Chapter 4, Problem 4.23P

(a)

Interpretation Introduction

Interpretation:

Whether the two given methods agree with each other at the $95\%$ confidence level for both rain water and drinking water has to be given.

Concept Introduction:

Comparing replicate measurements:

When the two standard deviations are not significantly different from each other, the equation used is:

$\begin{array}{l}t\text{ test for comparison of means: }\\ t=\frac{\left|\overline{x_1}-\overline{x_2}\right|}{\sqrt{(s_{\text{pooled}})}}\sqrt{\frac{n_1{n}_2}{n_1+n_2}}(\text{since }{s}_{\text{pooled}}=\sqrt{\frac{s_1^2(n_1-1)+s_2^2(n_2-1)}{n_1+n_2-2}})\end{array}$

To Give: Whether the two given methods agree with each other at the $95\%$ confidence level for both rain water and drinking water

Answer to Problem 4.23P

The difference between the two methods of rain water is not significant.

The difference between the two methods of drinking water is not significant.

Explanation of Solution

Given data:

The results for the measurement of nitrite $({\text{NO}}_{\text{2}}{}^{-})$ using the given two methods is given below in Figure 1 as follows:

Figure 1

Rainwater:

Let us assume,

The mean $\pm$ standard deviation by spectrophotometry as $\overline{x_1}\pm s_1=0.063\pm 0.008\text{ mg/L}$

The mean $\pm$ standard deviation by gas chromatography as $\overline{x_2}\pm s_2=0.069\pm 0.005\text{ mg/L}$

The number of measurement of spectrophotometry method is $n_1=5$

The number of measurement of spectrophotometry methods is $n_2=7$

F-Test:

First, calculate F-Test and find whether the standard deviations of the given methods are significant or not.

$\begin{array}{l}{\text{F}}_{\text{calculated}}=\frac{s_1^2}{s_2^2}\\ =\frac{{0.008}^2}{{0.005}^2}\\ =2.56\end{array}$

For $4$ degrees of freedom in the numerator and $6$ degrees of freedom in the denominator, the ${\text{F}}_{\text{table}}$ value is $4.53$

${\text{F}}_{\text{calculated}}{\text{<F}}_{\text{table}}$

Hence, the standard deviations of two sets of measurements are not significantly different.

Therefore, the equation used is:

$t=\frac{\left|\overline{x_1}-\overline{x_2}\right|}{\sqrt{(s_{\text{pooled}})}}\sqrt{\frac{n_1{n}_2}{n_1+n_2}}$

Calculate $s_{\text{pooled}}$ as follows,

$\begin{array}{l}{s}_{\text{pooled}}=\sqrt{\frac{{0.005}^2(7-1)+{0.008}^2(5-1)}{7+5-2}}\\ =\sqrt{\frac{{0.005}^2(6)+{0.008}^2(4)}{7+5-2}}\\ =0.00637\end{array}$

Calculate the t value:

$\begin{array}{l}t=\frac{\left|\overline{x_1}-\overline{x_2}\right|}{\sqrt{(s_{\text{pooled}})}}\sqrt{\frac{n_1{n}_2}{n_1+n_2}}\\ \text{ =}\frac{\left|0.069-0.063\right|}{0.00637}\sqrt{\frac{(7)(5)}{7+5}}\\ \text{ =1}\text{.61}\end{array}$

The value of ${\text{t}}_{\text{table}}$ is $2.228$

$t_{\text{calculated}}<t_{\text{table}}$

Here, $t_{\text{calculated}}$ is much smaller than $t_{\text{table}}$

Therefore, the difference is not significant.

Drinking water:

Let us assume,

The mean $\pm$ standard deviation by spectrophotometry as $\overline{x_1}\pm s_1=0.087\pm 0.008\text{ mg/L}$

The mean $\pm$ standard deviation by gas chromatography as $\overline{x_2}\pm s_2=0.078\pm 0.007\text{ mg/L}$

The number of measurement of spectrophotometry method is $n_1=5$

The number of measurement of spectrophotometry methods is $n_2=5$

F-Test:

First, calculate F-Test and find whether the standard deviations of the given methods are significant or not.

$\begin{array}{l}{\text{F}}_{\text{calculated}}=\frac{s_1^2}{s_2^2}\\ =\frac{{0.008}^2}{{0.007}^2}\\ =1.31\end{array}$

For $4$ degrees of freedom in the numerator and $4$ degrees of freedom in the denominator, the ${\text{F}}_{\text{table}}$ value is $6.39$

${\text{F}}_{\text{calculated}}{\text{<F}}_{\text{table}}$

Hence, the standard deviations of two sets of measurements are not significantly different.

