Confidence intervals 201-SN1-RE: revision sheet

The thread of the chapter

The standard deviation of the individuals is not that of the mean: the interval is built with the standard error $\frac{\sigma}{\sqrt{n}}$ (or $\sqrt{\frac{p(1-p)}{n}}$ ), its 95% describes the method and not the parameter, and it only holds under its conditions.

Topics covered

Sampling distribution
Standard error
Central limit theorem
Sampling with replacement
Confidence interval
Confidence level
Margin of error
Student's t distribution
Degrees of freedom
Estimating a proportion
Sample size

The chapter at a glance

The three intervals of the course, recognized at a glance

Read the row “What is estimated”, then the sample size: the column is chosen. The rest of the column goes onto your paper, conditions included.

Criterion	Mean, $n\ge 30$ large sample	Mean, $n<30$ small sample	Proportion, $n\ge 30$ successes out of n
What is estimated	the mean $\mu$ of a measurement (content, mass, concentration)	the mean $\mu$ of a measurement, with few data	the proportion $p$ of individuals with a characteristic
Conditions	random sample, independent observations, $n\ge 30$	random sample, population assumed normal (no strong skewness, no outlier)	random sample, $n\ge 30$ , $n\hat{p}\ge 5$ and $n(1-\hat{p})\ge 5$
Standard error	$\dfrac{s}{\sqrt{n}}$	$\dfrac{s}{\sqrt{n}}$	$\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$
Critical value	$z$ : $1.645$ , $1.96$ , $2.576$	Student's $t$ with $n-1$ degrees of freedom, read in the table	$z$ : $1.645$ , $1.96$ , $2.576$
Interval	$\overline{x}\pm z\dfrac{s}{\sqrt{n}}$	$\overline{x}\pm t\dfrac{s}{\sqrt{n}}$	$\hat{p}\pm z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$
Typical pitfall	dividing by $s$ instead of $\frac{s}{\sqrt{n}}$	taking $1.96$ : the interval is too narrow	the square root that does not cover the $n$

Mean,

n\ge 30

large sample

What is estimated: the mean $\mu$ of a measurement (content, mass, concentration)
Conditions: random sample, independent observations, $n\ge 30$
Standard error: $\dfrac{s}{\sqrt{n}}$
Critical value: $z$ : $1.645$ , $1.96$ , $2.576$
Interval: $\overline{x}\pm z\dfrac{s}{\sqrt{n}}$
Typical pitfall: dividing by $s$ instead of $\frac{s}{\sqrt{n}}$

Mean,

n<30

small sample

What is estimated: the mean $\mu$ of a measurement, with few data
Conditions: random sample, population assumed normal (no strong skewness, no outlier)
Standard error: $\dfrac{s}{\sqrt{n}}$
Critical value: Student's $t$ with $n-1$ degrees of freedom, read in the table
Interval: $\overline{x}\pm t\dfrac{s}{\sqrt{n}}$
Typical pitfall: taking $1.96$ : the interval is too narrow

Proportion,

n\ge 30

successes out of n

What is estimated: the proportion $p$ of individuals with a characteristic
Conditions: random sample, $n\ge 30$ , $n\hat{p}\ge 5$ and $n(1-\hat{p})\ge 5$
Standard error: $\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$
Critical value: $z$ : $1.645$ , $1.96$ , $2.576$
Interval: $\hat{p}\pm z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$
Typical pitfall: the square root that does not cover the $n$

Key: the words in the table

Random sample: Every individual in the population has the same chance of being drawn. Without it, no margin means anything.
Population assumed normal: Assumption that the values in the population follow a normal distribution; it is judged plausible from a graph of the data.
Student: Distribution of $T=\frac{\overline{X}-\mu}{S/\sqrt{n}}$ when the population is normal: more spread out than the normal, it gets closer to it as $n$ grows.
Degrees of freedom: $n-1$ for an interval on a mean: one degree is used up by $\overline{x}$ , which is needed to compute $s$ .

The case “mean, $n<30$ ” on very skewed data has no method in this course: say so, and enlarge the sample.

