Bicarbonate in replicate samples of horse blood was measured four times by each of two methods with the following results:

Method 1:31.40, 31.24, 31.18, 31.43 mM

Method 2:30.70, 29.49, 30.01, 30.15mM

(a) Find the mean, standard deviation, and standard uncertainty (= standard deviation of the mean) for each analysis.

(b) Are the standard deviations significantly different at the 95%confidence level?

Mean:$\overline{x}=\frac{{\displaystyle \sum _i}{x}_i}{n}$

The standard deviation (s) is given by the formula:

$s=\sqrt{\frac{{\displaystyle \sum _i{\left(x_i-\overline{x}\right)}^2}}{n-1}}$

Standard deviation of the mean of sets of n values is:

${\sigma }_n=\frac{\theta }{\sqrt{n}}$

(a)

Let $\overline{x_1}$, $s_1$, and $u_1$be the mean, standard deviation, and standard uncertainty of method 1.

Method 's mean, standard deviation, and standard uncertainty are$\overline{x_2},s_2\mathrm{and}{u}_2$respectively.

For Method 1 :

Mean:

localid="1663562989158" $$\begin{gathered} \overline{x_1}=\frac{{\displaystyle \sum _i}{x}_i}{n} \\ =\frac{31.40+31.24+31.18+31.43}{4} \\ =31.31 \\ =31.312\mathrm{mM}\left(\mathrm{rounded}\mathrm{off}\right) \end{gathered}$$

Therefore, the mean value for method 1 is$\overline{x_1}=31.31\mathrm{mM}$

Standard deviation:

$$\begin{gathered} s_1=\sqrt{\frac{{\displaystyle \sum _i}(x_i-x^2}{n-1}} \\ =\sqrt{\frac{{\left(31.40-31.312\right)}^2+.....{\left(31.43-31.312\right)}^2}{\left(4-1\right)}} \\ =0.121 \\ =0.12\mathrm{mM}\left(\mathrm{rounded}\mathrm{off}\right) \end{gathered}$$

Therefore, the standard deviation value for method 1 is $s_1=0.12\mathrm{mM}$.

Standard uncertainty:

role="math" localid="1663563250038" $$\begin{gathered} u_1=\frac{s_1}{\sqrt{n}} \\ =\frac{\left(0.121\mathrm{mM}\right)}{\sqrt{4}} \\ =0.061 \\ =0.06\left(\mathrm{rounded}\mathrm{off}\right) \end{gathered}$$

Therefore, the standard uncertainty value for method 1 is $u_1=0.06$

For Method2:

Mean:

$$\begin{gathered} \overline{x_2}=\frac{{\displaystyle \sum _i{x}_i}}{n} \\ =\frac{30.70+29.49+30.01+30.15}{4} \\ =30.088 \\ =30.08\mathrm{mM}\left(\mathrm{rounded}\mathrm{off}\right) \end{gathered}$$

Therefore, the mean value for method 2 is $\overline{x_2}=30.08\mathrm{mM}$.

Standard deviation:

$$\begin{gathered} s_2=\sqrt{\frac{{\displaystyle \sum _i{(x_i-x)}^2}}{n-1}} \\ =\sqrt{\frac{{\left(30.70-30.088\right)}^2+.....{\left(30.15-30.088\right)}^2}{\left(4-1\right)}} \\ =0.497 \\ =0.49\mathrm{mM}\left(\mathrm{rounded}\mathrm{off}\right) \end{gathered}$$

Therefore, the standard deviation value for method 2 is $s_2=0.49\mathrm{mM}$.

Standard uncertainty:

$$\begin{gathered} u_2=\frac{n}{\sqrt{n}} \\ =\frac{\left(0.497\mathrm{mM}\right)}{\sqrt{4}} \\ =0.249 \\ =0.24\left(\mathrm{rounded}\mathrm{off}\right) \end{gathered}$$

Therefore, the standard uncertainty value for method 2 is $u_2=0.24$.

(b)

It must be determined whether the standard deviation differs considerably at the 95% confidence level.

Comparison of Standard Deviation with F Test:

We determine if the standard deviations of two sets of measurements are "statistically different" when comparing mean values. The Ftest is used to do this.

The Ftest with quotient F is given as:

${\mathrm{F}}_{\mathrm{calculated}}=\frac{k_1^2}{x_2^2}$

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

The difference between${\mathrm{F}}_{\mathrm{calculated}}>{\mathrm{F}}_{\mathrm{table}}$ is substantial.

If${\mathrm{F}}_{\mathrm{calculated}}>{\mathrm{F}}_{\mathrm{table}}$, then there is a considerable difference..

Degrees of Freedom:

The degrees of freedom for aset of measurements will be$\left(n-1\right)$ .

The F-test is used to compare standard deviations, which is computed as follows:

$$\begin{gathered} F_{\mathrm{calculated}}=\frac{s_1^2}{s_2^2} \\ =\frac{0.{497}^2}{0.{121}^2} \\ =16.8 \\ =16\left(\mathrm{rounded}\mathrm{off}\right) \end{gathered}$$

The ${\mathrm{F}}_{\mathrm{table}}$is found as ${\mathrm{F}}_{\mathrm{table}}=9.28$for three degrees of freedom in both standard deviations.

$F_{computed}\left(16\right)>F_{table}=\left(9.28\right)$after comparing the two values.

As a result, the standard deviations varied dramatically.