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

BuyFind

Principles of Instrumental Analysis

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

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-

    ABCD
    61.25 3.27 12.06 1.9
    61.333.2612.141.5
    61.123.241.6
    3.241.4
    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¯=x1+x2+.........+xnN

Or

x¯=i=1NxiN

Here, degree of freedom = N

Expert Solution

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-

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

Degree for freedom for set D = 4

(b)

Interpretation Introduction

Interpretation:

The table for the replicate measurements is given below-

    ABCD
    61.25 3.27 12.06 1.9
    61.333.2612.141.5
    61.123.241.6
    3.241.4
    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= i=1Nxi2 ( i=1 N x i ) 2 NN1

Here,

Degree of freedom = N1

Expert Solution

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

For set A-

    Samples xixi2
    161.25 3776.1025
    261.333785.9409
    361.123760.1424
    xi=184.3xi2=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 xixi2
    13.27 10.6929
    23.2610.6276
    33.2410.4976
    43.2410.4976
    53.2810.7584
    63.2310.4329
    xi=19.52xi2=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 xixi2
    112.06 145.4436
    212.14147.3796
    xi=24.2xi2=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-

    Samplesxixi2
    11.9 7.29
    21.55.76
    31.66.76
    41.48.41
    xi=10.6xi2=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-

    ABCD
    61.25 3.27 12.06 1.9
    61.333.2612.141.5
    61.123.241.6
    3.241.4
    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=sx¯×100%

Here,

s = standard deviation

x¯ = mean

Expert Solution

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

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-

    ABCD
    61.25 3.27 12.06 1.9
    61.333.2612.141.5
    61.123.241.6
    3.241.4
    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-

sm=sN

Here,

Standard deviation = s

Degree of freedom = N

Expert Solution

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,

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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