FRM Exam Part I · Hypothesis Testing

Confidence Intervals and Standard Error for the FRM

Updated 11 October 2026 · Fact-checked

The standard error of the sample mean is s ÷ √n, the standard deviation of the sample mean across repeated samples. A confidence interval for a population mean is the sample mean ± critical value × standard error. Use z when the population variance is known or n is large; use t with n − 1 degrees of freedom otherwise.

Understand Confidence Intervals and Standard Error

A sample mean is a random variable. Draw another sample and you get a different mean. The distribution of these sample means is the sampling distribution of the mean. Its centre is the population mean μ, so the sample mean is unbiased.

The spread of that distribution is the standard error. It equals σ ÷ √n when σ is known. When σ is unknown, you replace it with the sample standard deviation s. Standard error is not the same as standard deviation. Standard deviation describes how individual observations vary. Standard error describes how the sample mean varies. Quadrupling the sample size only halves the standard error.

The Central Limit Theorem says that for large n the sampling distribution is approximately normal, whatever the shape of the underlying data (given finite variance). If the data are normal, the sample mean is exactly normal.

A confidence interval takes the estimate and adds a margin on each side: sample mean ± critical value × standard error. A 95% interval means that if you repeated the sampling many times, about 95% of the intervals built this way would contain the true μ. It does not mean there is a 95% probability that μ lies in one particular computed interval. μ is fixed. The interval is what varies.

Confidence intervals and hypothesis tests are linked. For a two-sided test at significance level α, you reject H0: μ = μ0 exactly when μ0 lies outside the (1 − α) confidence interval, provided both use the same critical value and standard error. So a 95% interval matches a 5% two-sided test.

Key formulas to remember

Standard error (σ known): SE = σ ÷ √n; Use when the population standard deviation is given.
Standard error (σ unknown): SE = s ÷ √n; s is the sample standard deviation, with n − 1 in the denominator.
Confidence interval for the mean: x̄ ± c × SE; c is the z critical value or the t critical value with n − 1 degrees of freedom.
Common z critical values (two-sided): 90%: 1.645; 95%: 1.96; 99%: 2.576; Memorise these. Each leaves α/2 in each tail.
Test statistic: (x̄ − μ0) ÷ SE; Compare with the same critical value used for the interval.
Degrees of freedom for t: df = n − 1; Use the t table for small samples with unknown variance.

How to solve Confidence Intervals and Standard Error questions

Use this order for any question on standard error or confidence intervals for a mean.

1Identify what is given: sample mean, standard deviation, sample size, and whether the standard deviation is the population value or the sample value.
2Compute the standard error: divide the standard deviation by √n.
3Choose the distribution. Use z if σ is known (or n is large and you are told to use z). Use t with n − 1 df if σ is unknown and n is small.
4Read the critical value for the confidence level, remembering that a two-sided interval puts α/2 in each tail.
5Compute the margin of error: critical value × standard error.
6Write the interval as x̄ − margin to x̄ + margin.
7If a test is asked, check whether the hypothesised mean lies inside the interval. Outside means reject H0 (two-sided).
8Check the answer: a higher confidence level or a smaller n must give a wider interval.

Quickest way: Margin-first shortcut

When to use it: Use when the options are intervals or widths and you need speed.

Compute SE = s ÷ √n once.
Multiply by 1.96 (or the given critical value) to get the margin.
Add and subtract the margin from x̄, or just check the width: the width is 2 × margin.
Eliminate options whose midpoint is not x̄ or whose width is clearly wrong.
To change sample size, scale: the margin shrinks by the square root of the ratio of sample sizes.

Common mistakes in Confidence Intervals and Standard Error

Using the standard deviation instead of the standard error in the interval.
The standard deviation is the number given, so it feels like the one to use.
Fix: Always divide by √n first. The interval for the mean uses SE.
Forgetting to take the square root of n.
Students divide by n by habit, as in an average.
Fix: Write SE = s ÷ √n on paper before substituting numbers.
Using a one-tailed critical value for a two-sided interval.
Confusing 1.645 and 1.96 at the 95% level.
Fix: For a 95% two-sided interval use 1.96. Use 1.645 for 90% two-sided.
Interpreting a 95% interval as a 95% chance that μ is in this interval.
It is a natural but incorrect reading.
Fix: Say the method captures μ in 95% of repeated samples. The parameter is fixed.
Using z when σ is unknown and the sample is small.
z values are more familiar than the t table.
Fix: If only s is given and n is small, use t with n − 1 df. It gives a wider interval.
Thinking doubling the sample size halves the standard error.
Treating the relationship as linear.
Fix: SE falls with √n. Doubling n divides SE by √2 (about 1.414). Quadrupling n halves it.

Worked examples

Example 1

A risk analyst records 64 daily returns with a sample mean of 0.08% and a sample standard deviation of 0.40%. Construct a 95% confidence interval for the mean daily return, using the normal critical value.

Show the solution

SE = s ÷ √n = 0.40% ÷ √64 = 0.40% ÷ 8 = 0.05%.
Critical value for 95% two-sided = 1.96.
Margin = 1.96 × 0.05% = 0.098%.
Lower = 0.08% − 0.098% = −0.018%. Upper = 0.08% + 0.098% = 0.178%.

Answer: The 95% confidence interval is −0.018% to 0.178%. It contains zero, so you cannot reject a zero mean return at the 5% level (two-sided).

Example 2

A sample of 16 monthly portfolio returns has a mean of 1.2% and a sample standard deviation of 3.0%. Returns are assumed normal. The t critical value for 95% with 15 degrees of freedom is 2.131. Construct the 95% confidence interval for the mean.

Show the solution

SE = 3.0% ÷ √16 = 3.0% ÷ 4 = 0.75%.
Variance is unknown and n is small, so use t with df = 15: critical value 2.131.
Margin = 2.131 × 0.75% = 1.598%.
Lower = 1.2% − 1.598% = −0.398%. Upper = 1.2% + 1.598% = 2.798%.

Answer: The 95% confidence interval is about −0.40% to 2.80%. Using z = 1.96 would give a narrower interval of 1.2% ± 1.47%, which would understate uncertainty.

Exam tips

Check whether the question gives a population or sample standard deviation. It decides z or t.
Many questions test direction, not arithmetic: a higher confidence level widens the interval, a larger n narrows it.
Use the interval to answer a hypothesis test question. If the hypothesised mean is outside, reject.
Keep the 1.645, 1.96 and 2.576 values memorised so you do not waste time on tables.
Watch units: keep returns in percent or decimals consistently.

Practice questions from Hypothesis Testing

Confidence Intervals and Standard Error in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

Confidence Intervals and Standard Error: frequently asked questions

What is the formula for the standard error of the sample mean?

It is σ ÷ √n when the population standard deviation is known, and s ÷ √n when you only have the sample standard deviation. It measures how much the sample mean varies from sample to sample.

When do I use the z distribution and when the t distribution?

Use z when the population variance is known, or when the sample is large and you are told to use the normal approximation. Use t with n − 1 degrees of freedom when the variance is unknown and the sample is small. The t distribution has fatter tails, so its critical values are larger.

How are confidence intervals related to hypothesis tests?

A two-sided test at significance level α rejects H0: μ = μ0 when μ0 falls outside the (1 − α) confidence interval. Both use the same standard error and critical value. So an interval gives you the test result and a range of plausible values.

Why does a larger sample give a narrower interval?

The standard error is divided by √n, so more data reduces the variability of the sample mean. The gain is slow: you need four times the data to halve the interval width.