The essentials

Two standard deviations never to confuse

• $\sigma$ (estimated by $s$ ) measures the spread of the INDIVIDUALS: one grain, one fish, one day.
• $\sigma_{\overline{X}}=\dfrac{\sigma}{\sqrt{n}}$ , the STANDARD ERROR, measures how much the MEAN of a sample of size $n$ spreads from one sample to another.
• Proof with replacement: $E(\overline{X})=\frac{1}{n}\sum E(X_{i})=\mu$ (linearity) and $V(\overline{X})=\frac{1}{n^{2}}\sum V(X_{i})=\frac{\sigma^{2}}{n}$ (independence).
• For a proportion, $\hat{P}=\frac{X}{n}$ with $X$ binomial: $E(\hat{P})=p$ and $V(\hat{P})=\frac{p(1-p)}{n}$ .

Same centre

5.0

ng, but the mean of

36

grains is six times less spread out than a single grain:

\frac{0.9}{\sqrt{36}}=0.15

ng.

Without replacement, the result still holds as long as n is a small fraction of the population (usual rule: less than 5%).

The central limit theorem, and what it does not say

• Independent observations, identically distributed, $n$ large enough ( $n\ge 30$ in practice): $\overline{X}$ approximately follows $N\left(\mu, \frac{\sigma^{2}}{n}\right)$ , whatever the shape of the population.
• It is about the MEAN. The data keep their shape: a skewed histogram stays skewed with $1{,}000$ observations.
• It requires independence: individuals taken together (same net, same days, same family) give a standard error that is too optimistic.
• Below $30$ observations, it is no longer enough: Student's method replaces the theorem by the assumption of a normal population.

What “95%” means

• $\mu$ is a FIXED number. What varies from one sample to another is the interval, through its centre $\overline{x}$ and its width, which depends on $s$ .
• 95% is the proportion of samples for which the METHOD produces an interval that contains $\mu$ .
• A computed interval contains $\mu$ or does not; we do not know which.
• More confidence, wider interval; more data, narrower interval, as $\frac{1}{\sqrt{n}}$ .

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Giving the parameter the probability of the method

1 to 2 marks, for every interpretation asked for

What not to write

“There is a 95% chance that the mean content is between $0.40$ and $0.52$ mg/kg.”

What to write

“At the 95% confidence level, the mean content lies between $0.40$ and $0.52$ mg/kg: the method used captures the true mean in 95% of samples.”

Why: The population mean does not move: once the interval is computed, there is no randomness left. The probability belonged to the drawing of the sample.

2. Taking the interval for the mean as the range of the individuals

the whole interpretation question

What not to write

“95% of the pollen grains weigh between $4.80$ and $5.39$ ng.”

What to write

“The MEAN mass of the grains lies between $4.80$ and $5.39$ ng at the 95% level; the grains themselves are about $\sqrt{36}=6$ times more spread out.”

The

36

grains spread from

3.3

6.9

ng; the 95% interval for the mean,

[4.80, 5.39]

, covers only a small part of them, and that is normal (grains grouped by

0.2

ng).

Why: The interval is built on the standard error, not on the standard deviation of the individuals. It is the thread of the chapter, and the figure makes it undeniable.

3. Dividing by the standard deviation instead of the standard error

2 marks and the whole rest of the exercise

What not to write

“ $z=\dfrac{0.5-0.42}{0.22}\approx 0.36$ , so $P(\overline{X}>0.5)\approx 0.36$ .”

What to write

“ $\sigma_{\overline{X}}=\dfrac{0.22}{\sqrt{40}}\approx 0.0348$ , so $z\approx 2.30$ and $P(\overline{X}>0.5)\approx 0.0107$ .”

Why: A question about $\overline{X}$ is standardized with the standard deviation of $\overline{X}$ . Writing the standard error alone, on its own line, before any margin, prevents the slip.

4. Taking 1.96 on a small sample

1 to 2 marks, plus the interpretation

What not to write

“ $n=12$ : $\overline{x}\pm 1.96\dfrac{s}{\sqrt{12}}$ at the 95% level.”

What to write

“ $\sigma$ unknown and $n=12<30$ , population assumed normal: $\overline{x}\pm t_{0.025, 11}\dfrac{s}{\sqrt{12}}$ , with $t=2.201$ .”

Why: With $1.96$ , the method captures $\mu$ only about 92.4% of the time with 11 degrees of freedom: it claims a precision it does not have. The $t$ is ALWAYS larger than $z$ .

