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The sample mean plus and minus a critical value times its standard error, and what each of the three inputs does to the width.
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 for a population mean when the population spread is known, computing the standard error, choosing the critical value from the level, and reporting both endpoints. You will also say what each input does to the width, and what a change in the sample size would have produced.
The standard error of a sample mean came out of the first lesson and the shape of a confidence interval out of the last. This lesson does nothing but put them together, in the case where the population spread is a number somebody already knows. The case where it is not is the next lesson.
A z interval is a confidence interval whose critical value comes from the standard normal distribution, which is appropriate when the population standard deviation $\sigma$ is known — or when the sample is large enough for the difference to stop mattering. Coverage is the fraction of samples for which the interval contains the parameter, and it is what the level promises.
For a sample of $n$ from a population with mean $\mu$ and known standard deviation $\sigma$, the central limit theorem makes
$$Z = \frac{\bar X - \mu}{\sigma/\sqrt n}$$
approximately standard normal — exactly so if the population is normal. That quantity is a pivot: its distribution involves no unknown parameter, which is precisely what allows it to be inverted. Writing $P(-z_{\alpha/2} \le Z \le z_{\alpha/2}) = 1 - \alpha$ and solving the double inequality for $\mu$ gives
$$\bar x \pm z_{\alpha/2}\,\frac{\sigma}{\sqrt n}.$$
Three numbers go in and one comes out. The critical value is chosen by the level alone; the standard error $\sigma/\sqrt n$ is chosen by the population and the sample size; their product is the margin of error.
The design consequences follow immediately. Quadrupling $n$ halves the margin. Moving from $95\%$ to $99\%$ multiplies it by $2.576/1.960 \approx 1.31$. And the population's own spread enters directly, which is why a pilot study that estimates $\sigma$ is the usual first step in planning a real one.
Another way: picture
The sampling distribution of $\bar X$ drawn as a bell centred on $\mu$, with cut points $1.96$ standard errors either side. The middle $95\%$ of that bell is the set of sample means that would produce an interval covering $\mu$; the two tails are the samples that would miss. Nothing about the picture moves when one sample is drawn.
Another way: steps
A population with $\sigma = 20$, a $95\%$ level, and the interval that results from each sample size.
| Sample size | Standard error | Margin of error | Width |
|---|---|---|---|
| 25 | 4 | 7.84 | 15.68 |
| 100 | 2 | 3.92 | 7.84 |
| 400 | 1 | 1.96 | 3.92 |
| 2500 | 0.4 | 0.784 | 1.568 |
The critical value $1.96$ multiplies every row and changes none of the comparisons between them. That is worth noticing: the shape of this table is set entirely by the square root, and choosing a different confidence level would rescale the whole column without reordering anything in it.
Dividing by $n$ instead of $\sqrt n$. This makes the interval far too narrow, and the error grows with the sample size rather than shrinking.
Using $\sigma$ where the standard error belongs. The interval is built on the spread of the estimator, not on the spread of the data.
Using this interval when $\sigma$ was estimated from the same small sample. That is the next lesson, and the interval it needs is wider.
Narrowing an interval by lowering the level. It works, and it works by making the procedure wrong more often. Reporting the new interval without the new level is the part that is dishonest.
$n = 64$, $\bar x = 52$, $\sigma = 16$ known, level $95\%$.
All three inputs present.
Standard error $16/8 = 2$; margin $1.96 \times 2 = 3.92$.
Square root of $64$ is $8$.
Interval $(48.08, 55.92)$.
Mean plus and minus the margin.
The same sample at $99\%$: the critical value becomes $2.576$.
Only the first factor changes.
Margin $2.576 \times 2 = 5.152$, interval $(46.848, 57.152)$.
Wider, as it must be.
Thirty-one per cent wider, in exchange for an error rate five times smaller.
The trade, in numbers.
The standard error is $30/10 = 3$.
Square root of a hundred is ten.
The critical value at $90\%$ is $1.645$, so the margin is $4.935$.
Critical value times standard error.
The interval is $(35.065, 44.935)$.
A sample of $9$ observations has mean $53$, and the population standard deviation is known to be $6$. Give the $98\%$ confidence interval for the population mean, with both endpoints included.
This task has no paper form; do it on a device.
A sample of $16$ observations comes from a population with standard deviation $4$. Using a critical value of $2$, what is the margin of error for the mean?
Answer:
The population standard deviation is $10$ and the level is $98\%$, so the critical value is $2.326$. Give the margin of error for the mean at each of these sample sizes.
| Margin of error | |
|---|---|
| A sample of $25$ | |
| A sample of $100$ | |
| A sample of $625$ |
A sample of $36$ observations is drawn from a population with standard deviation $12$, and a $99\%$ interval is wanted. Give the standard error of the mean, and the margin of error.
Standard error: e. Margin of error: g.
From the sample collected, a $95\%$ interval for the mean runs from $52$ to $60$. Suppose four times as many observations had been collected and the sample mean had come out the same. Put the marker at the lower end of that interval.
0 |——————————| 100
Mark the position with a cross, then write the value:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A $95\%$ interval for a mean, built from $9$ observations, has come out wider than the client wanted. Which change narrows it without weakening what it claims?
You can build an interval for a mean from a sample size, a known spread and a confidence level, and say which changes narrow it honestly. Say in your own words why the sample size enters under a square root.
9. Your turn: $n = 100$, $\bar x = 40$, $\sigma = 30$ known, at the $90\%$ level, step 3