Back to the on-screen lesson ·
An estimate plus and minus a margin of error, the level that describes the procedure rather than the interval, and the statements the level does not support.
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 build a confidence interval from an estimate, a critical value and a standard error, move between the width, the margin of error and the two endpoints, and state exactly what the confidence level claims. You will also recognise the statements about an interval that it does not support.
An estimate on its own says nothing about how far it might be from the truth. The standard error of the first unit says how far, typically. This lesson turns that typical distance into a stated range, and then says — carefully — what the range does and does not claim.
The confidence level is the fraction of samples for which the procedure produces an interval containing the parameter. The error rate is what is left of one, split evenly between the two tails for a two-sided interval. The critical value is the cut point leaving half the error rate beyond it, and the margin of error is the critical value times the standard error. Coverage is the property the level names.
A confidence interval at level $1 - \alpha$ is a pair of statistics $(L, U)$, both computed from the sample, with
$$P(L \le \theta \le U) = 1 - \alpha$$
before the data are seen. Everything turns on that qualification. In that probability statement $\theta$ is a fixed number and $L$ and $U$ are the random things; the statement is about where the random interval lands, not about where the fixed parameter is.
The usual shape is
$$\text{estimate} \pm (\text{critical value}) \times (\text{standard error}),$$
and the product on the right is the margin of error. The critical value comes from the sampling distribution and grows with the confidence level; the standard error shrinks like $1/\sqrt n$.
Once the numbers are in, the interval is two numbers and the parameter is one number. It covers or it does not. The honest report is therefore about the procedure: intervals built this way cover the parameter $95\%$ of the time, and this is one of them. That sentence is longer than the one people want to write, and it is the only one the mathematics supports.
Another way: picture
A vertical line marking the true parameter, and a hundred horizontal bars stacked beside it, one per sample. About ninety-five of the bars cross the line and about five miss it entirely. Nothing about any single bar is uncertain once it is drawn; what the level describes is the picture as a whole.
Another way: steps
One estimate, one standard error of $1$, four levels.
| Level | Error rate | Each tail | Critical value | Margin of error |
|---|---|---|---|---|
| 80% | 20% | 10% | 1.282 | 1.282 |
| 90% | 10% | 5% | 1.645 | 1.645 |
| 95% | 5% | 2.5% | 1.960 | 1.960 |
| 99% | 1% | 0.5% | 2.576 | 2.576 |
The error rate falls by a factor of twenty from the first row to the last and the width only doubles. That is the shape of a normal tail, and it is why the conventional levels sit where they do: below $90\%$ the interval is barely narrower, and above $99\%$ it grows without end.
Reading the level as a probability about the parameter. A confidence level is a property of the procedure, fixed before the data arrive. Once the numbers are in, the interval is a pair of numbers and the parameter is a number: it either covers or it does not, and no probability is left to report. The statement the level supports is about a long run of samples built the same way.
Reading the interval as a range for the data. An interval for a mean is narrower than the data by a factor of $\sqrt n$, and it keeps narrowing as data accumulate while the spread of the data does not.
Treating the middle as more likely than the ends. A confidence interval carries no distribution over the parameter. Every point inside it is a value the data do not rule out at this level, and that is all.
Confusing the margin with the width. The margin is half the width. Quoting one for the other doubles or halves the stated uncertainty.
An estimate of $48$ has standard error $2$; the level is $95\%$, so the critical value is $1.960$.
Level first, then the cut point.
Margin of error $1.960 \times 2 = 3.92$, so the interval is $(44.08, 51.92)$.
Estimate plus and minus the margin.
Reported as: intervals built this way cover the mean $95\%$ of the time, and this one runs from $44.08$ to $51.92$.
The procedure carries the claim.
A report gives a $99\%$ interval of $(12, 20)$ for a mean.
Centre $16$, margin $4$.
It does not say the mean is probably about $16$; it says a procedure that misses once in a hundred produced this range.
The level is about the recipe.
A $95\%$ interval from the same data would be narrower, and would miss more often.
Width and coverage trade.
The critical value at $90\%$ is $1.645$.
Five per cent in each tail.
The margin of error is $1.645 \times 5 = 8.225$.
Critical value times standard error.
So the interval is $(21.775, 38.225)$, and the claim is about the procedure that produced it.
An estimate is $52$ and the margin of error is $6$. Give the confidence interval, with both endpoints included.
This task has no paper form; do it on a device.
A $95\%$ confidence interval for a population mean has been computed from a sample, and it runs from $20$ to $24$. Which statement about it is correct?
A two-sided interval is built at the $92\%$ confidence level. Give the total percentage left outside the interval, and the percentage left in each tail.
Total outside: e. In each tail: k.
A two-sided interval for a mean is built around an estimate of $55$, with the population spread known. Match each confidence level to the standard normal critical value it needs.
| $1.282$ | $1.645$ | $1.960$ | $2.576$ | |
|---|---|---|---|---|
| $80\%$ confidence | ||||
| $90\%$ confidence | ||||
| $95\%$ confidence | ||||
| $99\%$ confidence |
A report quotes a $95\%$ confidence interval of $64$ to $68$ for a population mean, and then says four things about it. Mark the two sentences that misstate what the interval claims.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
An interval runs from $16$ to $30$. What is its margin of error?
Answer:
You can build an interval from an estimate and a margin, read a level as an error rate split between two tails, and write down what the level claims. Say in your own words why no probability is left to report once the numbers are in.
9. Your turn: an estimate of $30$, a standard error of $5$, and a $90\%$ level, step 3