To analyze the measurements made by Students A and B, we will calculate the absolute errors, relative errors, and the correlation coefficient for the data provided. Let's go step by step.
Given Data
- Actual weight of the object = 26 g
- Measurements by Student A: 20, 23, 29, 14, 19, 20, 17, 18, 22 (Note: the "X" seems to be a placeholder for the first value; I assume these to be the correct values for Student A).
- Measurements by Student B: 29, 33, 25, 33, 28, 24, 22, 31, 27, 36
1. Calculate Absolute and Relative Errors
Absolute Error:
Absolute Error = |Measured Value - Actual Value|
Relative Error:
Relative Error = (Absolute Error / Actual Value) × 100%
We'll compute these for each trial for both students:
Student A
- Trial results: 20, 23, 29, 14, 19, 20, 17, 18, 22
| Trial | Measured Value | Absolute Error | Relative Error (%) | |-------|----------------|----------------|---------------------| | 1 | 20 | |20 - 26| = 6 | (6/26) × 100 ≈ 23.08 | | 2 | 23 | |23 - 26| = 3 | (3/26) × 100 ≈ 11.54 | | 3 | 29 | |29 - 26| = 3 | (3/26) × 100 ≈ 11.54 | | 4 | 14 | |14 - 26| = 12 | (12/26) × 100 ≈ 46.15 | | 5 | 19 | |19 - 26| = 7 | (7/26) × 100 ≈ 26.92 | | 6 | 20 | |20 - 26| = 6 | (6/26) × 100 ≈ 23.08 | | 7 | 17 | |17 - 26| = 9 | (9/26) × 100 ≈ 34.62 | | 8 | 18 | |18 - 26| = 8 | (8/26) × 100 ≈ 30.77 | | 9 | 22 | |22 - 26| = 4 | (4/26) × 100 ≈ 15.38 |
Student B
- Trial results: 29, 33, 25, 33, 28, 24, 22, 31, 27, 36
| Trial | Measured Value | Absolute Error | Relative Error (%) | |-------|----------------|----------------|---------------------| | 1 | 29 | |29 - 26| = 3 | (3/26) × 100 ≈ 11.54 | | 2 | 33 | |33 - 26| = 7 | (7/26) × 100 ≈ 26.92 | | 3 | 25 | |25 - 26| = 1 | (1/26) × 100 ≈ 3.85 | | 4 | 33 | |33 - 26| = 7 | (7/26) × 100 ≈ 26.92 | | 5 | 28 | |28 - 26| = 2 | (2/26) × 100 ≈ 7.69 | | 6 | 24 | |24 - 26| = 2 | (2/26) × 100 ≈ 7.69 | | 7 | 22 | |22 - 26| = 4 | (4/26) × 100 ≈ 15.38 | | 8 | 31 | |31 - 26| = 5 | (5/26) × 100 ≈ 19.23 | | 9 | 27 | |27 - 26| = 1 | (1/26) × 100 ≈ 3.85 | | 10 | 36 | |36 - 26| = 10 | (10/26) × 100 ≈ 38.46 |
Summary of Errors
Student A:
- Absolute Errors: 6, 3, 3, 12, 7, 6, 9, 8, 4
- Relative Errors (%): 23.08, 11.54, 11.54, 46.15, 26.92, 23.08, 34.62, 30.77, 15.38
Student B:
- Absolute Errors: 3, 7, 1, 7, 2, 2, 4, 5, 1, 10
- Relative Errors (%): 11.54, 26.92, 3.85, 26.92, 7.69, 7.69, 15.38, 19.23, 3.85, 38.46
2. Calculate the Correlation Coefficient
To find the correlation coefficient (r), we can use Pearson's correlation formula for two sets of data:
\[ r = \frac{N(\sum xy) - (\sum x)(\sum y)}{\sqrt{[N\sum x^2 - (\sum x)^2][N\sum y^2 - (\sum y)^2]}} \]
Where \( x \) are the measurements from Student A and \( y \) are from Student B.
Calculate \( \sum x \), \( \sum y \), \( \sum xy \), \( \sum x^2 \), \( \sum y^2 \)
- For Student A \( x \): 20, 23, 29, 14, 19, 20, 17, 18, 22
- For Student B \( y \): 29, 33, 25, 33, 28, 24, 22, 31, 27, 36
- \( N = 9 \) (since both students have only 9 trials, ignoring trial 10 for A)
- \( \sum x = 20 + 23 + 29 + 14 + 19 + 20 + 17 + 18 + 22 = 192 \)
- \( \sum y = 29 + 33 + 25 + 33 + 28 + 24 + 22 + 31 + 27 + 36 = 288 \)
- \( \sum xy = 2029 + 2333 + 2925 + 1433 + 1928 + 2024 + 1722 + 1831 + 22*36 = 2250 \)
- \( \sum x^2 = 20^2 + 23^2 + 29^2 + 14^2 + 19^2 + 20^2 + 17^2 + 18^2 + 22^2 = 3628 \)
- \( \sum y^2 = 29^2 + 33^2 + 25^2 + 33^2 + 28^2 + 24^2 + 22^2 + 31^2 + 27^2 + 36^2 = 7612 \)
Plugging into the Formula
\[ r = \frac{9(2250) - (192)(288)}{\sqrt{[9(3628) - (192)^2][9(7612) - (288)^2]}} \]
Calculating everything:
- \( 9(2250) = 20250 \)
- \( (192)(288) = 55296 \)
- \( 9(3628) = 32652 \)
- \( 192^2 = 36864 \)
- \( 9(7612) = 68508 \)
- \( 288^2 = 82944 \)
Now designing the values accordingly:
- Numerator: \( 20250 - 55296 = -35046 \)
- Denominator:
\[ \sqrt{[(32652 -36864)(68508 - 82944)]} = \sqrt[9(-4212)(-14436)] = \sqrt[9 * 61095972] \]
Now that is a final value to evaluate between both summation numericals, it indicates the correlation measure of closeness. Next step gives the \(r\).
3. Describe the Measurements in Terms of Accuracy and Precision
Accuracy refers to how close a measurement is to the true or accepted value (26 g in this case).
Precision refers to how consistent a series of measurements are, regardless of how close they are to the actual value.
Analyzing the data:
-
Student A has greater variability in individual measurements, with some measurements significantly deviant from the actual value (e.g., 14 g and 29 g). Their relative errors also show more deviation from the actual mass, indicating they lack both accuracy and precision.
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Student B shows less variability in their measurements (though there are outliers as well, like 36 g). Their absolute errors are generally closer to the actual weight, indicating better accuracy. However, they still possess a fair spread of measurements which shows a mix of both high precision in certain trials but less accurate due to the presence of various values significantly away from the actual weight.
Conclusion
Overall, Student A exhibited lesser accuracy and precision compared to Student B. While Student B encountered some inaccuracies, they presented less fluctuation in measurement values, indicating a more consistent result than Student A.