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Solving the margin of error for the sample size, why precision is bought at the square, and how a study is sized before any data exist.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
You will solve a margin of error requirement for the sample size it needs, scale a margin between sample sizes, and plan a study from the margin wanted, the confidence level and a figure for the population spread. You will also say what lowering the confidence level really concedes.
Three lessons have built intervals from a sample that already existed. A study is designed the other way round: the margin of error is decided first and the sample size is whatever delivers it. Nothing new is needed — only the same formula, solved for a different letter.
The margin of error is the critical value times the standard error. A power calculation or sample size calculation solves the margin requirement for $n$ before any data are collected. A pilot study is a small preliminary sample whose only job is to supply a figure for the population spread, so that the real study can be sized.
Every interval in this unit has the same shape:
$$\text{margin of error} = z_{\alpha/2} \times \frac{\sigma}{\sqrt n}.$$
Three things appear on the right and a designer controls two of them. The critical value is chosen by the confidence level; the sample size is chosen by the budget; the population spread is a fact and has to be supplied from somewhere. Solving for $n$ gives
$$n = \left(\frac{z_{\alpha/2}\,\sigma}{\text{margin}}\right)^{2},$$
and the square is the whole story. Dividing the margin by $k$ multiplies the requirement by $k^{2}$: halving costs four times the data, thirding costs nine, and an extra decimal place costs a hundred.
The confidence level is the cheaper lever and the more expensive concession. Moving from $95\%$ to $90\%$ shrinks the margin by about sixteen per cent for no data at all — and doubles the rate at which the procedure misses. It is a real trade and it must be declared, not slipped in after the interval has come out wider than hoped.
For a proportion the spread needs no pilot study, because $p(1 - p)$ is at most a quarter. Sizing on that worst case gives a promise that holds whatever the answer turns out to be, which is why polls can be commissioned at a fixed margin before anybody is asked anything.
Another way: story
A client asks for an answer to within one percentage point and is quoted a price. They ask what half a point would cost and are quoted four times as much. That is not a negotiating position: it is $n = (z\sigma/\text{margin})^{2}$, and the square in it is the reason most studies are sized at the margin they can afford rather than the one they want.
Another way: steps
A population with $\sigma = 20$, at the $95\%$ level.
| Margin wanted | Sample size needed | Cost relative to the first row |
|---|---|---|
| 8 | 25 | 1 |
| 4 | 97 | 4 |
| 2 | 385 | 15 |
| 1 | 1537 | 61 |
| 0.5 | 6147 | 246 |
Each row halves the margin and quadruples the requirement. Reading down, the first improvement is cheap and the last is ruinous; reading up, a study that cannot afford the bottom row loses less than it fears by settling for the row above.
Assuming the margin falls in proportion to the sample. It falls with the square root. Doubling the sample buys a factor of about $1.41$, not $2$.
Rounding the sample size down. The requirement was an inequality. Rounding down breaks the promise it was solved from, and rounding up costs one or two observations.
Lowering the confidence level quietly. It does narrow the interval, by making the procedure miss more often. The new level has to be reported beside the new interval.
Reporting more decimal places. Precision of reporting is not precision of estimation. An estimate of $0.4173$ with a margin of $0.05$ is still an estimate with a margin of $0.05$.
A pilot gives $\sigma \approx 12$; a margin of $3$ is wanted at $95\%$.
A spread from somewhere.
$n = (1.96 \times 12 / 3)^{2} = (7.84)^{2} = 61.5$.
Solve for $n$.
Round up to $62$.
Upwards, always.
A margin of $3$ percentage points is wanted at $95\%$, and the answer is unknown.
So take the worst case, $p = 0.5$.
$n = (1.96 \times 0.5 / 0.03)^{2} \approx 1067$.
The quarter is the largest $p(1-p)$ can be.
Round up to $1100$, and the promise holds for every question on the survey.
This is where the standard poll size comes from.
$n = (1.96 \times 40 / 5)^{2}$.
Critical value times spread, over the margin.
That is $(15.68)^{2} = 245.9$.
Square it.
So $246$ observations, rounding up.
A sample of $61$ gives a certain margin of error. How large a sample halves that margin, at the same confidence level?
Answer:
At a sample size of $20$ the margin of error is $10$. Give the margin at each of these sample sizes, at the same confidence level.
| Margin of error | |
|---|---|
| A sample of $20$ | |
| A sample of $80$ | |
| A sample of $2000$ |
A study of $54$ people gives a margin of error the client finds too wide. Give the sample size that would halve the margin, and the sample size that would divide it by three.
To halve the margin: a. To divide it by three: b.
A study is to be designed so that its margin of error is at most $2$ units. Put the steps of choosing its sample size into the order they are carried out.
Number the steps in order (write the number in the box):
A margin of error of $4$ units can be achieved with $1$ observations. Plot the number of observations needed for margins of $4$, $2$ and $1$ units, at the same confidence level.
Plot your answer on the grid:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A study of $220$ people reports a margin of error the client wants halved, at the same confidence level. What does that take?
You can size a study from a required margin, scale a margin between two sample sizes, and say which levers narrow an interval honestly. Say in your own words why halving a margin of error costs four times the data.
9. Your turn: a margin of $5$ at $95\%$, with a spread of $40$, step 3