CFA Level I Exam · Estimation and Hypothesis Testing

Point Estimates and Confidence Intervals for CFA Level I

Updated 7 October 2026

A point estimate is one sample number, such as the sample mean, used to estimate a population parameter. A confidence interval is: point estimate ± reliability factor × standard error. Use z when σ is known. When σ is unknown, use t with n − 1 degrees of freedom; for a large sample, z is an acceptable approximation.

Understand Point Estimates and Confidence Intervals

A parameter is a fixed number that describes a population, such as the true mean return. You rarely know it. So you draw a sample and compute a statistic, such as the sample mean. The formula you use is the estimator. The number it gives you is the point estimate.

A good estimator has three properties. It is unbiased: its expected value equals the parameter. It is efficient: among unbiased estimators, it has the smallest sampling variance. It is consistent: as the sample size grows, the estimate gets closer to the parameter, because its sampling variance shrinks.

A point estimate is almost never exactly right. A confidence interval adds a range around it. A 95% interval means that if you repeated the sampling many times and built an interval each time, about 95% of those intervals would contain the true parameter. It does not mean there is a 95% chance the parameter sits in one particular interval you already computed.

The general form is: point estimate ± reliability factor × standard error. The reliability factor comes from the z or t table and depends on your confidence level. The standard error of the mean is s ÷ √n (or σ ÷ √n if σ is known). A higher confidence level, a smaller sample or a more volatile population all make the interval wider.

The t distribution has fatter tails than z, so its reliability factor is larger for the same confidence level. It approaches z as degrees of freedom grow. Use t when you must estimate the variance from the sample.

Key formulas to remember

Confidence interval, known variance: x̄ ± z(α/2) × σ ÷ √n; Use when the population is normal (or n is large) and σ is known.
Confidence interval, unknown variance: x̄ ± t(α/2, n−1) × s ÷ √n; Use t with n − 1 degrees of freedom. This is the usual case in practice.
Standard error of the sample mean: σx̄ = σ ÷ √n, estimated as sx̄ = s ÷ √n; Quadrupling n halves the standard error.
Common z reliability factors: 90%: 1.645; 95%: 1.96; 99%: 2.576; These are two-tailed values. Remember them; they appear often.
Degrees of freedom: df = n − 1; For a single sample mean.
Estimator properties: Unbiased: E(estimator) = parameter. Efficient: smallest variance among unbiased estimators. Consistent: accuracy improves as n rises.; Learn the definitions in these exact terms.

How to solve Point Estimates and Confidence Intervals questions

Use this sequence for any confidence interval question.

1Identify the parameter and the point estimate, usually the sample mean x̄.
2Check what you know: is σ known or only the sample standard deviation s? Is the population normal? Is n large?
3Choose the distribution: use z when σ is known. When σ is unknown, use t (df = n − 1); if the sample is large, z is an acceptable approximation. If the population is non-normal and n is small with unknown σ, the curriculum offers no standard method, so such a case is not tested.
4Convert the confidence level to α, then find the reliability factor for α/2 in each tail.
5Compute the standard error: σ ÷ √n or s ÷ √n.
6Compute the margin of error = reliability factor × standard error.
7State the interval as x̄ − margin to x̄ + margin. Check that the options are in the right order and magnitude.

Quickest way: Margin of error shortcut

When to use it: Use when the question gives numbers and three options; you only need the margin of error and the centre.

The interval is always centred on x̄. Eliminate any option whose midpoint is not x̄.
Compute the standard error s ÷ √n first, using your calculator's √ key.
Multiply by the factor: 1.645, 1.96 or 2.576 for z; the given t value if provided.
Add and subtract from x̄. Match the option.
If two options share the centre, the wider one is the higher confidence level or the t-based interval.

