Methods — Sampling & Confidence Intervals: Concepts & Equations

40 simple concept questions covering sections 9A–9F: sample sums and means, the central limit theorem, proportions, confidence intervals, sample size and assessing claims. Equations include plain-English explanations. Study the separate Fundamentals set first.

What is the sample sum Sₙ?

The total of n observations: Sₙ = X₁ + X₂ + … + Xₙ. Its value changes when the sample changes.

What are the mean and spread of a sample sum?

For n independent observations: E(Sₙ) = nμ, SD(Sₙ) = σ√n, and Var(Sₙ) = nσ². Totals become larger and more spread out as n increases.

How do you find the probability distribution of a small sample sum?

List the possible combinations, add their values, and multiply probabilities within each independent combination. Add the probabilities of combinations that give the same total.

What are the mean and spread of the sample-mean distribution?

E(X̄) = μ and SD(X̄) = σ/√n, so Var(X̄) = σ²/n. Sample averages are centred on the true mean but vary less than individual observations.

Why do larger samples give more consistent averages?

High and low observations tend to balance out. The standard error σ/√n decreases, while the centre stays at μ. Four times as many observations halves the standard error.

How is a random sample mean X̄ different from an observed mean x̄?

X̄ represents the average before the sample is chosen, so it can take different values. x̄ is the specific number calculated after collecting your sample.

What does the central limit theorem say?

For sufficiently large independent samples, sample means are approximately normally distributed, even if the original population is not normal. Their mean is μ and their standard error is σ/√n.

When are sample means and sample sums exactly normal?

When the original population is normal, both are normally distributed for any sample size. Otherwise, sufficiently large samples allow a normal approximation.

Does the central limit theorem make the original data normal?

No. It describes the distribution of sample averages (and sums). The individual observations can still have a skewed or otherwise non-normal distribution.

How large must a sample be for the central limit theorem?

The notes say sufficiently large; there is no single size that always works. More skewed populations usually need larger samples. Increasing n generally improves the approximation.

Which spread do you use for a probability about one value, an average or a total?

One value: σ. Average of n: σ/√n. Total of n: σ√n. Their centres are μ, μ and nμ respectively. Choose the quantity the question actually asks about.

How do you calculate probabilities for a sample average or total?

Use the appropriate normal distribution, then find the required area. For an average: z = (x̄ − μ)/(σ/√n). For a total s: z = (s − nμ)/(σ√n). Normality must be justified.

What does 95% confidence mean?

If we repeatedly took random samples and built intervals the same way, about 95% of those intervals would contain the true population value. One particular interval either contains it or does not.

Does a 95% confidence interval contain 95% of individual observations?

No. It estimates a population mean or proportion. It is not a range designed to contain 95% of the individual data values.

What does the critical value z∗ do in a confidence interval?

It sets how far the interval extends on either side of the estimate, measured in standard errors. E = z∗ × standard error. Higher confidence needs a larger z∗.

Which critical values are commonly used?

For two-sided normal intervals: 90% uses z∗ ≈ 1.645, 95% uses 1.96, and 99% uses 2.576. At 95%, 95% of the standard normal curve lies between −1.96 and +1.96.

How do you find z∗ for another confidence level?

Write the confidence level as C, then α = 1 − C. Each tail has area α/2. Use inverse normal with left-hand area 1 − α/2, mean 0 and standard deviation 1.

How do you construct a confidence interval for a population mean?

Calculate E = z∗σ/√n, then use [x̄ − E, x̄ + E]. The sample mean is the centre; E allows for sampling uncertainty. At 95%, z∗ = 1.96.

When is the normal confidence interval for a mean suitable?

Use an independent random sample. The population should be normal, or n large enough for the central limit theorem. The formula x̄ ± z∗σ/√n uses the population standard deviation σ.

What if a large-sample mean question gives raw data but no population standard deviation?

Calculate the sample mean x̄ and sample standard deviation s. For the large-sample normal approximation, estimate σ with s and use x̄ ± z∗s/√n. This adds approximation; it is not an exact small-sample rule.

