{array}\]

Now, to test the hypothesis that there is a relationship between two rates.

The null and alternative hypothesis is defined as follows,

The null hypothesis is that there is no relationship between the variables liberation rate and ${\rm{N}}{{\rm{o}}_x}$ emission rate.

The alternative hypothesis is that there is useful relationship between the variables liberation rate and ${\rm{N}}{{\rm{o}}_x}$ emission rate.

That is,

\[\begin{array}{l} {H_0}:{\beta _1} = 0\\ {H_a}:{\beta _1} \ne 0 \end{array}\]

Here, the hypothesis is two tailed.

The test statistics is,

\[t = \frac{{{{\hat \beta }_1} - {\beta _1}}}{{{s_{{{\hat \beta }_1}}}}} \sim {t_{\alpha ,n - 2}}\]

Now, the estimate of error standard deviation of slope coefficient is computed as follows,

\[\begin{array}{c} {s_{{{\hat \beta }_1}}} = \frac{{{S_{yx}}}}{{\sqrt {{S_{xx}}} }}\\ = \frac{{\sqrt {\frac{{\sum\limits_i {y_i^2} - {{\hat \beta }_0}\sum\limits_i {{y_i}} - {{\hat \beta }_1}\sum\limits_i {{x_i}{y_i}} }}{n}} }}{{\sqrt {\sum\limits_i {x_i^2} - \frac{{{{\left( {\sum\limits_i {{x_i}} } \right)}^2}}}{n}} }}\\ = \frac{{\sqrt {\frac{{2207100 - \left( { - 45.5443 \times 5010} \right) - \left( {1.7114 \times 1413500} \right)}}{{14}}} }}{{\sqrt {913750 - \frac{{{{\left( {3300} \right)}^2}}}{{14}}} }}\\ = \frac{{\sqrt {1158.0745} }}{{\sqrt {135892.8571} }} \end{array}\] \[ = 0.0923\]

Therefore, the estimate of error standard deviation of slope coefficient is ${s_{{{\hat \beta }_1}}} = 0.0923$ .

Thus, the test statistics is:

\[\begin{array}{c} t = \frac{{{{\hat \beta }_1} - {\beta _1}}}{{{s_{{{\hat \beta }_1}}}}}\\ = \frac{{1.7114 - 0}}{{0.0923}}\\ = 18.5417 \end{array}\]

Therefore, the value of test statistics is 18.5417.

Now, the critical value of t is computed as follows:

The degree of freedom is,

\[\begin{array}{c} df = n - 2\\ = 14 - 2\\ = 12 \end{array}\]

The level of significance is,

\[\begin{array}{c} \alpha = 1 - 0.99\\ = 0.01\\ \frac{\alpha }{2} = \frac{{0.01}}{2}\\ = 0.005 \end{array}\]

Using t-table, the value of t at 0.005 level of significance and 12 degrees of freedom for two tailed is $ \pm 3.428$ .

Therefore, the critical value of t is $ \pm 3.428$.

Decision rule:

Reject the null hypothesis, if $\left| t \right| > {t_\alpha }$.

Fail to reject the null hypothesis, if $\left| t \right| \le {t_\alpha }$ .

Conclusion:

Here, $t = 18.5417 > 3.428$.

Thus, the decision is to reject the null hypothesis.

So, it can be concluded that reject the null hypothesis at 5% level of significance

Therefore, the is sufficient evidence to conclude that there is useful relationship between the variables liberation rate and ${\rm{N}}{{\rm{o}}_x}$ emission rate at 1% level of significance.

 
Step 3:

b.

The Confidence interval for the slope regression line is,

\[CI = {\hat \beta _1} \pm {t_{\frac{a}{2},n - 2}} \times {s_{{{\hat \beta }_1}}}\]

Here, the confidence interval for the expected change in ${\rm{N}}{{\rm{o}}_x}$ emission rate associated with 10 $\left( {\frac{{{\rm{MBtu}}}}{{{\rm{hr - f}}{{\rm{t}}^{\rm{2}}}}}} \right)$ is computed as follows:

\[\begin{array}{c} CI = 10 \times \left( {{{\hat \beta }_1} \pm {t_{\frac{a}{2},n - 2}} \times {s_{{{\hat \beta }_1}}}} \right)\\ = 10 \times \left( {1.7114 \pm {t_{\frac{a}{2},n - 2}} \times 0.0923} \right) \end{array}\]

Now, the critical value of t is computed as follows:

The sample size is $n = 17$.

The level of significance is,

\[\begin{array}{c} 1 - \alpha = 1 - 0.95\\ = 0.05\\ \frac{\alpha }{2} = \frac{{0.05}}{2}\\ = 0.025 \end{array}\]

The degree of freedom is,

\[\begin{array}{c} df = n - 2\\ = 14 - 2\\ = 12 \end{array}\]

Using the t-table, the critical value of t at 0.025 level of significance and 12 degree of freedom for right tailed is 2.560.

Thus,

\[\begin{array}{c} CI = 10 \times \left( {1.7114 \pm {t_{\frac{a}{2},n - 2}} \times 0.0923} \right)\\ = 10 \times \left( {1.7114 \pm 2.560 \times 0.0923} \right)\\ = 10 \times \left( {1.7114 \pm 0.2363} \right)\\ = \left( {10 \times \left( {1.7114 - 0.2363} \right),10 \times \left( {1.7114 + 0.2363} \right)} \right) \end{array}\] \[\begin{array}{c} = \left( {10 \times 1.4751,10 \times 1.9477} \right)\\ = \left( {14.751,19.477} \right) \end{array}\]

Therefore, the 95% confidence interval for the expected change in ${\rm{N}}{{\rm{o}}_x}$ emission rate associated

with 10 $\left( {\frac{{{\rm{MBtu}}}}{{{\rm{hr - f}}{{\rm{t}}^{\rm{2}}}}}} \right)$ is lies between 14.751 to 19.477.