5. Believing that n ≥ 30 makes the data normal

1 mark of justification

What not to write

“Since $n=45\ge 30$ , the mercury contents follow a normal distribution.”

What to write

“Since $n=45\ge 30$ and the walleyes were drawn at random, the MEAN content $\overline{X}$ approximately follows a normal distribution.”

Why: The central limit theorem is about the mean. A skewed population stays skewed; it is the mean that becomes normal.

6. Rounding the sample size down

1 mark, the final answer

What not to write

“ $n\ge 150.06$ , so $n=150$ .”

What to write

“ $n\ge 150.06$ , so $n=151$ : with $150$ , the margin would be $2.0004$ mm, above the target.”

Why: The condition is an inequality: the first integer that satisfies it is above. ALWAYS round up, even 150.06.

7. Doubling n to halve the margin

1 mark

What not to write

“With twice as many days of measurement, the margin will be half as large.”

What to write

“The margin goes as $\dfrac{1}{\sqrt{n}}$ : dividing it by $2$ takes $4$ times as many data; twice as many only divides it by $\sqrt{2}\approx 1.41$ .”

Why: The precision of a mean improves like the square root of the effort; that is what makes large samples expensive.

8. Swapping the bounds after a decreasing transformation

1 mark, and an absurd interval

What not to write

“ $0.111\le p\le 0.189$ , so $\dfrac{600}{0.111}\le N\le\dfrac{600}{0.189}$ .”

What to write

“ $N=\dfrac{600}{p}$ decreases as $p$ increases: $\dfrac{600}{0.189}\le N\le\dfrac{600}{0.111}$ , that is about $3{,}170\le N\le 5{,}410$ .”

Why: A decreasing function swaps the bounds. The confidence level is preserved, since the two inequalities describe the same event.

Which method to choose

Which interval, z or t?

Look at what is being estimated, then at the sample size, then at what is known about the population.

With

4

degrees of freedom, Student's t distribution has much heavier tails: you must go out to

2.776

, and not

1.96

, to leave

2.5\%

in the tail (right half shown, the figure is symmetric).

If you count successes out of $n$ (a proportion) and $n\ge 30$ , $n\hat{p}\ge 5$ , $n(1-\hat{p})\ge 5$ → interval for a proportion, with $z$

Example: $85$ infested ash trees out of $250$ : $0.34\pm 1.96\times 0.0300$
If you measure a quantity (a mean) and $n\ge 30$ → $\overline{x}\pm z\frac{s}{\sqrt{n}}$ , $s$ in place of $\sigma$

Example: $36$ days of ozone: $45.9\pm 1.96\times\frac{10.9}{6}$
If a mean, $n<30$ , $\sigma$ unknown, and a graph of the data with no strong skewness and no outlier → Student's t distribution with $n-1$ degrees of freedom

Example: $12$ maple tapholes: $2.30\pm 2.201\times\frac{0.316}{\sqrt{12}}$
If a mean, $n<30$ , and very skewed data or an outlier → no method of the course applies: say so, and enlarge the sample

transforming the variable or using a nonparametric method goes beyond the course description

If $\sigma$ is truly known (rare: calibrated instrument, entire population already measured), keep $z$ with $\sigma$ , even on a small sample from a normal population.

Which sample size?

Look at what you want to estimate and at what you already know.

If a mean, with a standard deviation that is known or estimated by a pilot study → $n\ge\left(\frac{z\sigma}{E}\right)^{2}$ , rounded up

Example: $\sigma=12.5$ mm, $E=2$ mm: $n\ge 150.06$ , so $n=151$
If a proportion, with no information at all → $p=0.5$ , the worst case, because $p(1-p)\le\frac{1}{4}$

Example: $E=0.05$ : $n\ge 384.16$ , so $n=385$
If a proportion, with a reliable earlier estimate $p_{0}$ → $n\ge\frac{z^{2}p_{0}(1-p_{0})}{E^{2}}$ , rounded up

Example: $p_{0}=0.2$ , $E=0.05$ : $n=246$

The size of the population does not enter the calculation as long as the sample remains a small fraction of it.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Writing up a complete confidence interval

When to use it: “Estimate with a confidence interval”, “build the 95% interval”, “what can be said about the mean?”

