Interpretation:
The table for sets of the replicate measurements is given below-
A | B | C | D | E | F |
3.5 | 70.24 | 0.812 | 2.7 | 70.65 | 0.514 |
3.1 | 70.22 | 0.792 | 3.0 | 70.63 | 0.503 |
3.1 | 70.10 | 0.794 | 2.6 | 70.64 | 0.486 |
3.3 | 0.900 | 2.8 | 70.21 | 0.497 | |
2.5 | 3.2 | 0.472 |
The value of mean and standard deviation is to be calculated for the above data sets. The 95% confidence interval for each data is to be determined and the value of interval mean is also to be determined.
Concept introduction:
The formula for the mean is given as-
The formula for the standard deviation is given as-
Answer to Problem A1.12QAP
For set A −
Mean= 3.1
Standard deviation = 0.37
95% degree interval =
For set B −
Mean= 70.19
Standard deviation = 0.08
95% degree interval =
For set C −
Mean= 0.82
Standard deviation = 0.05
95% degree interval =
For set D −
Mean= 2.86
Standard deviation = 0.25
95% degree interval =
For set E −
Mean= 70.53
Standard deviation = 0.22
95% degree interval =
For set F −
Mean= 0.494
Standard deviation = 0.016
95% degree interval =
Explanation of Solution
The formula that will be used-
Therefore, mean value of set A −
The standard deviation for the set A is calculated as-
Set A | ||
1 | 3.5 | 12.25 |
2 | 3.1 | 9.61 |
3 | 3.1 | 9.61 |
4 | 3.3 | 10.89 |
5 | 2.5 | 6.25 |
From above table-
The standard deviation for the set A is calculated as-
Similarly for set B-
Therefore mean value of set B −
The standard deviation for the set A is calculated as-
Set A | ||
1 | 70.24 | 4933.6576 |
2 | 70.22 | 4930.8484 |
3 | 70.10 | 4914.01 |
From above table-
The standard deviation for the set A is calculated as-
Similarly for set C-
Therefore mean value of set C −
The standard deviation for the set C is calculated as-
Set C | ||
1 | 0.812 | 0.659344 |
2 | 0.792 | 0.627264 |
3 | 0.794 | 0.630436 |
4 | 0.900 | 0.81 |
From above table-
The standard deviation for the set C is calculated as-
Similarly for set D-
Therefore mean value of set D −
The standard deviation for the set D is calculated as-
Set D | ||
1 | 2.7 | 7.29 |
2 | 3.0 | 9.00 |
3 | 2.6 | 6.79 |
4 | 2.8 | 7.64 |
5 | 3.2 | 10.24 |
From above table-
The standard deviation for the set A is calculated as-
Similarly for set E-
Therefore mean value of set E −
The standard deviation for the set E is calculated as-
Set E | ||
1 | 70.65 | 4991.4225 |
2 | 70.63 | 4988.5969 |
3 | 70.64 | 4990.0096 |
4 | 70.21 | 4929.4441 |
From above table-
The standard deviation for the set A is calculated as-
Similarly for set F-
Therefore mean value of set F −
The standard deviation for the set F is calculated as-
Set A | ||
1 | 0.514 | 0.264196 |
2 | 0.503 | 0.253009 |
3 | 0.486 | 0.236196 |
4 | 0.497 | 0.247009 |
5 | 0.472 | 0.222784 |
From above table-
The standard deviation for the set F is calculated as-
The 95% confidence interval for each data is calculated as-
Where,
N = degree of freedom
s = standard deviation
The value of t-
Degree of freedom | 95% |
1 | 12.7 |
2 | 4.30 |
3 | 3.18 |
4 | 2.78 |
5 | 2.57 |
For set A-
Given that-
N = 5
s = 0.37
t = 2.78
Put the above values in Equ (3)
For set B-
Given that-
N = 3
s = 0.08
t = 4.30
Put the above values in Equ (3)
For set C-
Given that-
N = 4
s = 0.05
t = 3.18
Put the above values in Equ (3)
For set D-
Given that-
N = 5
s = 0.25
t = 2.78
Put the above values in Equ (3)
For set E-
Given that-
N = 4
s = 0.22
t = 3.18
Put the above values in Equ (3)
For set F-
Given that-
N = 5
s = 0.016
t = 2.78
Put the above values in Equ (3)
Thus, this can be concluded that the results for each set are obtained by using the mean standard deviation and 95% confidence interval formula. Therefore,
For set A −
Mean= 3.1
Standard deviation = 0.37
95% degree interval =
For set B −
Mean= 70.19
Standard deviation = 0.08
95% degree interval =
For set C −
Mean= 0.82
Standard deviation = 0.05
95% degree interval =
For set D −
Mean= 2.86
Standard deviation = 0.25
95% degree interval =
For set E −
Mean= 70.53
Standard deviation = 0.22
95% degree interval =
For set F −
Mean= 0.494
Standard deviation = 0.016
95% degree interval =
Want to see more full solutions like this?
Chapter A1 Solutions
Principles of Instrumental Analysis
- Hexane (C6H14, density = 0.766 g/cm3), perfluoro-hexane (C6F14, density = 1.669 g/cm3), and water are immiscible liquids; that is, they do not dissolve in one another. You place 10 mL of each in a graduated cylinder, along with pieces of high-density polyethylene (HDPE, density = 0.97 g/cm3), polyvinyl chloride(PVC, density = 1.36 g/cm3), and Teflon (density = 2.3 g/cm3). None of these common plastics dissolves in these liquids. Describe what you expect to see.arrow_forwardThe gold content of an ore is being measured. The results of these analyses are: Content (ppt) 20.065 19.999 21.054 20.533 20.088 20.112 What is the average value, standard deviation and confidence interval for this set of data?Show each step of the calculation. Using your results calculate the %accuracy and %precision (relative standard deviation) if the true value is 20.045%.arrow_forwardYou measure 35 turtles' weights, and find they have a mean weight of 72 ounces. Assume the population standard deviation is 3.1 ounces. Based on this, construct a 99% confidence interval for the true population mean turtle weight. Give your answers as decimals, to two places ouncesarrow_forward
- Principles of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
- Principles of Modern ChemistryChemistryISBN:9781305079113Author:David W. Oxtoby, H. Pat Gillis, Laurie J. ButlerPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning