Estimation, Confidence Intervals and Tests using the Normal Distribution
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- Statistics & estimators: using sample data to estimate unknown population parameters (mean, variance), and what makes an estimator "unbiased."
- Combining samples: merging two samples' data to get a single, better estimate.
- Standard error of the mean: how "wobbly" your sample mean is as an estimate — and why bigger samples wobble less.
- Confidence intervals: building a range of plausible values for the true population mean, at a chosen confidence level (e.g. 95%, 99%).
- Central Limit Theorem (CLT): the reason all of this works — sample means become normally distributed even if the original population isn't.
- Hypothesis testing for the mean: using a sample to test whether a claim about the population mean is credible.
The apple problem
Imagine you want to know the true average weight of every apple in a massive orchard — millions of apples. You can't weigh them all. So instead, you grab a random sample of, say, 10 apples, weigh them, and use that to estimate the true average.
This is the entire chapter in one sentence: we use small, manageable samples to make honest, mathematically justified guesses about huge, unknowable populations.
Key vocabulary — read this slowly, it matters
- Population parameter — the real, true value for the whole population (e.g. the true mean weight μ, or true variance σ² of all apples). We almost never know this. It's the thing we're trying to estimate.
- Random sample — a set of observations
X₁, X₂, X₃, ... Xₙtaken independently from the population. EachXᵢis itself a random variable and follows the same distribution as the population variable X. - Statistic — any function of the sample data that does NOT contain an unknown parameter. For example, the sample mean
x̄is a statistic — you can calculate it purely from your data. But something like "x̄ − μ" is NOT a statistic, because μ is unknown. - Estimator — a statistic that is being used specifically to estimate a population parameter. The sample mean x̄ is an estimator for μ. Since a statistic is a random variable (it changes every time you take a new sample), it has its own probability distribution — this is called the sampling distribution.
- Unbiased estimator — an estimator T is unbiased for parameter θ if
E(T) = θ. In plain English: if you repeated the sampling process infinitely many times and averaged all your estimates together, you'd land exactly on the true value. No systematic over- or under-estimating. - Bias — when
E(T) ≠ θ, the estimator is biased, and the bias is the differenceE(T) − θ.
n − 1 (not n) when calculating sample variance as an estimate of population variance — dividing by n on its own gives a biased (slightly too small) estimate. Dividing by n − 1 corrects this and gives you the unbiased estimator, often written s².
The formulas you need
Unbiased variance: s² = ( Σx² − n·x̄² ) / (n − 1) n = sample size. Note the denominator is n − 1, NOT n — this is what makes it "unbiased."
The table shows the lowest temperature of the day, x (°C), recorded on 25 random days:
| x | 3, 4, 5, 6, 7, 8 |
| Frequency | 5, 4, 3, 5, 4, 4 |
| Step 1: Find Σx and Σx² | Σx = (3×5)+(4×4)+(5×3)+(6×5)+(7×4)+(8×4) = 136 Σx² = (3²×5)+(4²×4)+(5²×3)+(6²×5)+(7²×4)+(8²×4) = 816 |
| Step 2: Unbiased mean | x̄ = 136 / 25 = 5.44 |
| Step 3: Unbiased variance | s² = (816 − 25×5.44²) / 24 = 238/75 = 3.17 (3 s.f.) |
A random sample of 12 observations gives Σx = 96 and Σx² = 850. Find the unbiased estimates of the population mean and variance.
Combining two samples
Sometimes you're given summary statistics (mean & variance) for two separate samples and asked to combine them into one bigger, more reliable estimate. The trick: convert everything back into Σx and Σx² first, then merge, then recalculate.
Σx = n × x̄
Σx² = s²(n−1) + n×x̄²
This comes directly from rearranging the unbiased variance formula. Memorise this rearrangement — it's the key move in every "combine two samples" question.
A second, independent sample of 20 days has mean 6.22 and variance 4.07. Combine both samples to find the new unbiased mean and variance.
| Recover Σy, Σy² for sample 2 | Σy = 6.22 × 20 = 124.4 Σy² = 4.07×19 + 20×6.22² = 851.098 |
| Combine: Σw, Σw² (w = combined data) | Σw = Σx + Σy = 136 + 124.4 = 260.4 Σw² = Σx² + Σy² = 816 + 851.098 = 1667.098 |
| New combined n | 25 + 20 = 45 |
| New unbiased mean | w̄ = 260.4 / 45 = 5.79 |
| New unbiased variance | s² = (1667.098 − 45×5.79²) / 44 = 3.60 |
Sample A: n = 10, mean = 12, variance = 5. Sample B: n = 15, mean = 14, variance = 6. Find the combined unbiased mean and variance.
Every time you take a new sample, you'll get a slightly different sample mean. The standard error of the mean measures how much these sample means typically vary — i.e., how much "wobble" there is in your estimate of μ.
The key intuition: a bigger sample gives you a more stable, trustworthy estimate. Averaging 1000 apples is far less likely to be thrown off by one weirdly heavy apple than averaging just 3. So as n increases, the standard error decreases — your estimator becomes more precise.
