Finding a wrongly recorded reading

Finding a wrongly recorded reading

Secondary 3

When one reading has been recorded wrongly, its point lies clearly off the straight line formed by the others. You find it from the graph, draw the line through the remaining points and read the corrected value from that line.

Why: the law predicts that all correct points lie on one straight line. A point far from that line, when all the others are close to it, cannot be explained by small measuring errors.

Method

  1. Plot all the points, including the suspect one.
  2. Draw the line of best fit through the points that follow the trend; ignore the one that is clearly off.
  3. State which reading is wrong, giving both its \(x\)-value and its recorded value.
  4. Read the line at that \(X\)-value, then change \(Y\) back to the original variable for the corrected reading.
  5. Justify in words: the point lies well away from the line while every other point lies close to it.

Example

Readings should satisfy \(y=a+\dfrac{b}{x}\), so \(y\) is plotted against \(\dfrac{1}{x}\). One reading of \(y\) is wrong.

\(x\)\(1\)\(2\)\(3\)\(4\)\(6\)
\(\dfrac{1}{x}\)\(1\)\(0.5\)\(0.333\)\(0.25\)\(0.167\)
\(y\)\(15.0\)\(9.1\)\(6.9\)\(7.2\)\(5.0\)

Four points lie close to the line \(y=12\left(\dfrac{1}{x}\right)+3\). The point \((0.25,\,7.2)\) is \(1.2\) above the line, while the others are within \(0.1\). So \(y=7.2\) at \(x=4\) is wrong. At \(\dfrac{1}{x}=0.25\) the line gives \(y=12(0.25)+3=6.0\), the corrected reading. Hence \(a=3\) and \(b=12\).

Exam check Do not change a reading just because its point is not exactly on the line: every real reading scatters a little. Only a point clearly separated from an otherwise straight pattern is treated as wrong.

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