How does increasing the confidence level change an interval?

It makes the interval wider if the sample stays the same. You need a wider range to be more confident of capturing the true population value.

How does increasing sample size change an interval?

It makes the interval narrower when confidence and variability stay the same. Width is proportional to 1/√n, so halving the width requires four times the sample size.

How does greater population variability affect a mean interval?

A larger σ gives a wider interval at the same confidence and sample size. More variable data make the population mean harder to estimate precisely.

How do you recover the estimate and margin of error from interval endpoints?

For [L, U], estimate = (L + U)/2, margin of error E = (U − L)/2, and width w = U − L. These intervals are symmetric about the estimate.

How do you choose a sample size for estimating a population mean?

For a maximum margin E, use n ≥ (z∗σ/E)². If given full width w, use n ≥ (2z∗σ/w)². Round UP to a whole number so the precision requirement is met.

How do you use a confidence interval to assess a claimed mean or proportion?

If the claimed value is outside the interval, reject the claim at that confidence level. If it is inside, there is not enough evidence to reject it. Being inside does not prove the claim true.

How do you write a confidence-interval interpretation in context?

State the confidence level, population, quantity and endpoints. For example: “We are 95% confident that the mean mass of all items is between L and U grams.” For proportions, the endpoints can be percentages.

When does the number of successes have a binomial distribution?

When there are n independent trials, each with two outcomes and the same success probability p. “Success” simply means the characteristic being counted. Then X ~ B(n, p).

What are the mean and standard deviation of a binomial count?

E(X) = np and SD(X) = √[np(1 − p)]. These describe the number of successes, not the proportion.

Why can a sample proportion be treated as a sample mean?

Code each success as 1 and each failure as 0. Their average is the number of successes divided by n, which is p̂. This connects proportions to the central limit theorem.

What are the mean and standard error of sample proportions?

E(p̂) = p and SE(p̂) = √[p(1 − p)/n]. Sample proportions are centred on the population proportion and become less variable as n increases.

When can you use a normal approximation for sample proportions?

The rule in these notes is np ≥ 5 AND n(1 − p) ≥ 5. You need enough expected successes and failures; n alone is not enough, especially when p is close to 0 or 1.

How do you find a probability involving a sample proportion?

Use an approximately normal distribution with mean p and SD √[p(1 − p)/n]. Standardise with z = (proportion boundary − p)/√[p(1 − p)/n], then find the relevant area.

When should you use p, p̂ or p∗ in a proportion formula?

Use p when the true population proportion is given for a probability. Use observed p̂ to estimate the standard error for a confidence interval. Use a preliminary estimate p∗ when planning sample size.

How do you construct a confidence interval for a population proportion?

First find p̂ = x/n. Then E = z∗√[p̂(1 − p̂)/n] and the interval is [p̂ − E, p̂ + E]. At 95%, z∗ = 1.96.

What should you check before using a normal proportion interval?

Check random, independent sampling and enough successes and failures. Using the notes’ rule with the sample estimate: np̂ ≥ 5 and n(1 − p̂) ≥ 5. Very small counts make this approximation unreliable.

How do you choose a sample size for estimating a population proportion?

With preliminary estimate p∗, use n ≥ (z∗/E)²p∗(1 − p∗). For full width w, use n ≥ (2z∗/w)²p∗(1 − p∗). Convert percentages to decimals and round n UP.

Why use p∗ = 0.5 when planning a sample with no preliminary proportion?

It makes p∗(1 − p∗) as large as possible: 0.25. This gives the largest required sample size, so it is the conservative choice within this formula.

How should you compare two confidence intervals?

Compare their centres, widths and confidence levels. At the same confidence level, a narrower interval gives a more precise estimate. Sampling variation can shift the centres; overlap alone does not prove equal population values.

Does a larger sample remove bias?

No. More observations reduce random sampling variation, but a sample that systematically favours one group can still give a misleading estimate. A narrow interval does not fix poor sampling.