*Note: Typical student and "expert" response rates are in brackets
after each answer. Correct answers are in bold.*

1) Rank the following measurements in order from the most precise to
the least precise based on the relative uncertainty implied by each value:
9.7 m, 13 m, 1.45 m, 2.1 m,
0.005 m

(A > B means A is more precise than B, and A = B indicates equal precision)

a) 0.005 > 1.45 > 9.7 = 2.1 > 13 [33%,2) A group of students are told to use a meter stick to find the length of a hallway. They make 6 independent measurements: 4.402 m, 4.217 m, 4.345 m, 4.925 m, 4.372 m, 4.289 m. How should they report their best estimate of the length of the hallway?33%]

b) 0.005 > 1.45 > 2.1 > 9.7 > 13 [4%,0%]

c) 1.45 > 9.7 = 13 = 2.1 > 0.005 [48%,25%]d) 1.45>9.7>2.1>13>0.005[0%,30%]Explanation: Many people associate higher

precisionwith a smallerabsolute uncertainty(i.e. more decimal places, as in option A above), but therelative uncertaintyis a more general indicator of precision, and is specifically mentioned in this question. The relative uncertainty of a measurement roughly corresponds to the number of significant figures reported, and measurements with the same number of significant figures differ in precision depending on the value of the leftmost digit. For example, the value 9.7 implies an absolute uncertainty of ± 0.1, or a relative uncertainty of ± 0.1/9.7 = 1%, which is less than the relative uncertainty implied by 2.1, which is ± 5%. (It could be argued that 9.7 and 2.1 each imply an uncertainty of ± 0.05, but the conclusion remains the same.) This connection between significant figures and relative uncertainty is a difficult concept for many students to understand, but it is important because it underlies the purpose of proper reporting of significant figures in measured and calculated values.

3) A student uses a protractor to measure an angle to be A = 82a) L = 4.33 ± 0.03 m[0%,20%]

b) L = 4.43 ± 0.25 m [7%,10%]

c) L = 4.325 ± 0.073 m [5%,0%]

d) L = 4.425 ± 0.104 m [25%,40%]Explanation: Even though Option D is the most popular response among both students and experts, the best estimate of the length of the hallway is L = 4.33 ± 0.03 m because this is the average and standard error of the individual measurements, excluding the extreme outlier: 4.925 m. Option A is also properly rounded so that the number of decimal places reported for the measurement is consistent with the uncertainty, unlike the other options which have extra digits that are unnecessary. Options B and D include and are skewed by the outlier, which is most likely a mistake since it is over 8 standard deviations away from the other 5 measurements. Option C is reasonable (average of 5 good measurements ± standard deviation), but in this situation, the standard error is the best estimate of the uncertainty associated with the average.

a) sin(A) = 1.0 ± 0.2 [5%,7%]

b) sin(A) = 0.99 ± 0.02 [7%,13%]c) sin(A) = 0.990 ± 0.002[25%,50%]

d) sin(A) = 0.9903 ± 0.0024 [10%,27%]Explanation: The uncertainty in sin(A) can be found either from half the difference between the extreme values of sin(83

^{o}) and sin(81^{o}) or from applying the propagation of error equation. This problem is unique because the correctly-reported result has 3 significant figures even though the original measurement only has two. Many students mistakenly believe that the rule for rounding a result to the least number of significant figures of any individual factor when multiplying and dividing apply to other functions as well, but this is not always the case.

4) Which period is more *accurate*, and why?

a) Student A's period of 1.25 s because a digital stopwatch is more reliable. [5%,5) Which measurement is more0%]

b) Student A's period of 1.25 s because the stopwatch can measure to 0.01 s. [15%,0%]c) Student B's period of 1.6 s because it is closer to the known period than A's value.[80%,100%]

d) There is not enough information to answer this question. [0%,0%]Explanation: Accuracy is how close a measurement is to a known value. The situation presented in this problem is somewhat unusual because the more accurate measurement was made with a less precise measuring device. A precise measuring device does not guarantee accurate measurements!

a) Student A's period of 1.25 s because a digital stopwatch is more reliable. [5%,6) What is the most probable source of error that could explain the difference in the results?0%]b) Student A's period of 1.25 s because the stopwatch can measure to 0.01 s.[70%,100%]

c) Student B's period of 1.6 s because it is closer to the known period than A's value. [10%,0%]

d) There is not enough information to answer this question. [5%,0%]Explanation: Student A's result of 1.25 s is more precise because it has more significant figures (and decimal places) than Student B's result of 1.6 s.

a) Human reaction time in starting and stopping the timing devices. [80%,30%]

b) The stopwatch may run too fast; not calibrated properly. [10%,20%]

c) The amplitude of oscillation may have been too large for one pendulum. [5%,10%]d) Student A mistakenly measured 4 oscillations instead of the intended 5.[5%,20%]Explanation: Here is sample data that could explain the measurements reported by the students:

Student A: The time I measured for 5 oscillations was t = 8 s , with an estimated uncertainty of about 0.5 s based on the instrument precision of my analog wristwatch. This gives an average period of T = 1.6 s ± 0.1 s , which agrees with the known period of 1.55 s.

