Principles of Instrumental Analysis, 6th Edition
Principles of Instrumental Analysis, 6th Edition
6th Edition
ISBN: 9788131525579
Author: Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher: Cenage Learning
Question
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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:

x¯=x1+x2+.........+xnN

x¯=i=1NxiN

s=i=1Nxi2(i=1Nxi)2NN1

Expert Solution
Check Mark

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-

x¯=x1+x2+.........+xnN

x¯=i=1NxiN

For the set 1 -

So, the mean is −

x¯=0.99+0.84+1.023=0.95

Sample xi xi2
1 0.99 0.9801
2 0.84 0.7056
3 1.02 1.0404
xi=2.85 xi2=2.7261

(xi)2N=(2.85)23=2.7075

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

s=i=1Nxi2(i=1Nxi)2NN1

s=2.72612.707531

s=0.01862=0.096

For the set 2 -

Sample xi xi2
1 1.02 1.0404
2 1.13 1.2769
3 1.17 1.3689
4 1.02 1.0404
xi=4.34 xi2=4.7266

So the mean is −

x¯=1.02+1.13+1.17+1.024=4.34

(xi)2N=(4.34)24=4.7089

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

s=i=1Nxi2(i=1Nxi)2NN1

s=4.72664.708941

s=0.01773=0.077

For the set 3 -

Sample xi xi2
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
xi=6.02 xi2=7.2762

So the mean is −

x¯=1.25+1.32+1.13+1.20+1.125=1.204

(xi)2N=(6.02)25=7.2480

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

s=i=1Nxi2(i=1Nxi)2NN1

s=7.27627.248051

s=0.02824=0.084

For the set 4 -

Sample xi xi2
1 0.72 0.5184
2 0.77 0.5929
3 0.61 0.3721
4 0.58 0.3364
xi=2.68 xi2=1.8198

So the mean is −

x¯=0.72+0.77+0.61+0.584=0.67

(xi)2N=(2.68)24=1.7956

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

s=i=1Nxi2(i=1Nxi)2NN1

s=1.81981.795641

s=0.02424=0.090

For the set 5 -

Sample xi xi2
1 0.90 0.81
2 0.92 0.8464
3 0.73 0.5329
xi=2.55 xi2=2.1893

So the mean is −

x¯=0.90+0.92+0.733=0.85

(xi)2N=(2.55)23=2.1675

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

s=i=1Nxi2(i=1Nxi)2NN1

s=2.18932.167531

s=0.02182=0.104

For the set 6 -

Sample xi xi2
1 0.70 0.49
2 0.88 0.7744
3 0.72 0.5184
4 0.73 0.5329
xi=3.03 xi2=2.3157

So the mean is −

x¯=0.70+0.88+0.72+0.734=0.7575

(xi)2N=(3.03)24=2.2952

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

s=i=1Nxi2(i=1Nxi)2NN1

s=2.31572.295241

s=0.02053=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:

spooled=i=1N1(xix¯1)2+i=1N1(xix¯1)2+...N1+N2+....

Expert Solution
Check Mark

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-

spooled=i=1N1(xix¯1)2+i=1N1(xix¯1)2+...N1+N2+....

For bottle 1-

Mean = 0.95

=(0.990.95)2+(0.840.95)2+(1.020.95)2

Sum of squares = 0.0186

For bottle 2-

Mean = 1.085

=(1.021.085)2+(1.131.085)2+(1.171.085)2+(1.021.085)2

Sum of squares = 0.0177

For bottle 3-

Mean = 1.024

=(1.251.204)2+(1.322.04)2+(1.131.204)2+(1.201.204)2+(1.121.204)2

Sum of squares = 0.02812

For bottle 4-

Mean = 0.67

=(0.720.67)2+(0.770.67)2+(0.610.67)2+(0.580.67)2

Sum of squares = 0.0242

For bottle 5-

Mean = 0.85

=(0.900.85)2+(0.920.85)2+(0.730.85)2

Sum of squares = 0.02812

For bottle 6-

Mean = 0.7575

=(0.700.7575)2+(0.880.7575)2+(0.720.7575)2+(0.730.7575)2

Sum of squares = 0.020475

spooled=0.130895236

spooled=0.087=0.88%

Pooled value = 0.88%

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