91Ó°ÊÓ

A laboratory instructor gives a sample of amino-acid powder to each of four students, I, II, III, and IV, and they weigh the samples. The true value is \(8.72 \mathrm{~g}\). Their results for three trials are I: \(8.72 \mathrm{~g}, 8.74 \mathrm{~g}, 8.70 \mathrm{~g}\) II: \(8.56 \mathrm{~g}, 8.77 \mathrm{~g}, 8.83 \mathrm{~g}\) III: \(8.50 \mathrm{~g}, 8.48 \mathrm{~g}, 8.51 \mathrm{~g} \quad \mathrm{IV}: 8.41 \mathrm{~g}, 8.72 \mathrm{~g}, 8.55 \mathrm{~g}\) (a) Calculate the average mass from each set of data and tell which set is the most accurate. (b) Precision is a measure of the average of the deviations of each piece of data from the average value. Which set of data is the most precise? Is this set also the most accurate? (c) Which set of data is both the most accurate and the most precise? (d) Which set of data is both the least accurate and the least precise?

Step by step solution

01

- Calculate Average Mass for Each Student

First, calculate the average mass for each set of data. The formula for the average is: \[ \text{Average} = \frac{\text{Sum of all measurements}}{\text{Number of measurements}} \]- For Student I: Average = \( \frac{8.72 + 8.74 + 8.70}{3} = 8.72 \, \text{g} \)- For Student II: Average = \( \frac{8.56 + 8.77 + 8.83}{3} = 8.72 \, \text{g} \)- For Student III: Average = \( \frac{8.50 + 8.48 + 8.51}{3} = 8.50 \, \text{g} \)- For Student IV: Average = \( \frac{8.41 + 8.72 + 8.55}{3} = 8.56 \, \text{g} \)

02

- Determine Accuracy

Accuracy is determined by how close the average mass is to the true value of 8.72 g.- Student I: Average = 8.72 g (Most accurate)- Student II: Average = 8.72 g- Student III: Average = 8.50 g- Student IV: Average = 8.56 g

03

- Calculate Precision

Calculate the precision for each student by finding the average of the deviations from the average value.- For Student I: Deviations: |8.72 - 8.72|, |8.74 - 8.72|, |8.70 - 8.72| \[ \text{Average deviation} = \frac{0 + 0.02 + 0.02}{3} = 0.013 \, \text{g} \]- For Student II: Deviations: |8.56 - 8.72|, |8.77 - 8.72|, |8.83 - 8.72| \[ \text{Average deviation} = \frac{0.16 + 0.05 + 0.11}{3} = 0.107 \, \text{g} \]- For Student III: Deviations: |8.50 - 8.50|, |8.48 - 8.50|, |8.51 - 8.50| \[ \text{Average deviation} = \frac{0 + 0.02 + 0.01}{3} = 0.01 \, \text{g} \] - For Student IV: Deviations: |8.41 - 8.56|, |8.72 - 8.56|, |8.55 - 8.56| \[ \text{Average deviation} = \frac{0.15 + 0.16 + 0.01}{3} = 0.107 \, \text{g} \]

04

- Determine Most Precise and Accurate Data Set

Compare the precisions:- Student I: Precision = 0.013 g- Student II: Precision = 0.107 g- Student III: Precision = 0.01 g (Most precise)- Student IV: Precision = 0.107 gStudent III is the most precise but not the most accurate. The most accurate data set is from Student I and II. Thus, none of the sets is both the most accurate and the most precise.

05

- Determine Least Accurate and Precise Data Set

Compare accuracies and precisions:- Least accurate: Student III (8.50 g)- Least precise: Student II and IV (precision = 0.107 g)Student III has the lowest accuracy and a relatively higher precision, but considering both criteria, Student II and IV both are less precise and have lower average accuracies compared to Student I.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

average mass calculation

Calculating the average mass is a fundamental step in determining the central value of a set of measurements. The average mass is calculated by summing up all the individual measurements and then dividing by the number of measurements. This process helps to find a value that represents the entire data set.
The formula to calculate the average mass is: \[ \text{Average} = \frac{\text{Sum of all measurements}}{\text{Number of measurements}} \] In our exercise:

• For Student I, the average is calculated as: \( \frac{8.72 + 8.74 + 8.70}{3} = 8.72 \) g
• For Student II, the average is: \( \frac{8.56 + 8.77 + 8.83}{3} = 8.72 \) g
• For Student III, the average is: \( \frac{8.50 + 8.48 + 8.51}{3} = 8.50 \) g
• For Student IV, the average is: \( \frac{8.41 + 8.72 + 8.55}{3} = 8.56 \) g

These calculations provide a clear picture of the central tendency of each student's measurements.

measurement deviations

Understanding measurement deviations is crucial for evaluating precision. A deviation is the difference between each measurement and the average value of those measurements. The benefit of examining deviations is that it allows us to see how spread out the individual measurements are from the average.
In our example, the deviations are calculated as follows:

• For Student I: \( |8.72 - 8.72|, |8.74 - 8.72|, |8.70 - 8.72| \), leading to an average deviation of \( \frac{0 + 0.02 + 0.02}{3} = 0.013 \) g
• For Student II: \( |8.56 - 8.72|, |8.77 - 8.72|, |8.83 - 8.72| \), leading to an average deviation of \( \frac{0.16 + 0.05 + 0.11}{3} = 0.107 \) g
• For Student III: \( |8.50 - 8.50|, |8.48 - 8.50|, |8.51 - 8.50| \), leading to an average deviation of \( \frac{0 + 0.02 + 0.01}{3} = 0.01 \) g
• For Student IV: \( |8.41 - 8.56|, |8.72 - 8.56|, |8.55 - 8.56| \), leading to an average deviation of \( \frac{0.15 + 0.16 + 0.01}{3} = 0.107 \) g

These deviations illustrate how consistent or inconsistent the measurements are relative to their average.

precision analysis

Precision analysis involves examining the consistency and repeatability of measurements. It is quantified by calculating the average deviation from the average value. The smaller the average deviation, the more precise the measurements are.
In this exercise, the precision for each student was calculated as follows:

• Student I: Precision = 0.013 g
• Student II: Precision = 0.107 g
• Student III: Precision = 0.01 g (most precise)
• Student IV: Precision = 0.107 g

Student III's data set is the most precise as it has the smallest average deviation. However, it is not the most accurate as the average mass is farther from the true value of 8.72 g. Conversely, Student I has a highly accurate average mass (exactly 8.72 g) but is slightly less precise than Student III. This exercise demonstrates that precision and accuracy, though related, are distinct characteristics of measurements.