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 =
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