Nitrite$\left({\mathrm{NO}}_2^{-}\right)$was measured by two methods in rainwater and unchlorinated drinking water. The results$\mathbf{\pm }$standard deviation (number of samples) are

SOURCE. Data from I. Sarudi and I. Nagy, Talanta 1995,42,1099

(a) Do the two methods agree with each other at theconfidence level for both rainwater and drinking water?

(b) For each method, does the drinking water contain significantly more nitrite than the rainwater (at the 95%confidence level)?

A standard deviation is a measurement of a collection of the variance or dispersion of values. A low standard deviation implies that the values are close to a set's mean, whereas a high standard deviation shows that the values are dispersed throughout a larger range.

(a) To know whether the two methods agree with each other atconfidence level for both rainwater and drinking water, we will perform theF and t-tests for both the methods,for rainwater and drinking water.

For rainwater,

TheF -testis done as shown below:

Let$s_1$ and$s_2$be equal to the standard deviations of spectrophotometry and gas chromatography respectively, since$s_1>s_2$.

$$\begin{gathered} F_{calc}=\frac{s_1^2}{s_2^2} \\ =\frac{{\left(0.0008\right)}^2}{{\left(0.005\right)}^2} \\ =2.56 \end{gathered}$$

$F_{table}=4.53$(based on Table 4-3; degrees of freedom for$s_1=4$ and$s_2=6$).

Since$F_{calc}<F_{table},$ the variances and standard deviations of the two techniques are not significantly different.

The t-test calculates the${\mathrm{s}}_{\mathrm{pooled}}$ as a prerequisite for solving the t -test

$$\begin{gathered} s_{pooled}=\sqrt{\frac{s_1^2\left(n_1-1\right)+s_2^2\left(n_2-1\right)}{n_1+n_2-n_t}} \\ =\sqrt{\frac{{\left(0.008\right)}^2\left(5-1\right)+{\left(0.005\right)}^2\left(7-1\right)}{5+7-2}} \\ =6.371812929\times {10}^{-3} \end{gathered}$$

Now, calculate for$t_{calc}$ and compare it with$t_{table}$

$$\begin{gathered} t_{calc}=\frac{\left|{\overline{x}}_1-{\overline{x}}_2\right|}{s_{pooled}}\sqrt{\frac{n_1{n}_2}{n_1+n_2}} \\ =\frac{\left|0.063-0.0691\right|}{6.371812929\times {10}^{-3}}\sqrt{\frac{\left(5\right)\left(7\right)}{5+7}} \\ =1.608. \end{gathered}$$

$t_{table}=2.228$(based on Table 4-4).

The degree of freedom is calculated by adding $n_1+n_2-n_t,$thus we have:

$$\begin{gathered} ={\mathrm{n}}_1+{\mathrm{n}}_2-{\mathrm{n}}_{\mathrm{t}} \\ =5+7-2 \\ =10. \end{gathered}$$

(95% Hconfidence level)

Since $t_{calc}<t_{table},$there is no discernible difference between the two approaches' means.

For drinking water, use the same process that was used in rainwater.

The F-test is done as shown below:

Let$s_1$ and$s_2$be equal to the standard deviations of spectrophotometry and gas chromatography respectively, since$s_1>s_2$.

$$\begin{gathered} F_{calc}=\frac{s_1^2}{s_2^2} \\ =\frac{{\left(0.008\right)}^2}{{\left(0.007\right)}^2} \\ =1.31. \end{gathered}$$

$F_{table}=6.39$(based on Table 4-3; degrees of freedom for$s_1=4$ and $s_2=4$).

Since $F_{calc}<F_{table},$the variances and standard deviations of the two techniques are not significantly different.

The t -test is done as shown below.

We concluded in our-test result that$F_{calc}<F_{table}.$ Now, perform Case 2a of a t-test. We will first calculate the$s_{pooled}$ as a prerequisite for solving the t-test.

$$\begin{gathered} {\mathrm{s}}_{\mathrm{pooled}}=\sqrt{\frac{{\mathrm{s}}_1^2\left({\mathrm{n}}_1-1\right)+{\mathrm{s}}_2^2\left({\mathrm{n}}_2-1\right)}{{\mathrm{n}}_1+{\mathrm{n}}_2-{\mathrm{n}}_{\mathrm{t}}}} \\ =\sqrt{\frac{{\left(0.008\right)}^2\left(5-1\right)+{\left(0.007\right)}^2\left(5-1\right)}{5+5-2}} \\ =5.88217648\times {10}^{-3} \end{gathered}$$

Now, calculate for$t_{calc}$and compare it with$t_{table}$.

$$\begin{gathered} t_{calc}=\frac{\left|{\overline{x}}_1-{\overline{x}}_2\right|}{s_{pooled}}\sqrt{\frac{n_1{n}_2}{n_1+n_2}} \\ =\frac{\left|0.087-0.0781\right|}{5.882176468\times {10}^{-3}}\sqrt{\frac{\left(5\right)\left(5\right)}{5+5}} \\ =1.893. \end{gathered}$$

${\mathrm{t}}_{\mathrm{table}}=2.306$(based on Table 4-4).

