Chapter A1, Problem A1.9QAP

Interpretation Introduction

(a)

Interpretation:

The table for the analysis of the six bottles of wine of same variety for residue sugar is given below-

Bottle	Percent (w/v) Residual sugar
1	0.99,0.84,1.02
2	1.02,1.13,1.17,1.02
3	1.25,1.32,1.13,1.20,1.12
4	0.72,0.77,0.61,0.58
5	0.90,0.92,0.73
6	0.70,0.88,0.72,0.73

The value of standard deviation for each set is to be determined.

Concept introduction:

The formula for the calculation of mean and standard deviation can be calculated as follows:

$\overline{x}=\frac{x_1+x_2+\mathrm{.........}+x_n}{N}$

$\overline{x}=\frac{\sum _{i=1}^N{x}_i}{N}$

$s=\sqrt{\frac{{\displaystyle \sum }_{i=1}^N{x}_i^2-\frac{{\left({\displaystyle \sum }_{i=1}^N{x}_i\right)}^2}{N}}{N-1}}$

Answer to Problem A1.9QAP

For bottle 1-

Standard deviation = 0.096

For bottle 2-

Standard deviation = 0.077

For bottle 3-

Standard deviation = 0.084

For bottle 4-

Standard deviation = 0.090

For bottle 5-

Standard deviation = 0.104

For bottle 6 −

Standard deviation = 0.083

Explanation of Solution

The formula for the calculation of the mean is given below-

$\overline{x}=\frac{x_1+x_2+\mathrm{.........}+x_n}{N}$

$\overline{x}=\frac{\sum _{i=1}^N{x}_i}{N}$

For the set 1 -

So, the mean is −

$\overline{x}=\frac{0.99+0.84+1.02}{3}=0.95$

Sample	$x_i$	$x_i^2$
1	0.99	0.9801
2	0.84	0.7056
3	1.02	1.0404
	$\sum x_i=2.85$	$\sum x_i^2=2.7261$

$\frac{{\left(\sum x_i\right)}^2}{N}=\frac{{\left(2.85\right)}^2}{3}=2.7075$

The formula for the calculation of standard deviation is given as-

$s=\sqrt{\frac{{\displaystyle \sum }_{i=1}^N{x}_i^2-\frac{{\left({\displaystyle \sum }_{i=1}^N{x}_i\right)}^2}{N}}{N-1}}$

$s=\sqrt{\frac{2.7261-2.7075}{3-1}}$

$s=\sqrt{\frac{0.0186}{2}}=0.096$

For the set 2 -

Sample	$x_i$	$x_i^2$
1	1.02	1.0404
2	1.13	1.2769
3	1.17	1.3689
4	1.02	1.0404
	$\sum x_i=4.34$	$\sum x_i^2=4.7266$

So the mean is −

$\overline{x}=\frac{1.02+1.13+1.17+1.02}{4}=4.34$

$\frac{{\left(\sum x_i\right)}^2}{N}=\frac{{\left(4.34\right)}^2}{4}=4.7089$

The formula for the calculation of standard deviation is given as-

$s=\sqrt{\frac{{\displaystyle \sum }_{i=1}^N{x}_i^2-\frac{{\left({\displaystyle \sum }_{i=1}^N{x}_i\right)}^2}{N}}{N-1}}$

$s=\sqrt{\frac{4.7266-4.7089}{4-1}}$

$s=\sqrt{\frac{0.0177}{3}}=0.077$

For the set 3 -

Sample	$x_i$	$x_i^2$
1	1.25	1.5625
2	1.32	1.7424
3	1.13	1.2769
4	1.20	1.44
5	1.12	1.2544
	$\sum x_i=6.02$	$\sum x_i^2=7.2762$

So the mean is −

$\overline{x}=\frac{1.25+1.32+1.13+1.20+1.12}{5}=1.204$

$\frac{{\left(\sum x_i\right)}^2}{N}=\frac{{\left(6.02\right)}^2}{5}=7.2480$

The formula for the calculation of standard deviation is given as-

$s=\sqrt{\frac{{\displaystyle \sum }_{i=1}^N{x}_i^2-\frac{{\left({\displaystyle \sum }_{i=1}^N{x}_i\right)}^2}{N}}{N-1}}$

$s=\sqrt{\frac{7.2762-7.2480}{5-1}}$

$s=\sqrt{\frac{0.0282}{4}}=0.084$

For the set 4 -

Sample	$x_i$	$x_i^2$
1	0.72	0.5184
2	0.77	0.5929
3	0.61	0.3721
4	0.58	0.3364
	$\sum x_i=2.68$	$\sum x_i^2=1.8198$

So the mean is −

$\overline{x}=\frac{0.72+0.77+0.61+0.58}{4}=0.67$

$\frac{{\left(\sum x_i\right)}^2}{N}=\frac{{\left(2.68\right)}^2}{4}=1.7956$

The formula for the calculation of standard deviation is given as-

$s=\sqrt{\frac{{\displaystyle \sum }_{i=1}^N{x}_i^2-\frac{{\left({\displaystyle \sum }_{i=1}^N{x}_i\right)}^2}{N}}{N-1}}$