Common mistakes in Point Estimates and Confidence Intervals

Using z when σ is unknown and the sample is small
Students remember 1.96 and use it by default.
Fix: Ask first: is σ given? If only s is given, use t with n − 1 degrees of freedom.
Forgetting to divide by √n
Students use the standard deviation directly as the standard error.
Fix: Always compute s ÷ √n before multiplying by the reliability factor.
Using the wrong tail area, such as the 5% value for a 95% interval
Confusing one-tailed and two-tailed values.
Fix: For a 95% interval, α = 0.05 and each tail holds 0.025. Use z = 1.96 or t at 2.5% in one tail.
Interpreting a 95% interval as a 95% probability that the parameter is inside this specific interval
It sounds natural, but the parameter is fixed, not random.
Fix: Think in terms of repeated sampling: 95% of such intervals would capture the parameter.
Using n instead of n − 1 for degrees of freedom
Rushing when reading the t table.
Fix: For one sample mean, df = n − 1. With n = 25, df = 24.
Mixing up bias and inefficiency
Both mean the estimator is imperfect.
Fix: Bias is about the centre of the sampling distribution; efficiency is about its spread.

Worked examples

Example 1

An analyst draws 36 monthly returns from a fund. The sample mean is 1.2% and the sample standard deviation is 3.0%. Assuming the population variance is unknown and using a z-value of 1.96 as an approximation for a large sample, what is the approximate 95% confidence interval for the mean monthly return? Options: A) −0.28% to 2.68%; B) −0.18% to 2.58%; C) 0.22% to 2.18%.

Show the solution

Point estimate: x̄ = 1.2%.
Standard error = s ÷ √n = 3.0 ÷ √36 = 3.0 ÷ 6 = 0.5%.
Margin of error = 1.96 × 0.5 = 0.98%.
Lower limit = 1.2 − 0.98 = 0.22%.
Upper limit = 1.2 + 0.98 = 2.18%.
Distractors: all three options are centred on 1.2%, so centring alone cannot separate them. Option A uses a margin of 1.48 and option B uses a margin of 1.38. Neither margin follows from the standard error calculation above, so only the margin of 0.98 in option C is correct.

Answer: C) 0.22% to 2.18%.

Example 2

A sample of 16 observations from a normal population has a mean of €50.00 and a standard deviation of €8.00. The population variance is unknown. The t-value for df = 15 at the 2.5% one-tail level is 2.131. What is the 95% confidence interval for the mean? Options: A) €33.00 to €67.00; B) €45.74 to €54.26; C) €46.00 to €54.00.

Show the solution

Unknown variance, so use t with df = 16 − 1 = 15.
Standard error = 8 ÷ √16 = 8 ÷ 4 = 2.00.
Margin of error = 2.131 × 2 = 4.262.
Lower limit = 50 − 4.262 = 45.738, about €45.74.
Upper limit = 50 + 4.262 = 54.262, about €54.26.
Option C would come from using a factor of 2.0 (50 ± 4). Option A wrongly skips the division by √n and uses the standard deviation as the standard error (2.131 × 8 = 17.05, giving roughly €33.00 to €67.00 after rounding).

Answer: B) €45.74 to €54.26.

Exam tips

Options are listed smallest to largest. Find the centre (x̄) first and eliminate any option that is not symmetric around it.
Look for the phrase 'population variance is unknown' or 'sample standard deviation'. It signals t.
Know the three z values (1.645, 1.96, 2.576) from memory. They save time when a question gives no table.
For property questions, match the keyword: 'expected value equals parameter' is unbiased, 'smallest variance' is efficient, 'improves with sample size' is consistent.
Expect conceptual items on width: higher confidence or smaller n gives a wider interval; t is wider than z.

Practice questions from Estimation and Hypothesis Testing

Point Estimates and Confidence Intervals: frequently asked questions

When should I use the t distribution instead of z?

Use t when the population variance is unknown and you estimate it with the sample standard deviation. Use z when σ is known. As n grows, t and z values get very close.

What are the properties of a good estimator?

A good estimator is unbiased, efficient and consistent. Unbiased means its expected value equals the parameter. Efficient means it has the smallest variance among unbiased estimators. Consistent means it gets closer to the parameter as the sample size increases.

How do I build a confidence interval for a population mean?

Take the sample mean, then add and subtract the reliability factor times the standard error. The standard error is the standard deviation divided by the square root of n. Choose z or t based on whether σ is known.

Why is a t-based interval wider than a z-based one?

The t distribution has fatter tails, which reflects extra uncertainty from estimating the variance. So its reliability factor is larger at the same confidence level. The gap shrinks as degrees of freedom rise.