1 Name the parameter in context: “ $\mu$ , the mean mercury content of the adult walleyes in the lake”.
2 Choose the case and check its conditions with one sentence each: random sample, independence, $n\ge 30$ (or plausible normality, or $n\hat{p}\ge 5$ and $n(1-\hat{p})\ge 5$ ).
3 Write the standard error alone, on its own line: $\frac{s}{\sqrt{n}}$ or $\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ .
4 Write the critical value with its source: $z_{0.025}=1.96$ , or $t_{0.025, n-1}$ read in the table.
5 Compute the margin, then the two bounds, rounded to the precision of the data.
6 Conclude with the standard sentence, in the units and the vocabulary of the problem.

Concluding sentence

“At the 95% confidence level, the mean mercury content of the adult walleyes in the lake lies between 0.40 and 0.52 mg/kg.”

The trap: Skipping the conditions: without them, the 95% is not justified, and the grader takes off the mark even if the bounds are right.

Marking: Usually: conditions 1 mark, standard error and critical value 1 mark, bounds 1 mark, concluding sentence 1 mark.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The centre is the point estimate

Average the two bounds: you should get $\overline{x}$ or $\hat{p}$ back. Half the difference is the margin.

$[42.3, 49.5]$ has centre $45.9=\overline{x}$ : a good sign. A centre at $46.5$ reveals a badly copied bound.
The standard error is much smaller than s

Compare the standard error with $s$ : their ratio should be $\sqrt{n}$ , so $6$ for $n=36$ , $10$ for $n=100$ .

$s=10.9$ and a “standard error” of $10.9$ : you forgot to divide by $\sqrt{36}=6$ , the margin is six times too large.
The widths come in the right order

With the same data: 90% interval narrower than 95%, narrower than 99%; and Student interval wider than the one with $z$ .

A 99% interval narrower than the 95% one signals swapped critical values ( $1.96$ and $2.576$ exchanged).
A proportion stays between 0 and 1, a size is a whole number rounded up

Reread the bounds of a proportion and substitute the size found back into the margin formula.

$n=150$ gives back a margin of $2.0004>2$ : $151$ was needed.

The typical problem, taken apart

Ten ragweed pollen grains: a Student interval from start to finish

An allergen monitoring station weighs ten ragweed pollen grains taken at random from its samplers. Masses, in nanograms: $5.6$ , $4.8$ , $5.1$ , $6.0$ , $5.3$ , $4.6$ , $5.5$ , $5.0$ , $5.8$ , $5.3$ .

Estimate the mean mass $\mu$ of the grains with a 95% confidence interval. You are given $t_{0.025, 9}=2.262$ .

The ten masses spread from

4.6

6.0

ng with no isolated value and no strong skewness: the assumption of a normal population is plausible.

Step 1

$\overline{x}=\dfrac{53.0}{10}=5.30$ ng; sum of squared deviations $1.74$ , so $s=\sqrt{\dfrac{1.74}{9}}\approx 0.440$ ng.

Why

Divide by $n-1=9$ : $s$ estimates the standard deviation of the population, and dividing by $10$ would underestimate it.

Step 2

Parameter: $\mu$ , mean mass of the ragweed pollen grains collected. Case: mean, $n=10<30$ , $\sigma$ unknown.

Why

Naming the case BEFORE computing chooses the critical value; it is the mark students lose most often.

Step 3

Conditions: grains taken at random; the figure shows no strong skewness and no outlier, the population is assumed normal.

Why

Below 30 observations, the central limit theorem is not enough: it is the assumed normality that justifies Student's t distribution.

Step 4

Standard error: $\dfrac{0.440}{\sqrt{10}}\approx 0.139$ ng; critical value $t_{0.025, 9}=2.262$ ; margin $2.262\times 0.139\approx 0.315$ ng.

Why

Setting the standard error down alone prevents forgetting the square root; the $t$ with $9$ degrees of freedom, larger than $1.96$ , pays for the uncertainty in $s$ .

Step 5

$[5.30-0.315, 5.30+0.315]\approx[4.99, 5.61]$ ng

Why

The bounds are rounded to the hundredth, one digit more than the data.

Step 6

Check: centre $\frac{4.99+5.61}{2}=5.30=\overline{x}$ ; with $1.96$ we would get $\pm 0.27$ , narrower, as expected.