Student B: I measured the total time for 5 oscillations to be t = 6.25 s , with an uncertainty of about 0.2 s based on my estimated reaction time. This gives an average period of T = 1.25 s ± 0.04 s , which clearly does not agree with the known period of 1.55 s. Evidently I made a mistake. When I started counting the 5 swings of the pendulum, I remember saying "one" when I started the timer and "five" when I stopped. This resulted in a total time for only 4 oscillations instead of the intended 5. If this is true, then the average period should be (6.25 s ± 0.2s)/4 = 1.56 s ± 0.04 s, which does agree with the known period.

h = 5.25 ± 0.15 m, t = 1.14 ± 0.06 s.

7) Use the equation a = 2h/t^{2} to determine the average acceleration
and its uncertainty.

a) 8.08 ± 0.1 m/s8) Comment on the accuracy of the acceleration result. Do you think the student made any mistakes?^{2}[5%,0%]

b) 8.08 ± 0.88 m/s^{2}[15%,20%]

c) 8.08 ± 0.06 m/s^{2}[15%,0%]d) 8.1 ± 0.9 m/s[3%,^{2}20%]Explanation: By applying the propagation of uncertainty equation, we find that the relative uncertainty for the height is ± 2.9% and for the time is ± 2(5.3%), so that the total relative uncertainty is ± 10.9%. This yields an average acceleration of 8.08 ± 0.88 m/s

^{2}, which when properly rounded is 8.1 ± 0.9 m/s^{2}.

a) The uncertainty is high; probably a mistake in height measurement or reaction time with stopwatch. [35%,9) What one suggestion would you tell this student to improve the accuracy of the experimental result?30%]b) Although a < g, the result seems reasonably accurate since air resistance would reduce the ball's acceleration.[5%,10%]

c) The result does not agree with 9.8 m/s^{2}, so the student must have made a mistake. [40%,90%]

d) The result can only be as accurate as the measurements; cannot tell if a mistake was made. [3%,0%]Explanation: While it is impossible to know if the student's result is accurate without knowing detailed information about the drag factor, it is plausible that the average acceleration of the falling ball is about 0.8 g over the fall distance of about 3 stories. The important point of this question is that the objective of the experiment was to find the average acceleration of a falling object, which is not the same as measuring g. With this objective, the systematic effect of the air resistance is not an error, and just because the average acceleration is significantly less than g does not necessarily indicate that a procedural mistake was made by the student.

a) Measure height better, maybe use a long tape measure that does not stretch. [12%,10) A student measures the ratio of the circumference of a circle divided by its diameter to be 3.3 ± 0.1, where the reported uncertainty is the standard error found from repeated measurements. Does this result agree with the known value of PI = 3.14159...?40%]b) Improve the precision of the time with more trials or automatic device.[50%,40%]

c) Get an assistant to time when the ball hits the ground [10%,0%]

d) Reduce or eliminate air resistance. [8%,0%]Explanation: The uncertainty in the time measurement contributes the most to the overall uncertainty in the average acceleration, so improving the precision of this measurement will most significantly reduce the uncertainty of the result, and hopefully give a more accurate value. Even though improving the timing precision does not guarantee improved accuracy, it is a necessary condition, and is a better option than the other alternatives.

a) No, the uncertainty range of the measured result does not include the value of PI.b) The probability of agreement is about 5%.c) The relative error is 5%, which is small enough to be considered acceptable.d) We need to know how the measurements were made to answer this question.Explanation: All of the above answers could be considered correct. When it comes to deciding whether two measurements agree or not, there is often a gray judgement area. This is the nature of science and the interpretation of data. For this problem, the level of confidence for the reported uncertainty is not stated, so it is impossible to make a definitive judgement about agreement. If the reported measurement follows the ISO Guidelines for reporting uncertainty, then we can assume that 0.1 is the standard uncertainty, which means that there is about a 68% probability that the true value of PI lies within the interval of 3.3 ± 0.1. While the known value of 3.14... is not within this interval, there is a chance that the measurement is consistent with the known value, but it is impossible to know the probability of agreement without more detailed knowledge about how the measurements were made.