Chapter A1, Problem A1.1QAP

(a)

Interpretation Introduction

Interpretation:

The table for the replicate measurements is given below-

A	B	C	D
61.25	3.27	12.06	1.9
61.33	3.26	12.14	1.5
61.12	3.24		1.6
	3.24		1.4
	3.28
	3.23

The value of mean and number of degree of freedoms associated with the calculation of the mean needs to be determined.

Concept Introduction :

The mean can be calculated using the following formula:

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

Or

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

Here, degree of freedom = N

Answer to Problem A1.1QAP

For set A −

Mean = 61.43

Degree of freedom = 3

For set B −

Mean = 3.25

Degree of freedom = 6

For set C −

Mean = 12.1

Degree of freedom = 2

For set D −

Mean = 2.65

Degree of freedom = 4

Explanation of Solution

The following formula will be used for the calculation of the mean-

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

Or

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

Now the mean value for set A-

$\overline{x}=\frac{61.25+61.33+61.12}{3}=61.43$

Degree of freedom for set A = 3

Mean value for set B-

$\overline{x}=\frac{3.27+3.26+3.24+3.24+3.28+3.23}{6}=3.25$

Degree of freedom for set B = 6

Mean value for set C-

$\overline{x}=\frac{12.06+12.14}{2}=12.1$

Degree of freedom for set C = 2

Mean value for set D −

$\overline{x}=\frac{1.9+1.5+1.6=1.4}{4}=2.65$

Degree for freedom for set D = 4

(b)

Interpretation Introduction

Interpretation:

The table for the replicate measurements is given below-

A	B	C	D
61.25	3.27	12.06	1.9
61.33	3.26	12.14	1.5
61.12	3.24		1.6
	3.24		1.4
	3.28
	3.23

The value of standard deviation and number of degree of freedoms associated with the calculation of the standard deviation needs to be determined.

Concept Introduction :

The value of standard deviation can be calculated as follows:

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

Here,

Degree of freedom = $N-1$

Answer to Problem A1.1QAP

For set A −

Standard deviation = 0.11

Degree of freedom = 2

For set B −

Standard deviation = 0.02

Degree of freedom = 5

For set C −

Standard deviation = 0.06

Degree of freedom = 1

For set D −

Standard deviation = 0.21

Degree of freedom = 3

Explanation of Solution

The following formula will be used for the calculation of standard deviation −

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

For set A-

Samples	$x_i$	$x_i^2$
1	61.25	3776.1025
2	61.33	3785.9409
3	61.12	3760.1424
	$\sum x_i=184.3$	$\sum x_i^2=11322.1858$

$\overline{x}=\frac{\sum x_i}{N}=61.43$$\frac{{\left(\sum x_i\right)}^2}{N}=\frac{{\left(184.3\right)}^2}{3}=11322.1633$

Put the values,

$s=\sqrt{\frac{11322.1858-11322.1633}{3-1}}$

$s=\sqrt{\frac{0.0225}{2}}=0.11$

Degree of freedom for standard deviation = $N-1=2$

For set B-

Samples	$x_i$	$x_i^2$
1	3.27	10.6929
2	3.26	10.6276
3	3.24	10.4976
4	3.24	10.4976
5	3.28	10.7584
6	3.23	10.4329
	$\sum x_i=19.52$	$\sum x_i^2=63.507$

$\overline{x}=\frac{\sum x_i}{N}=3.25$$\frac{{\left(\sum x_i\right)}^2}{N}=\frac{{\left(19.52\right)}^2}{6}=63.505$

Put the values,

$s=\sqrt{\frac{63.507-63.505}{6-1}}$

$s=\sqrt{\frac{0.002}{5}}=0.02$

Degree of freedom for standard deviation = $N-1=5$

For set C-

Samples	$x_i$	$x_i^2$
1	12.06	145.4436
2	12.14	147.3796
	$\sum x_i=24.2$	$\sum x_i^2=292.8232$

