Chapter 4, Problem 4.GE

(a)

Interpretation Introduction

Interpretation:

Whether the concentration $\text{pg/g}$ refers to parts per million, parts per billion or something else has to be given.

Explanation of Solution

Given data:

The sediments in North Sea which are traces of toxic, man-made hexachlorohexanes were extracted by two new procedures and by a known process.

The concentration of sediments measured by chromatography are given in the below Figure 1.

Figure 1

Expansion of $pg/g$:

The given concentration $\text{pg/g}$ corresponds to parts per trillion and which is equal to ${10}^{-12}\text{ g/g}$

Conclusion

The concentration $\text{pg/g}$ corresponds to ${10}^{-12}\text{ g/g}$ and is parts per trillion.

(b)

Interpretation Introduction

Interpretation:

Whether the standard deviations of procedure B and conventional procedure differ significantly has to be given.

Concept Introduction:

Comparison of Standard Deviation with F Test:

For the comparison of mean values of two sets of measurements, we decide whether the standard deviations of the two sets are “statistically different”. This is done by F test.

The F test with quotient F is given as:

${\text{F}}_{\text{calculated}}=\frac{s_1^2}{s_2^2}$

Where, $s_1$ and $s_2$ are standard deviations for the set of measurements using original instrument and substitute instrument.

If ${\text{F}}_{\text{calculated}}>{\text{F}}_{\text{table}}$ , then the difference is significant.

Degrees of Freedom:

For a n set of measurement, the degrees of freedom will be $(n-1)$

To Give: Whether the standard deviations of procedure B and conventional procedure differ significantly.

Answer to Problem 4.GE

The standard deviation of procedure B and conventional procedure are not significantly different.

Explanation of Solution

Given data:

The sediments in North Sea which are traces of toxic, man-made hexachlorohexanes were extracted by two new procedures and by a known process.

The concentration of sediments measured by chromatography are given in the below Figure 1.

Figure 1

Comparison of Standard deviation:

From the given data, let us assume

The standard deviation for procedure B as $s_1=4.6$

The standard deviation for conventional procedure as $s_2=3.6$

The number of measurements for procedure B as $n_1=6$

The number of measurements for conventional procedure as $n_2=6$

The significance in standard deviation is found using F test.

$\begin{array}{l}{\text{F}}_{\text{calculated}}=\frac{{4.6}^2}{{3.6}^2}\\ =1.63\\ =1.6(\text{rounded to correct significant number)}\end{array}$

The ${\text{F}}_{\text{table}}$ value corresponds to five degrees of freedom both in numerator and denominator is $5.05$

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

Hence, standard deviations are not significantly different at $95\%$ confidence level.

Conclusion

The standard deviation of procedure B and conventional procedure are found to be not significantly different.

(c)

Interpretation Introduction

Interpretation:

Whether the mean concentration of procedure B and conventional procedure differ significantly 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 mean concentration of procedure B and conventional procedure differ significantly

Answer to Problem 4.GE

The standard deviation of procedure B and conventional procedure are not significantly different.

Explanation of Solution

Given data:

The sediments in North Sea which are traces of toxic, man-made hexachlorohexanes were extracted by two new procedures and by a known process.

The concentration of sediments measured by chromatography are given in the below Figure 1.

Figure 1

Comparison of Standard deviation:

From the given data, let us assume

The standard deviation for procedure B as $s_1=4.6$

The standard deviation for conventional procedure as $s_2=3.6$

The number of measurements for procedure B as $n_1=6$

The number of measurements for conventional procedure as $n_2=6$

The result of F- test from part (b) indicates that the standard deviations are not significant.

Hence, the equation used to compare mean is as follows,

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

Calculate $s_{\text{pooled}}$ as below:

$\begin{array}{l}{s}_{\text{pooled}}=\sqrt{\frac{s_1^2(n_1-1)+s_2^2(n_2-1)}{n_1+n_2-2}}\\ =\sqrt{\frac{{4.6}^2(6-1)+{3.6}^2(6-1)}{6+6-2}}\\ =4.13\\ =4.1(\text{rounded off)}\end{array}$

The $t_{\text{calculated}}$ is found as follows,

$\begin{array}{l}{t}_{\text{calculated}}=\frac{\left|\overline{x_1}-\overline{x_2}\right|}{\sqrt{(s_{\text{pooled}})}}\sqrt{\frac{n_1{n}_2}{n_1+n_2}}\\ =\frac{\left|51.1-34.4\right|}{4.13}\sqrt{\frac{(6)(6)}{6+6}}\\ =7.0\end{array}$

The $t_{\text{table}}$ value corresponds to ten degrees of freedom is $2.228$

Thus, $t_{\text{calculated}}>t_{\text{table}}$

Therefore, the difference in mean concentrations are significant at the $95\%$ confidence level.

