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Lesson 6 of 6

Random and systematic error.

About 25 minutes · 6 practice questions

Every repeated measurement gives a slightly different reading: the spread of one reading, the standard error of the mean and its √n, and a systematic error that no number of measurements fixes.

Step 1 of 6 · The spread of one reading

Step 1 of 6

The spread of one reading

A steel ball hangs from an electromagnet, 1.000 m above a contact pad. Cut the current, and a timer measures how long the ball takes to fall. The time gives the acceleration of free fall: h=12gt2h = \tfrac{1}{2}gt^2⁠, so g=2ht2g = \dfrac{2h}{t^2}⁠. Every drop gives a slightly different reading, and each reading becomes a ball on the board, in the column of its value: a histogram lying on its side.

Errors that change from drop to drop, sometimes up and sometimes down, are random errors. The spread ss⁠ measures their size: roughly how far one reading lies from the mean gˉ\bar g⁠. At Spread the board fits a bell curve to the pile and prints gˉ\bar g⁠ and ss⁠. Timing jitter and Magnet delay are fixed. Predict what will happen to the width of the pile with four times the drops.

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Step 2 of 6

The standard error

What the experiment wants is gg⁠, and its best estimate is the mean. The mean is far steadier than any single reading, because the random errors in it partly cancel: some readings are high, some low, and the more of them there are, the better they balance. The uncertainty of the mean is its standard error:

sn\frac{s}{\sqrt{n}}⁠

The board does not show it yet. Work it out from the spread on the board.

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Step 3 of 6

The square root of n

At Error bar a bar of one standard error each way appears beside the pile, with thin whiskers out to two. The amber ticks beside it are the means of 2424⁠ more runs of the same number of drops: the standard error is roughly how far another run’s mean would land from this one. Change Drops and find how many it takes to halve the bar.

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Step 4 of 6

The true value

At True value a rose line marks the accepted value of gg⁠, 9.81 m/s², which a real experimenter would not know in advance. With 2525⁠ drops it is inside the bar. Predict what will happen at 100100⁠ drops, when the bar halves.

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Step 5 of 6

Systematic error

The reason is the Magnet delay: leftover magnetism holds the ball for an instant after the current is cut, so every fall is timed too long. An error that pushes every reading the same way is a systematic error. It does not cancel in the mean, because it does not change from drop to drop. Now the delay is 5 ms, and Work out the true value is fixed at Error bar, so the board does not give the answer away.

Try it: after checking, step Work out the true value to True value. The whole pile slides down by the amount you worked out, and the bar misses the line.

Back to the experiment

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Step 6 of 6

Precision and accuracy

A careful timer, Timing jitter at 2 ms, shrinks the random error: the pile is half as wide. But the 1.5 ms delay is still there. Random error spoils precision, how close the readings are to one another, and more measurements cure it. Systematic error spoils accuracy, how close they are to the truth, and more measurements do not help. Finding it takes a change of method: drop from several heights, or calibrate against something whose answer is known.

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Still of the Models and uncertainty 3D scene
Interactive scene

Loading the interactive version…

Each step sets up the Scene for you, and a control marked “fixed” stays put for that step.

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Practice · Predict, then look

With 2525⁠ drops, the readings’ Spread is s = 0.20 m/s². What will happen to the width of the pile when we go to 100100⁠ drops?

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Practice · Calculate

2525⁠ drops, and the readings’ spread is s = 0.20 m/s². What is the standard error of the mean?

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Practice · Reach the target

With 2525⁠ drops the standard error is 0.04 m/s². Change Drops until it halves, to about 0.02 m/s², and press Check.

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Practice · Predict, then look

With 2525⁠ drops the true value, 9.81 m/s², is inside the error bar. We go to 100100⁠ drops. Will it still agree?

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Practice · Calculate

The magnet lets go of the ball 5 ms late, so every fall is timed 5 ms too long. Without the delay, a 1.000 m fall takes 0.4515 s. By how much does the delay alone lower every reading of gg⁠?

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Practice · Choose

With Timing jitter at 2 ms, the pile is narrow and the error bar small, but the true value lies outside two error bars. How would you describe this measurement?