Question

Two students determine the percentage of lead in a sample as a laboratory exercise. The true percentage is \(22.52 \%\). The students' results for three determinations are as follows: (1) \(22.52,22.48,22.54\) (2) \(22.64,22.58,22.62\) (a) Calculate the average percentage for each set of data and state which set is the more accurate based on the average. (b) Precision can be judged by examining the average of the deviations from the average value for that data set. (Calculate the average value for each data set; then calculate the average value of the absolute deviations of each measurement from the average.) Which set is more precise?

Short answer

In conclusion, for the two sets of data, Set 1 has an average percentage of 22.5133 and Set 2 has an average percentage of 22.6133. Set 1 is more accurate as it has a lower absolute difference between the average percentage and the true value (22.52%). However, both sets have equal precision, as the average deviation for both sets is 0.0222.

Step-by-step solution

Calculate the average percentage for each set of data

To calculate the average percentage for each set of data, we sum the values and divide by the number of values in the set. For Set 1: \(Average = \frac{22.52+22.48+22.54}{3} = \frac{67.54}{3} = 22.5133\) For Set 2: \(Average = \frac{22.64+22.58+22.62}{3} = \frac{67.84}{3} = 22.6133\)

Compare the average percentages to the true percentage

Now, we compare the average percentages of each set to the true percentage (22.52%) to determine which set is more accurate. For Set 1: \(Accuracy = |22.52 - 22.5133| = 0.0067\) For Set 2: \(Accuracy = |22.52 - 22.6133| = 0.0933\) Set 1 has a lower absolute difference between the average percentage and the true value, so Set 1 is more accurate.

Calculate the average deviation for each set of data

To calculate the average deviation for each set of data, we find the absolute difference between each measurement and the average, sum these differences, and then divide by the number of values in the set. For Set 1: \(Average\ Deviation = \frac{|(22.52-22.5133)|+|(22.48-22.5133)|+|(22.54-22.5133)|}{3} = \frac{0.0067+0.0333+0.0267}{3} = \frac{0.0667}{3} = 0.0222\) For Set 2: \(Average\ Deviation = \frac{|(22.64-22.6133)|+|(22.58-22.6133)|+|(22.62-22.6133)|}{3} = \frac{0.0267+0.0333+0.0067}{3} = \frac{0.0667}{3} = 0.0222\)

Compare the average deviations to determine which set is more precise

Since both Set 1 and Set 2 have the same average deviation of 0.0222, both sets are equally precise. In conclusion, Set 1 is more accurate, while both sets are equally precise.

Key concepts

Accuracy vs Precision

Understanding accuracy and precision is essential in laboratory data analysis. Accuracy refers to how close a measured value is to the true or accepted value. Precision, on the other hand, indicates the reproducibility of measurements – that is, how close the measurements are to one another, regardless of whether they're close to the true value or not.

Considering the exercise where two students determine the percentage of lead in a sample, we compare their average results to the true value to assess accuracy. Student Set 1 has a very small difference from the true value, signifying high accuracy. Meanwhile, Student Set 2 shows a larger discrepancy, indicating lower accuracy. Yet, when we look at the spread of the measurements within each set, known as the range, we can assess precision.

Interestingly, although Student Set 2 was less accurate, it doesn't necessarily mean it was less precise. Precision is about consistency, and if the measurements are consistently off from the true value in the same way, they can still be highly precise. In our example, the precision is the same for both sets, as evidenced by the identical average deviations, showing that both sets are equally consistent in their measurements.

Percentage Calculation

Percentage calculation is a fundamental skill in data analysis, particularly in chemistry labs where concentration and purity are frequently calculated in percentages. To master this concept, you need to divide the part by the whole and then multiply by 100. In the context of our exercise, students are calculating the percentage of lead in a sample.

For instance, to calculate the average percentage for Student Set 1, you sum up their measurements \(22.52, 22.48, 22.54\) and divide by the number of values, three in this case. This results in an average of \(22.5133\text{%}\). Similarly, the average percentage for Student Set 2 is calculated and found to be \(22.6133\text{%}\).

When working with percentages, it's vital to ensure that all your measurements are using the same units to maintain consistency and accuracy. Moreover, when presenting data or answers, remember to include the percentage sign to clearly indicate that the value is a percentage.

Average Deviation

The concept of average deviation is used to measure precision within a set of data points. It involves calculating the average of the absolute differences between each data point and the mean of the data set. An important step in understanding average deviation is to never simply sum the differences without taking the absolute value, as the negative and positive differences could cancel each other out, giving a misleading impression of precision.

In our exercise, the average deviation is calculated by taking the absolute difference between each student's measurements and the average value of their set, then averaging these differences. For both sets in the example, the average deviation turns out to be \(0.0222\). This indicates that both sets of measurements are equally spread around their mean – in other words, they are equally precise.

Knowing how to calculate and interpret average deviation is imperative for students and researchers as it is indicative of the reliability of their measurements. The lower the average deviation, the more consistent the results, which is an essential aspect of conducting accurate scientific research.