91Ó°ÊÓ

(a)

Two innovative processes and a well-known technology were used to recover quantities of harmful, man-made hexachlorohexanes from sediments in the North Sea.

The chromatographic concentrations of sediments are shown in the diagram below 1.50.

Expansion of pg/g:

The concentration in question pg/g stands for parts per trillion and is equivalent to ${10}^{-12}\mathrm{pg}/\mathrm{g}$.

(b)

It must be determined whether the standard deviations of methodB and the traditional technique differ considerably.

Comparison of Standard Deviation with F Test:

It is determined if the standard deviations of two sets of measurements are "statistically different" when comparing mean values. The F test is used to do this.

The F test with quotient F is given as:

${\mathrm{F}}_{\mathrm{calculated}}=\frac{{\mathrm{s}}_1^2}{{\mathrm{s}}_2^2}$

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

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

Comparison of Standard deviation:

Let’s assume

The standard deviation for procedureB 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.

$$\begin{gathered} F_{calculated}=\frac{4.6^2}{3.6^2} \\ =1.63 \\ =1.6 \end{gathered}$$

(Roundedto correct significant number)

The value of 5.05 in the numerator and denominator corresponds to five degrees of freedom in both the numerator and denominator.

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

As a result, at the 95% confidence level, standard deviations are not statistically different.

(c)

It must be determined whether the mean concentration of procedure B and the usual technique differs considerably.

Comparing replicate measurements:

The equation to utilise when the two standard deviations are not statistically different is:

t test for comparison of means:

$t=\frac{\left|x_1-n_n\right|}{\sqrt{\left(x_{mand}\right)}}\sqrt{\frac{n_{1122}}{n_1+n_2}}\left(\sin ce{s}_{poolod}=\sqrt{\frac{s_3^2\left(n_1-1\right)+s_2^2\left(n_2-1\right)}{n_1+n_2-2}}\right)$

Comparison of Standard deviation:

Let’s 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.

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

As a result, the equation for comparing means is as follows:

t test for comparison of means:

Calculate$s_{\mathrm{poolod}}$as below:

$$\begin{gathered} s_{poolod}=\sqrt{\frac{3_1^2\left(n_1-1\right)+s_2^2\left(m_2-1\right)}{n_1+m_2-2}} \\ =\sqrt{\frac{4.6^2\left(6-1\right)+3.6^2\left(6-1\right)}{6+6-2}} \\ =4.13 \\ =4.1\left(\mathrm{rounded}\mathrm{off}\right) \end{gathered}$$

The $t_{table}$is found as follows,

$$\begin{gathered} t_{calculated}=\frac{\left|\sqrt{x_1}-x_2\right|}{\sqrt{\left(x_{\mathrm{ponent}}\right)}}\sqrt{\frac{n_1{m}_2}{n_1+m_2}} \\ =\frac{\left|51.1-34.4\right|}{4.13}\sqrt{\frac{\left(6\right)\left(6\right)}{6+6}} \\ =7.0 \end{gathered}$$

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

Thus, $t_{\mathrm{calculated}}>t_{\mathrm{table}}$.

As a result, at the 95% confidence level, the difference in mean concentrations is considerable.

(d)

The chromatography-measured sediment concentrations are listed below.

Comparison of Standard deviation:

Let’s 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 usingtest.

role="math" localid="1663565947175" $$\begin{gathered} {\mathrm{F}}_{\mathrm{calculated}}=\frac{3.6^2}{1.2^2} \\ =9.00 \\ =9.0 \end{gathered}$$

(rounded to correct significant number)

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

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

As a result, standard deviations differ dramatically at the 95% confidence level.

The difference in standard deviations is considerable, according to the F- test result.

As a result, the equation for comparing means is as follows:

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

$$\begin{gathered} Degreesoffreedom=\frac{{\left(s_1^2/n_1+s_2^2/n_2\right)}^2}{{\displaystyle \frac{\left({\left(3l_2^2{n}_1\right)}^2\right)}{n_1-1}}+{\displaystyle \frac{\left({\left(2l_2^2{m}_2\right)}^2\right)}{n_2-1}}} \\ =\frac{{\left(3.6^2/6+1.2^2/6\right)}^2}{{\displaystyle \frac{{\left(.6^2\%\right)}^2}{G-1}}+{\displaystyle \frac{{\left({12}^2\%\right)}^2}{G-1}}} \\ =6.10 \\ ≈6\left(\mathrm{rounded}\mathrm{off}\right) \end{gathered}$$

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

$$\begin{gathered} t_{\mathrm{calculated}}=\frac{\left|34.4-42.9\right|}{\sqrt{\left(3.6^2/6+1.2^2/6\right)}} \\ =5.49 \\ =5.5\left(\mathrm{rounded}\mathrm{off}\right) \end{gathered}$$

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

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

As a result, at the 95% confidence level, the difference in mean concentrations is considerable.