Science

ChemistryPrinciples of Instrumental AnalysisThe 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 = NStart your trial now! First week only $4.99!*arrow_forward*

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7th Edition

Douglas A. Skoog + 2 others

Publisher: Cengage Learning

ISBN: 9781305577213

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:

Or

Here, 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

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

Or

Now the mean value for set A-

Degree of freedom for set A = 3

Mean value for set B-

Degree of freedom for set B = 6

Mean value for set C-

Degree of freedom for set C = 2

Mean value for set D −

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:

Here,

Degree of freedom =

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

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

For set A-

Samples | ||

1 | 61.25 | 3776.1025 |

2 | 61.33 | 3785.9409 |

3 | 61.12 | 3760.1424 |

Put the values,

Degree of freedom for standard deviation =

For set B-

Samples | ||

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 |

Put the values,

Degree of freedom for standard deviation =

For set C-

Samples | ||

1 | 12.06 | 145.4436 |

2 | 12.14 | 147.3796 |

Put the values,

Degree of freedom for standard deviation =

For set D-

Samples | ||

1 | 1.9 | 7.29 |

2 | 1.5 | 5.76 |

3 | 1.6 | 6.76 |

4 | 1.4 | 8.41 |

Put the values,

Degree of freedom for standard deviation =

**(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-

Here,

s = standard deviation

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%

For set A −

Given that-

s = 0.11

Put the above value,

For set B −

Given that-

s = 0.02

Put the above value,

For set C −

Given that-

s = 0.06

Put the above value,

For set D −

Given that-

s = 0.21

Put the above value,

**(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-

Here,

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

For set A-

Given that-

Standard deviation = 0.11

Degree of freedom = 3

Put the above values,

For set B-

Given that-

Standard deviation = 0.02

Degree of freedom = 6

Put the above values,

For set C-

Given that-

Standard deviation = 0.06

Degree of freedom = 2

Put the above values,

For set D-

Given that-

Standard deviation = 0.21

Degree of freedom = 4

Put the above values,

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