Why

Two checks in ten seconds: the symmetry about the estimate and the order of the widths between $t$ and $z$ .

The conclusion, written out

“At the 95% confidence level, the mean mass of the ragweed pollen grains collected by the station lies between 4.99 and 5.61 ng.”

The classic mistake on this problem: Taking $1.96$ because the formula “looks like” the one for a large sample: the interval obtained, $[5.03, 5.57]$ , is too narrow and does not deserve its 95%.

Learn by heart

• Standard error of a mean: $\dfrac{\sigma}{\sqrt{n}}$ , estimated by $\dfrac{s}{\sqrt{n}}$ ; of a proportion: $\sqrt{\dfrac{p(1-p)}{n}}$ .
• $z_{0.05}=1.645$ (90%), $z_{0.025}=1.96$ (95%), $z_{0.005}=2.576$ (99%).
• Mean, $n<30$ : Student with $n-1$ degrees of freedom, population assumed NORMAL; $t>z$ always.
• Proportion: $n\ge 30$ , $n\hat{p}\ge 5$ and $n(1-\hat{p})\ge 5$ .
• Size: $n\ge\left(\frac{z\sigma}{E}\right)^{2}$ or $\frac{z^{2}p(1-p)}{E^{2}}$ , $p=0.5$ with no information, ALWAYS rounded up.
• Margin divided by $2$ : $n$ multiplied by $4$ .
• 95% describes the METHOD: $\mu$ is fixed, the interval varies.

Frequently asked questions

When should Student's t distribution be used instead of the normal distribution for a confidence interval?

To estimate a mean when the standard deviation of the population is unknown and the sample has fewer than 30 observations. The population is then assumed normal, which is checked on a graph of the data, and the critical value is read in the Student table with n minus 1 degrees of freedom. With 30 observations or more, the course convention keeps the critical value of the normal distribution, 1.96 at 95 percent.

What does a 95% confidence interval mean?

That the method used produces, in 95 percent of possible samples, an interval that contains the true value of the parameter. The population mean is a fixed number: an interval already computed contains it or does not. So we write “at the 95 percent confidence level, the mean lies between...”, and not “there is a 95 percent chance that the mean is in the interval”.

How do you compute the sample size for a given margin of error?

For a mean, divide the product of the critical value and the standard deviation by the target margin, and square the result. For a proportion, multiply the square of the critical value by p times 1 minus p, then divide by the square of the margin, with p equal to 0.5 if nothing is known. In both cases, always round up to the next whole number, otherwise the target margin is not met.

Does the central limit theorem make the data normal when n exceeds 30?

No. It says that the mean of a sample of independent observations approximately follows a normal distribution when the sample is large enough, whatever the shape of the population. The data themselves keep their distribution: skewed contents stay skewed with 1,000 measurements. That is why a large sample allows the interval for the mean even with non-normal data.

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Corrected exercises: Confidence intervals: mean, Student's t and proportion, 201-SN1-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

11 corrected exercises
110 points
150 minutes

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-sn1-confidence-intervals. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

Revision sheet: confidence intervals (201-SN1-RE)

The three intervals of the course, recognized at a glance

Key: the words in the table

The essentials

Two standard deviations never to confuse

The central limit theorem, and what it does not say

What “95%” means

The mistakes that cost marks

1. Giving the parameter the probability of the method

2. Taking the interval for the mean as the range of the individuals

3. Dividing by the standard deviation instead of the standard error

4. Taking 1.96 on a small sample

5. Believing that n ≥ 30 makes the data normal

6. Rounding the sample size down

7. Doubling n to halve the margin

8. Swapping the bounds after a decreasing transformation

Which method to choose

Which interval, z or t?

Which sample size?

How the answer is expected to be written

Writing up a complete confidence interval

Check before you hand in

The typical problem, taken apart

Ten ragweed pollen grains: a Student interval from start to finish

Learn by heart

Frequently asked questions

When should Student's t distribution be used instead of the normal distribution for a confidence interval?

What does a 95% confidence interval mean?

How do you compute the sample size for a given margin of error?

Does the central limit theorem make the data normal when n exceeds 30?

Corrected exercises: Confidence intervals: mean, Student's t and proportion, 201-SN1-RE

This chapter comes back in

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