$\overline{x}=\frac{\sum x_i}{N}=12.1$$\frac{{\left(\sum x_i\right)}^2}{N}=\frac{{\left(24.2\right)}^2}{2}=292.82$

Put the values,

$s=\sqrt{\frac{292.8232-292.82}{2-1}}$

$s=\sqrt{\frac{0.0032}{1}}=0.06$

Degree of freedom for standard deviation = $N-1=1$

For set D-

Samples	$x_i$	$x_i^2$
1	1.9	7.29
2	1.5	5.76
3	1.6	6.76
4	1.4	8.41
	$\sum x_i=10.6$	$\sum x_i^2=28.22$

$\overline{x}=\frac{\sum x_i}{N}=2.65$$\frac{{\left(\sum x_i\right)}^2}{N}=\frac{{\left(10.6\right)}^2}{4}=28.09$

Put the values,

$s=\sqrt{\frac{28.22-28.09}{4-1}}$

$s=\sqrt{\frac{0.13}{3}}=0.207\approx 0.21$

Degree of freedom for standard deviation = $N-1=3$

(c)

Interpretation Introduction

Interpretation:

The table for the replicate measurements is given below-

A	B	C	D
61.25	3.27	12.06	1.9
61.33	3.26	12.14	1.5
61.12	3.24		1.6
	3.24		1.4
	3.28
	3.23

The coefficient of variation for each set is to be determined.

Concept Introduction :

The following formula will be used for the calculation of the coefficient of variation-

$CV=\frac{s}{\overline{x}}\times 100\%$

Here,

s = standard deviation

$\overline{x}$ = mean

Answer to Problem A1.1QAP

The coefficient of variation for set A = 0.17%

The coefficient of variation for set B = 0.61%

The coefficient of variation for set C = 0.49%

The coefficient of variation for set D = 7.9%

Explanation of Solution

For set A −

Given that-

s = 0.11

$\overline{x}$ = 61.43

Put the above value,

$CV=\frac{0.11}{61.43}\times 100\%$

$CV=0.17\%$

For set B −

Given that-

s = 0.02

$\overline{x}$ = 3.25

Put the above value,

$CV=\frac{0.02}{3.25}\times 100\%$

$CV=0.61\%$

For set C −

Given that-

s = 0.06

$\overline{x}$ = 12.1

Put the above value,

$CV=\frac{0.06}{12.1}\times 100\%$

$CV=0.49\%$

For set D −

Given that-

s = 0.21

$\overline{x}$ = 2.65

Put the above value,

$CV=\frac{0.21}{2.65}\times 100\%$

$CV=7.9\%$

(d)

Interpretation Introduction

Interpretation:

The table for the replicate measurements is given below-

A	B	C	D
61.25	3.27	12.06	1.9
61.33	3.26	12.14	1.5
61.12	3.24		1.6
	3.24		1.4
	3.28
	3.23

The standard error of mean for each set is to be determined.

Concept Introduction:

The following formula will be used for the calculation of the standard error for the mean-

$s_m=\frac{s}{\sqrt{N}}$

Here,

Standard deviation = s

Degree of freedom = N

Answer to Problem A1.1QAP

The standard error for mean for set A = 0.063

The standard error for mean for set B = 0.008

The standard error for mean for set C = 0.042

The standard error for mean for set D = 0.10

Explanation of Solution

For set A-

Given that-

Standard deviation = 0.11

Degree of freedom = 3

Put the above values,

$s_m=\frac{0.11}{\sqrt{3}}=0.063$

For set B-

Given that-

Standard deviation = 0.02

Degree of freedom = 6

Put the above values,

$s_m=\frac{0.02}{\sqrt{6}}=0.008$

For set C-

Given that-

Standard deviation = 0.06

Degree of freedom = 2

Put the above values,

$s_m=\frac{0.06}{\sqrt{2}}=0.042$

For set D-

Given that-

Standard deviation = 0.21

Degree of freedom = 4

Put the above values,

$s_m=\frac{0.21}{\sqrt{4}}=0.10$