# 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: x ¯ = x 1 + x 2 + ......... + x n N Or x ¯ = ∑ i = 1 N x i N Here, degree of freedom = N

### Principles of Instrumental Analysis

7th Edition
Douglas A. Skoog + 2 others
Publisher: Cengage Learning
ISBN: 9781305577213

### Principles of Instrumental Analysis

7th Edition
Douglas A. Skoog + 2 others
Publisher: Cengage Learning
ISBN: 9781305577213

#### Solutions

Chapter A1, Problem A1.1QAP

(a)

Interpretation Introduction

## Interpretation: The table for the replicate measurements is given below-ABCD61.25 3.27 12.06 1.9 61.333.2612.141.561.123.241.63.241.43.283.23The 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:x¯=x1+x2+.........+xnNOrx¯=∑i=1NxiNHere, degree of freedom = N

Expert Solution

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-

x¯=x1+x2+.........+xnN

Or

x¯=i=1NxiN

Now the mean value for set A-

x¯=61.25+61.33+61.123=61.43

Degree of freedom for set A = 3

Mean value for set B-

x¯=3.27+3.26+3.24+3.24+3.28+3.236=3.25

Degree of freedom for set B = 6

Mean value for set C-

x¯=12.06+12.142=12.1

Degree of freedom for set C = 2

Mean value for set D −

x¯=1.9+1.5+1.6=1.44=2.65

(b)

Interpretation Introduction

### Interpretation: The table for the replicate measurements is given below-ABCD61.25 3.27 12.06 1.9 61.333.2612.141.561.123.241.63.241.43.283.23The 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=∑ i=1Nxi2− ( ∑ i=1 N x i ) 2 NN−1Here, Degree of freedom = N−1

Expert Solution

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= i=1Nxi2 ( i=1 N x i ) 2 NN1

For set A-

 Samples xi xi2 1 61.25 3776.1025 2 61.33 3785.9409 3 61.12 3760.1424 ∑xi=184.3 ∑xi2=11322.1858

x¯=xiN=61.43( x i )2N=( 184.3)23=11322.1633

Put the values,

s=11322.185811322.163331

s=0.02252=0.11

Degree of freedom for standard deviation = N1=2

For set B-

 Samples xi xi2 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 ∑xi=19.52 ∑xi2=63.507

x¯=xiN=3.25( x i )2N=( 19.52)26=63.505

Put the values,

s=63.50763.50561

s=0.0025=0.02

Degree of freedom for standard deviation = N1=5

For set C-

 Samples xi xi2 1 12.06 145.4436 2 12.14 147.3796 ∑xi=24.2 ∑xi2=292.8232

x¯=xiN=12.1( x i )2N=( 24.2)22=292.82

Put the values,

s=292.8232292.8221

s=0.00321=0.06

Degree of freedom for standard deviation = N1=1

For set D-

 Samples xi xi2 1 1.9 7.29 2 1.5 5.76 3 1.6 6.76 4 1.4 8.41 ∑xi=10.6 ∑xi2=28.22

x¯=xiN=2.65( x i )2N=( 10.6)24=28.09

Put the values,

s=28.2228.0941

s=0.133=0.2070.21

Degree of freedom for standard deviation = N1=3

(c)

Interpretation Introduction

### Interpretation: The table for the replicate measurements is given below-ABCD61.25 3.27 12.06 1.9 61.333.2612.141.561.123.241.63.241.43.283.23The 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=sx¯×100%Here,s = standard deviationx¯ = mean

Expert Solution

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

x¯ = 61.43

Put the above value,

CV=0.1161.43×100%

CV=0.17%

For set B −

Given that-

s = 0.02

x¯ = 3.25

Put the above value,

CV=0.023.25×100%

CV=0.61%

For set C −

Given that-

s = 0.06

x¯ = 12.1

Put the above value,

CV=0.0612.1×100%

CV=0.49%

For set D −

Given that-

s = 0.21

x¯ = 2.65

Put the above value,

CV=0.212.65×100%

CV=7.9%

(d)

Interpretation Introduction

### Interpretation: The table for the replicate measurements is given below-ABCD61.25 3.27 12.06 1.9 61.333.2612.141.561.123.241.63.241.43.283.23The 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-sm=sNHere, Standard deviation = s Degree of freedom = N

Expert Solution

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,

sm=0.113=0.063

For set B-

Given that-

Standard deviation = 0.02

Degree of freedom = 6

Put the above values,

sm=0.026=0.008

For set C-

Given that-

Standard deviation = 0.06

Degree of freedom = 2

Put the above values,

sm=0.062=0.042

For set D-

Given that-

Standard deviation = 0.21

Degree of freedom = 4

Put the above values,

sm=0.214=0.10

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