Conclusion

The mean concentration of procedure B and conventional procedure are found to be significantly different.

(d)

Interpretation Introduction

Interpretation:

Whether the standard deviation and mean concentration of procedure A and conventional procedure differ significantly has to be given.

Concept Introduction:

Comparison of Standard Deviation with F Test:

For the comparison of mean values of two sets of measurements, we decide whether the standard deviations of the two sets are “statistically different”. This is done by F test.

The F test with quotient F is given as:

${\text{F}}_{\text{calculated}}=\frac{s_1^2}{s_2^2}$

Where, $s_1$ and $s_2$ are standard deviations for the set of measurements using original instrument and substitute instrument.

If ${\text{F}}_{\text{calculated}}>{\text{F}}_{\text{table}}$ , then the difference is significant.

Degrees of Freedom:

For a n set of measurement, the degrees of freedom will be $(n-1)$

When the standard deviations of two sets of measurements are significantly different, the equations used are:

$\begin{array}{l}{t}_{\text{calculated}}=\frac{\left|\overline{x_1}-\overline{x_2}\right|}{\sqrt{(s_1^2/n_1+s_2^2/n_2)}}\\ \text{Degrees of freedom}\text{=}\frac{{(s_1^2/n_1+s_2^2/n_2)}^2}{\frac{{(s_1^2/n_1)}^2}{n_1-1}+\frac{{(s_2^2/n_2)}^2}{n_2-1}}\\ =\frac{{(u_1^2+u_2^2)}^2}{\frac{u_1^4}{n_1-1}+\frac{u_2^4}{n_2-1}}\end{array}$

To Give: Whether the standard deviation and mean concentration of procedure A and conventional procedure differ significantly

Answer to Problem 4.GE

The standard deviation of procedure A and conventional procedure are significantly different from each other.

The mean concentration of procedure A and conventional procedure are significantly different from each other at the $95\%$ confidence level.

Explanation of Solution

Given data:

The sediments in North Sea which are traces of toxic, man-made hexachlorohexanes were extracted by two new procedures and by a known process.

The concentration of sediments measured by chromatography are given in the below Figure 1.

Figure 1

Comparison of Standard deviation:

From the given data, let us assume

The standard deviation for conventional procedure as $s_1=3.6$

The standard deviation for procedure A as $s_2=1.2$

The number of measurements for conventional procedure as  $n_1=6$

The number of measurements for procedure A as $n_2=6$

The significance in standard deviation is found using F test.

$\begin{array}{l}{\text{F}}_{\text{calculated}}=\frac{{3.6}^2}{{1.2}^2}\\ =9.00\\ =9.0(\text{rounded to correct significant number)}\end{array}$

The ${\text{F}}_{\text{table}}$ value corresponds to five degrees of freedom both in numerator and denominator is $5.05$

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

Hence, standard deviations are significantly different at $95\%$ confidence level.

The result of F- test indicates that the difference in standard deviations are significant.

Hence, the equation used to compare mean is as follows,

$t_{\text{calculated}}=\frac{\left|\overline{x_1}-\overline{x_2}\right|}{\sqrt{(s_1^2/n_1+s_2^2/n_2)}}$

Calculate degrees of freedom as below:

$\begin{array}{l}\text{Degrees of freedom}\text{=}\frac{{(s_1^2/n_1+s_2^2/n_2)}^2}{\frac{{(s_1^2/n_1)}^2}{n_1-1}+\frac{{(s_2^2/n_2)}^2}{n_2-1}}\\ =\frac{{({3.6}^2/6+{1.2}^2/6)}^2}{\frac{{({3.6}^2/6)}^2}{6-1}+\frac{{({1.2}^2/6)}^2}{6-1}}\\ =6.10\\ \approx 6(\text{rounded off)}\end{array}$

The $t_{\text{calculated}}$ is found as follows,

$\begin{array}{l}{t}_{\text{calculated}}=\frac{\left|\overline{x_1}-\overline{x_2}\right|}{\sqrt{(s_1^2/n_1+s_2^2/n_2)}}\\ =\frac{\left|34.4-42.9\right|}{\sqrt{({3.6}^2/6+{1.2}^2/6)}}\\ =5.49\\ =5.4(\text{rounded off)}\end{array}$

The $t_{\text{table}}$ value corresponds to six degrees of freedom is $2.447$

Thus, $t_{\text{calculated}}>t_{\text{table}}$

Therefore, the difference in mean concentrations are significant at the $95\%$ confidence level.

Conclusion

The standard deviation of procedure A and conventional procedure are found out as significantly different from each other.

The mean concentration of procedure A and conventional procedure are found out as significantly different from each other at the $95\%$ confidence level.