The degrees of freedom are calculated by adding$n_1+n_2-n_t,$ thus we have:

$$\begin{gathered} ={\mathrm{n}}_1+{\mathrm{n}}_2-{\mathrm{n}}_{\mathrm{t}} \\ =5+5-2 \\ =8. \end{gathered}$$

(95% confidence level)

Since $t_{calc}<t_{table}$, there is no discernible difference between the two approaches' means.

PerformF and tests to see if the drinking water has substantially more nitrite than the rainwater at 95% confidence level for each technique.

Ftest regarding Gas Chromatography

The F test is done as shown below.

Let$s_1$ and$s_2$be equal to the standard deviations of drinking water and rainwater respectively, since$s_1>s_2.$

$$\begin{gathered} F_{calc}=\frac{s_1^2}{s_2^2} \\ =\frac{{\left(0.007\right)}^2}{{\left(0.005\right)}^2} \\ =1.96 \end{gathered}$$

$F_{table}=4.53$(based on Table 4-3; degrees of freedom for $s_1=4$and $s_2=6$).

Since $F_{calc}<F_{table},$the variances and standard deviations of the two distinct water samples are not significantly different.

t test regarding Gas Chromatography

Thetest calculates the$s_{pooled}$as a prerequisite for solving the ttest.

$$\begin{gathered} s_{pooled}=\sqrt{\frac{s_1^2\left(n_1-1\right)+s_2^2\left(n_2-1\right)}{n_1+n_2-n_t}} \\ =\sqrt{\frac{{\left(0.007\right)}^2\left(5-1\right)+{\left(0.005\right)}^2\left(7-1\right)}{5+7-2}} \\ =5.882176468\times {10}^{-3}. \end{gathered}$$

Now, calculate for$t_{calc}$and compare it with$t_{table}$

$$\begin{gathered} t_{calc}=\frac{\left|{\overline{x}}_1-{\overline{x}}_2\right|}{s_{pooled}}\sqrt{\frac{n_1{n}_2}{n_1+n_2}} \\ =\frac{\left|0.078-0.0691\right|}{5.882176468\times {10}^{3-}}\sqrt{\frac{\left(5\right)\left(7\right)}{5+7}} \\ =2.613. \end{gathered}$$

$t_{table}=2.228$(based on Table 4-4).

The degrees of freedom are calculated by adding$n_1+n_2-n_t,$ thus we have:

$$\begin{gathered} =n_1+n_2-n_t \\ =5+7-2 \\ =10. \end{gathered}$$

(95% confidence level)

Since$t_{calc}>t_{table},$ the means of the two distinct water samples varies by a substantial amount.

F test regarding Spectrophotometry

Now, use the same process that we did in (b-1) Gas Chromatography.

TheF test is done as shown below:

Let$s_1$ and$s_2$be equal to the standard deviations of drinking water and rainwater, respectively. Now, randomly assign the standard deviations in this case since their standard deviations are equally the same.

$$\begin{gathered} F_{calc}=\frac{s_1^2}{s_2^2} \\ =\frac{{\left(0.008\right)}^2}{{\left(0.008\right)}^2} \\ =1.00 \end{gathered}$$

$F_{table}=6.39$(based on Table 4-3; degrees of freedom for$s_1=4$ and$s_2=4$).

Since$F_{calc}<F_{table}$,the variances and standard deviations of the two distinct water samples are not significantly different.

ttest regarding Spectrophotometry

Thetest is done as shown below:

We concluded in ourFtest result that$F_{calc}<F_{table}$, so, perform a Case 2a of thet test. First calculate the$s_{pooled}$ as a prerequisite for solving thet test.

$$\begin{gathered} s_{pooled}=\sqrt{\frac{s_1^2\left(n_1-1\right)+s_2^2\left(n-1\right)}{n_1+n_2-n_t}} \\ =\sqrt{\frac{{\left(0.008\right)}^2\left(5-1\right)+{\left(0.008\right)}^2\left(5-1\right)}{5+5-2}} \\ =8.00\times {10}^{-3} \end{gathered}$$

Now, calculate for$t_{calc}$ and compare it with$t_{table}$

$$\begin{gathered} t_{calc}=\frac{\left|{\overline{x}}_1-{\overline{x}}_2\right|}{s_{pooled}}\sqrt{\frac{n_1{n}_2}{n_1+n_2}} \\ =\frac{\left|0.087-0.063\right|}{8.00\times {10}^{-3}}\sqrt{\frac{\left(5\right)\left(5\right)}{5+5}} \\ =4.743. \end{gathered}$$

$t_{table}=2.306$(based on Table 4-4)

The degrees of freedom are calculated by adding$n_1+n_2-n_t,$ thus we have:

$$\begin{gathered} =n_1+n_2-n_t \\ =5+5-2 \\ =8. \end{gathered}$$

(95% confidence level)

Since $t_{calc}>t_{table}$, the means of two distinct water samples varies by a substantial amount.