Judging evidence against a hypothesis

A hypothesis is a statement from which testable numerical predictions follow; the scientific question fixes exactly which prediction we compare with the data.

To judge consistency, compare the prediction $x_{\text{theor}}$ with the measurement result $x_{\text{meas}} \pm u$, where $u$ is the standard uncertainty of the measurement. The key quantities are the discrepancy (residual) $\Delta = x_{\text{meas}} - x_{\text{theor}}$ and its ratio to the uncertainty, $|\Delta|/u$. If $|\Delta| \lesssim u$, the data are consistent with the hypothesis; if $|\Delta| > 3u$, the hypothesis is under serious tension. For several measurements, first find the mean and the standard uncertainty of the mean $u = \frac{s}{\sqrt{n}},$ and then compare it with the prediction. Agreement between data and hypothesis never proves it: agreement only means the hypothesis has not yet been refuted, whereas a large discrepancy is a strong argument against it. Before concluding that a hypothesis is refuted, check systematic effects and whether the uncertainty itself is correct.

A hypothesis predicts a pendulum period $T_{\text{theor}} = 2.00$ s. One measurement gives $T_{\text{meas}} = 2.04$ s with standard uncertainty $u = 0.01$ s. Are the data consistent with the hypothesis?

Three independent measurements of the fall time from height $h = 5.00$ m gave $1.01$, $0.99$ and $1.00$ s. The hypothesis states free fall with no initial velocity: $t_{\text{theor}} = \sqrt{2h/g}$ with $g = 9.81$ m/s$^2$. Are the data consistent with the hypothesis?

There are two competing hypotheses about the pendulum period: A) $T \propto L$; B) $T \propto \sqrt{L}$. Measurements: at $L = 1.00$ m, $T = 2.00 \pm 0.02$ s; at $L = 4.00$ m, $T = 4.01 \pm 0.04$ s. Which hypothesis agrees better with the data?

Self-check of a hypothesis in a pendulum simulation (for example, the PhET Pendulum Lab).

1. State the hypothesis: $T = 2\pi\sqrt{L/g}$ with $g = 9.81$ m/s$^2$, and the question: "Are the data consistent with this formula?"
2. Choose three lengths, for example $0.50$, $1.00$ and $1.50$ m.
3. For each length, measure the time of 10 full oscillations and divide by 10 — this reduces random error; repeat the measurement 3 times and find the mean and $u = s/\sqrt{n}$.
4. Compute the prediction, the discrepancy $\Delta$ and the ratio $|\Delta|/u$ for each length.
5. Write the conclusion in words: where the data agree, where the discrepancy exceeds $3u$; explain why agreement does not prove the hypothesis.

Use the simulation rather than household measurements: a hand-held stopwatch is imprecise, while the simulation lets you separate random scatter from a systematic discrepancy.

Show that judging a hypothesis depends not only on the size of the discrepancy but also on the measurement uncertainty: the same $\Delta$ can be either "agreement" or "conflict".

A diagram on a number line: the prediction $x_{\text{theor}}$ is marked as a vertical line (with no uncertainty band of its own), and three measurement results are drawn as points with $\pm u$ bars. Case 1: the bar comfortably covers $x_{\text{theor}}$, $|\Delta| \ll u$ — agreement, $|\Delta|/u \ll 1$. Case 2: the bar only just touches $x_{\text{theor}}$ with its edge, $|\Delta| \approx u$ — borderline agreement, $|\Delta|/u \approx 1$. Case 3: the bar is far from $x_{\text{theor}}$, $|\Delta| > 3u$ — conflict, $|\Delta|/u > 3$. Beside it, the formula $|\Delta|/u$ with labelled values for each case.

A coach claims the runner averages $8.0\ \text{m/s}$ over $100\ \text{m}$. Three electronic timings give $12.6\ \text{s}$, $12.8\ \text{s}$, and $12.7\ \text{s}$. The mean is $12.7\ \text{s}$, so the mean speed is $100/12.7 = 7.87\ \text{m/s}$. Repeated runs scatter by about $\pm 0.1\ \text{s}$.

The judgement rule: the discrepancy $8.0-7.87=0.13\ \text{m/s}$ must be compared with the uncertainty. A timing spread of $\pm 0.1\ \text{s}$ at $12.7\ \text{s}$ corresponds to roughly $\pm 0.06\ \text{m/s}$ in speed. The discrepancy is about twice the uncertainty, so the claim is not clearly supported and more trials are needed.

A phone times 10 swings of a $0.50\ \text{m}$ pendulum: $T=1.42\ \text{s}$ per swing. The model $T=2\pi\sqrt{L/g}$ with $g=9.81\ \text{m/s}^2$ predicts $T=1.42\ \text{s}$. Rearranging gives $g=4\pi^2 L/T^2 = 9.8\ \text{m/s}^2$. Reaction-time scatter in timing gives $\pm 0.02\ \text{s}$ in $T$.

Judging evidence here uses the model prediction versus measurement and uncertainty propagation. The measured $g=9.8\ \text{m/s}^2$ differs from $9.81\ \text{m/s}^2$ by $0.01\ \text{m/s}^2$. A $\pm 0.02\ \text{s}$ uncertainty in $T$ gives about $\pm 0.3\ \text{m/s}^2$ in $g$; the discrepancy is far smaller, so the hypothesis is consistent with the data. If $T$ were $1.60\ \text{s}$, $g$ would be $7.7\ \text{m/s}^2$, a discrepancy of $2.1\ \text{m/s}^2$ that would force rejection or search for a systematic error.

A kitchen thermometer reads the boiling water as $98.2\ ^\circ\text{C}$, $98.5\ ^\circ\text{C}$, and $98.1\ ^\circ\text{C}$; mean $98.3\ ^\circ\text{C}$. The thermometer's stated accuracy is $\pm 0.5\ ^\circ\text{C}$. The local altitude is about $500\ \text{m}$.

Judging evidence against the hypothesis $T=100\ ^\circ\text{C}$: the discrepancy is $100-98.3=1.7\ ^\circ\text{C}$, which is $3.4$ times the $\pm 0.5\ ^\circ\text{C}$ uncertainty, so the simple hypothesis is rejected. Because altitude is about $500\ \text{m}$, the expected boiling point is near $98.3\ ^\circ\text{C}$; the evidence supports the revised hypothesis that pressure, not just heat, sets the boiling temperature.