$s=\sqrt{\frac{1.8198-1.7956}{4-1}}$

$s=\sqrt{\frac{0.0242}{4}}=0.090$

For the set 5 -

Sample	$x_i$	$x_i^2$
1	0.90	0.81
2	0.92	0.8464
3	0.73	0.5329
	$\sum x_i=2.55$	$\sum x_i^2=2.1893$

So the mean is −

$\overline{x}=\frac{0.90+0.92+0.73}{3}=0.85$

$\frac{{\left(\sum x_i\right)}^2}{N}=\frac{{\left(2.55\right)}^2}{3}=2.1675$

The formula for the calculation of standard deviation is given as-

$s=\sqrt{\frac{{\displaystyle \sum }_{i=1}^N{x}_i^2-\frac{{\left({\displaystyle \sum }_{i=1}^N{x}_i\right)}^2}{N}}{N-1}}$

$s=\sqrt{\frac{2.1893-2.1675}{3-1}}$

$s=\sqrt{\frac{0.0218}{2}}=0.104$

For the set 6 -

Sample	$x_i$	$x_i^2$
1	0.70	0.49
2	0.88	0.7744
3	0.72	0.5184
4	0.73	0.5329
	$\sum x_i=3.03$	$\sum x_i^2=2.3157$

So the mean is −

$\overline{x}=\frac{0.70+0.88+0.72+0.73}{4}=0.7575$

$\frac{{\left(\sum x_i\right)}^2}{N}=\frac{{\left(3.03\right)}^2}{4}=2.2952$

The formula for the calculation of standard deviation is given as-

$s=\sqrt{\frac{{\displaystyle \sum }_{i=1}^N{x}_i^2-\frac{{\left({\displaystyle \sum }_{i=1}^N{x}_i\right)}^2}{N}}{N-1}}$

$s=\sqrt{\frac{2.3157-2.2952}{4-1}}$

$s=\sqrt{\frac{0.0205}{3}}=0.083$

Interpretation Introduction

(b)

Interpretation:

The pooled data is to be determined for the absolute standard deviation for the given method.

Concept introduction:

The pooled value for the given data can be calculated using the following formula:

$s_{pooled}=\sqrt{\frac{{\displaystyle \sum }_{i=1}^{N_1}{\left(x_i-{\overline{x}}_1\right)}^2+{\displaystyle \sum }_{i=1}^{N_1}{\left(x_i-{\overline{x}}_1\right)}^2+\mathrm{...}}{N_1+N_2+\mathrm{....}}}$

Answer to Problem A1.9QAP

The pooled value is 0.88%.

Explanation of Solution

The following formula will be used for the calculation of pooled value-

$s_{pooled}=\sqrt{\frac{{\displaystyle \sum }_{i=1}^{N_1}{\left(x_i-{\overline{x}}_1\right)}^2+{\displaystyle \sum }_{i=1}^{N_1}{\left(x_i-{\overline{x}}_1\right)}^2+\mathrm{...}}{N_1+N_2+\mathrm{....}}}$

For bottle 1-

Mean = 0.95

$={\left(0.99-0.95\right)}^2+{\left(0.84-0.95\right)}^2+{\left(1.02-0.95\right)}^2$

Sum of squares = 0.0186

For bottle 2-

Mean = 1.085

$={\left(1.02-1.085\right)}^2+{\left(1.13-1.085\right)}^2+{\left(1.17-1.085\right)}^2+{\left(1.02-1.085\right)}^2$

Sum of squares = 0.0177

For bottle 3-

Mean = 1.024

$={\left(1.25-1.204\right)}^2+{\left(1.32-2.04\right)}^2+{\left(1.13-1.204\right)}^2+{\left(1.20-1.204\right)}^2+{\left(1.12-1.204\right)}^2$

Sum of squares = 0.02812

For bottle 4-

Mean = 0.67

$={\left(0.72-0.67\right)}^2+{\left(0.77-0.67\right)}^2+{\left(0.61-0.67\right)}^2+{\left(0.58-0.67\right)}^2$

Sum of squares = 0.0242

For bottle 5-

Mean = 0.85

$={\left(0.90-0.85\right)}^2+{\left(0.92-0.85\right)}^2+{\left(0.73-0.85\right)}^2$

Sum of squares = 0.02812

For bottle 6-

Mean = 0.7575

$={\left(0.70-0.7575\right)}^2+{\left(0.88-0.7575\right)}^2+{\left(0.72-0.7575\right)}^2+{\left(0.73-0.7575\right)}^2$

Sum of squares = 0.020475

$s_{pooled}=\sqrt{\frac{0.130895}{23-6}}$

$s_{pooled}=0.087=0.88\%$

Pooled value = 0.88%