Therefore, the equation used is:

$t=\frac{\left|\overline{x_1}-\overline{x_2}\right|}{\sqrt{(s_{\text{pooled}})}}\sqrt{\frac{n_1{n}_2}{n_1+n_2}}$

Calculate $s_{\text{pooled}}$ as follows,

$\begin{array}{l}{s}_{\text{pooled}}=\sqrt{\frac{{0.007}^2(5-1)+{0.008}^2(5-1)}{5+5-2}}\\ =\sqrt{\frac{{0.007}^2(4)+{0.008}^2(4)}{5+5-2}}\\ =0.00752\end{array}$

Calculate the t value:

$\begin{array}{l}t=\frac{\left|\overline{x_1}-\overline{x_2}\right|}{\sqrt{(s_{\text{pooled}})}}\sqrt{\frac{n_1{n}_2}{n_1+n_2}}\\ \text{ =}\frac{\left|0.087-0.078\right|}{0.00752}\sqrt{\frac{(5)(5)}{5+5}}\\ \text{ =1}\text{.89}\end{array}$

The value of ${\text{t}}_{\text{table}}$ is $2.306$

$t_{\text{calculated}}<t_{\text{table}}$

Here, $t_{\text{calculated}}$ is much smaller than $t_{\text{table}}$

Therefore, the difference is not significant.

Conclusion

The difference between the two methods of rain water is found out to be not significant.

The difference between the two methods of drinking water is found out to be not significant.

(b)

Interpretation Introduction

Interpretation:

Whether the drinking water contain more nitrite than the rain water at the $95\%$ confidence level has to be given

Concept Introduction:

Comparing replicate measurements:

When the two standard deviations are not significantly different from each other, the equation used is:

$\begin{array}{l}t\text{ test for comparison of means: }\\ t=\frac{\left|\overline{x_1}-\overline{x_2}\right|}{\sqrt{(s_{\text{pooled}})}}\sqrt{\frac{n_1{n}_2}{n_1+n_2}}(\text{since }{s}_{\text{pooled}}=\sqrt{\frac{s_1^2(n_1-1)+s_2^2(n_2-1)}{n_1+n_2-2}})\end{array}$

To Give: Whether the drinking water contain more nitrite than the rain water at the $95\%$ confidence level

Answer to Problem 4.23P

At the $95\%$ confidence level, the difference between the rain water and drinking water is significant.

Explanation of Solution

Given data:

The results for the measurement of nitrite $({\text{NO}}_{\text{2}}{}^{-})$ using the given two methods is given below in Figure 1 as follows:

Figure 1

Gas chromatography:

The equation used is:

$t=\frac{\left|\overline{x_1}-\overline{x_2}\right|}{\sqrt{(s_{\text{pooled}})}}\sqrt{\frac{n_1{n}_2}{n_1+n_2}}$

Calculate $s_{\text{pooled}}$ as follows,

$\begin{array}{l}{s}_{\text{pooled}}=\sqrt{\frac{{0.005}^2(6)+{0.007}^2(4)}{7+5-2}}\\ =0.00588\end{array}$

Calculate the t value:

$\begin{array}{l}t=\frac{\left|\overline{x_1}-\overline{x_2}\right|}{\sqrt{(s_{\text{pooled}})}}\sqrt{\frac{n_1{n}_2}{n_1+n_2}}\\ \text{ =}\frac{\left|0.078-0.069\right|}{0.00588}\sqrt{\frac{(7)(5)}{7+5}}\\ \text{ =2}\text{.61}\end{array}$

The value of ${\text{t}}_{\text{table}}$ is $2.228$

$t_{\text{calculated}}<t_{\text{table}}$

Here, $t_{\text{calculated}}$ is much smaller than $t_{\text{table}}$

Therefore, the difference is not significant.

Spectrophotometry:

The equation used is:

$t=\frac{\left|\overline{x_1}-\overline{x_2}\right|}{\sqrt{(s_{\text{pooled}})}}\sqrt{\frac{n_1{n}_2}{n_1+n_2}}$

Calculate $s_{\text{pooled}}$ as follows,

$\begin{array}{l}{s}_{\text{pooled}}=\sqrt{\frac{{0.008}^2(4)+{0.008}^2(4)}{5+5-2}}\\ =0.00800\end{array}$

Calculate the t value:

$\begin{array}{l}t=\frac{\left|\overline{x_1}-\overline{x_2}\right|}{\sqrt{(s_{\text{pooled}})}}\sqrt{\frac{n_1{n}_2}{n_1+n_2}}\\ \text{ =}\frac{\left|0.087-0.063\right|}{0.00800}\sqrt{\frac{(5)(5)}{5+5}}\\ \text{ =4}\text{.74}\end{array}$

The value of ${\text{t}}_{\text{table}}$ is $2.306$

$t_{\text{calculated}}<t_{\text{table}}$

Here, $t_{\text{calculated}}$ is much smaller than $t_{\text{table}}$

Therefore, the difference is not significant.

Conclusion

At the $95\%$ confidence level, the difference between the rain water and drinking water is found out to be significant.