# Data about the concentration of sulfur dioxide (linear regression)

At a measurement station in Munich for 14 days the sulfur dioxide concentration was measured. Analyze the influence of the mean temperature per day in grad Celsius ($X_1$) on the $SO_2$-concentration $(Y)$. Is there a special effect on weekends? The variable $X_2$ is $1$ if it was measured on a saturday or sunday otherwise it is $0$. The following data were measured: $$\begin{array}{c|c|c|c|c|c|c|c} \hline y & -3.15 & -2.83 & -3.02 & -3.08 & -3.54 & -2.98 & -2.78\\ x_1 & 16.47 & 16.02 & 16.81 & 22.87 & 21.68 & 21.23 & 20.55\\ x_2 & 0 & 0 & 0 & 1 & 1 & 0 & 0\\ \hline \end{array}$$ \begin{array}{c|c|c|c|c|c|c|c} \hline y & -3.35 & -2.76 & -1.90 & -2.12 & -2.45 & -1.97 & -2.23\\ x_1 & 18.32 & 15.96 & 15.36 & 12.47 & 12.46 & 11.77 & 11.72\\ x_2 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \end{array} (a) Estimate the regression coefficients of the multiple linear model and comment your results. (b) As $R^2$ one gets $R^2=0.5805$. Does the regressors have any role by explaining the $SO_2$-concentration? Make a global F-test of level $\alpha=0.01$. (c) The estimated variances are $\hat{\sigma}_1=0.0267$ and $\hat{\sigma}_2=0.2166$. Test the hypothesis $\theta_i=0$ for $i=2,3$ to the level $\alpha=0.05$ by confidence intervalls. Remove the covariable that does not have an influence and then make a simple linear regression. Compare the multiple and the simple regression graphically.

Hello and good day!

First of all I examined by ANOVA if there is a weekend effect. I chose $\alpha=0.05$. For the $F$-statstic I got $F=0.1$ and a p-value of $0.76$. So the null hypothesis that $\theta_1=\theta_2$ can not be fefused and so there is no weekend effect.

(a) The estimated coefficients are $$\hat{\theta}_1=-1.0019,~~~\hat{\theta}_2=-0.1032,~~~\hat{\theta}_3=-0.0025,$$ where the first coefficient is the intercept, the second coefficient belongs to $X_1$ and the thirds one to $X_2$.

Now I do not know what I should comment here? What could be meant? What is to be commented?

(b)

The global $F$-test says that the hypothesis $H_0: \theta_2=\theta_3=0$ can be refused. So the regressors play a role for explaining the $SO_2$-concentration.

(c)

For $\theta_2$ the confidence interval is $$C_2^{0.05}=(-0.156863,-0.049537),$$ so because $0\notin C_{2}^{0.05}$, the hypothesis $H_0^2: \theta_2=0$ can be refused.

For $\theta_3$ the confidence intervall is given by $$C_3^{0.05}=(-0.437834,0.432834),$$ so because of $0\in C_3^{0.05}$, the hypothesis $H_0^3: \theta_3=0$ cannot be refused, so the covariable $X_3$ does not have an influence and can be removed from the model.

I now made the simple linear regression, again using R, getting now $$\hat{\theta}_1=-1.0020,~~~\hat{\theta}_2=-0.1033.$$

What I see is, that the coefficients are nearly the same compared to those of the multiple regression.

But I do not know how to compare the simple linear regression and the multiple linear regression graphically. Of course, I can plot the simple linear regression:

but I do not know how to plot the multiple regression in order to compare it then with the simple linear regression. Can you help me with that?

So to sum it up:

Concerning (a) I do not know what is to be commented and concerning (c) I do not know how I can compare the multiple and the simple regression graphically.

Would be great if you could help me with that!

See you

Miroslav

• I'm stuck just trying to understand the data. What do you suppose those negative "concentrations" mean? – whuber Jun 10 '14 at 18:37
• The data of $Y$ were logarithmized. – mathfemi Jun 10 '14 at 19:21

I'm wondering if the negative value is related to the idea of pH. If there is a logarithm of a concentration between 0 and 100% then a negative prefix would be required for anything but pure sulfur dioxide. For all $x<1$ the value of $log_{10}\left(x\right)$ is negatively valued. When I check this (link) then I find that 800 micrograms per cubic meter is an alert-level of sulfur dioxide. If I compute $10^{-y}$ I can get on the order of ~800 units for some of the values. The measurement is one per this, so around 1 part per 800. I think that they compute how many micrograms per cubic meter, then take the $log_{10}$ of that.