Using the combined sample from before (n = 45, s² = 3.60), find the standard error of the mean.
| Use the largest, most accurate variance | s² = 3.60 |
| Standard error = √s² / √n | √3.60 / √45 = 0.283 (3 s.f.) |
A sample of size 64 has an unbiased variance estimate of 12.96. Find the standard error of the mean.
What a confidence interval actually means
A confidence interval is a range of values that we believe, with a stated level of confidence (e.g. 95%), contains the true population mean μ. It's built by taking your sample mean and adding/subtracting a "margin of error" on either side.
The formula
The value of z depends on how confident you want to be:
- 90% confidence → z = 1.6449
- 95% confidence → z = 1.96
- 99% confidence → z = 2.5758
The width of the interval (upper limit minus lower limit) is simply:
A population is normally distributed with variance 16. A random sample of size 9 has a mean of 140.
| (a) Find the 95% CI for μ | z = 1.96, σ = √16 = 4, n = 9 Upper: 140 + 1.96×(4/√9) = 142.6 Lower: 140 − 1.96×(4/√9) = 137.4 CI = (137.4, 142.6) |
| (b) Find the width of the 99% CI | Width = 2 × 2.5758 × (4/√9) = 6.8688 |
| (c) Smallest n so 99% CI width < 5 | 5 > 2 × 2.5758 × (4/√n) √n > 20.6064/5 = 4.121 n > 16.98 → round UP → n = 17 |
A normal population has standard deviation 6. A sample of size 25 has mean 52. (a) Find the 90% confidence interval for μ. (b) Find the minimum sample size needed for a 95% CI with width less than 4.
This is the quiet hero that makes everything else in this chapter legal. It says:
n gets larger, the distribution of the sample mean X̄ approaches a normal distribution.
X̄ ~ N( μ, σ²/n ) (approximately, for large n)
Why does this matter so much? Every confidence interval and every hypothesis test in this chapter relies on X̄ being normally distributed, because that's the only way we're allowed to use the standard normal z-tables. The CLT is the theoretical justification that lets us do that — even when we don't know or can't assume the original population itself is normal.
A random variable X has an unknown, non-normal distribution with mean 40 and variance 100. A sample of size 50 is taken. Explain why it is still valid to use X̄ ~ N(40, 100/50) to construct a confidence interval.
The core idea
Hypothesis testing answers the question: "is my sample result surprising enough, given what's claimed to be true, that I should doubt the claim?" You compare a claimed value of μ against what your sample actually shows.
The 4-step process, every time
- State your hypotheses.
- H₀ (null hypothesis): μ = [claimed value] — this is what we assume true unless the evidence says otherwise.
- H₁ (alternative hypothesis): depends on the wording —
μ < value→ one-tailed test (suspect the value is LOWER)μ > value→ one-tailed test (suspect the value is HIGHER)μ ≠ value→ two-tailed test (suspect it's just DIFFERENT, direction unknown)
- Calculate the test statistic Z using the formula below.
- Find the critical value(s) from the z-table for your chosen significance level, and identify the critical region.
- Compare and conclude — if Z falls inside the critical region, reject H₀ (there IS evidence for the claim). If not, you don't have enough evidence to reject H₀.
| Significance level | One-tailed critical z | Two-tailed critical z |
| 10% | ±1.2816 | ±1.6449 |
| 5% | ±1.6449 | ±1.96 |
| 1% | ±2.3263 | ±2.5758 |
A factory's energy drink volume is normally distributed with standard deviation 5mL. They claim the mean volume is 250mL. Jason suspects it's actually less. He samples 16 bottles and finds a mean of 248.2mL. Test his claim at the 5% significance level.
| Step 1: Hypotheses | H₀: μ = 250 H₁: μ < 250 (one-tailed — "less than" is the suspicion) |
| Step 2: Test statistic | Z = (248.2 − 250) / (5/√16) = −1.8 / 1.25 = −1.44 |
| Step 3: Critical region (5%, one-tailed, lower) | Z < −1.6449 (using the more precise table value) |
| Step 4: Compare | −1.44 is NOT less than −1.6449, so it's not in the critical region. |
| Conclusion | Insufficient evidence to reject H₀. There is not enough evidence to support Jason's claim that the volume is below 250mL. |
A machine fills bags of sugar with a mean of 500g and a known standard deviation of 8g. The manager suspects the machine is now overfilling. A sample of 25 bags has a mean of 503.5g. Test at the 1% significance level whether there is evidence of overfilling.
Dividing by n instead of n − 1
The single most common slip in this whole topic. Unbiased variance ALWAYS divides by (n − 1), never n. Double check this every single time before you submit an answer.
Rounding sample sizes the wrong way
When solving for "minimum n," you must round UP even if the decimal is tiny (e.g. n > 16.02 still needs n = 17). Rounding to the nearest whole number will cost you the mark.
Misreading one-tailed vs two-tailed wording
Words like "less than," "lower," "decreased," "overestimate" → one-tailed. Words like "different," "changed," "not equal to," "inaccurate" (without direction) → two-tailed. Getting this wrong flips your critical value and can flip your entire conclusion.
Misinterpreting a confidence interval
Never say "there's a 95% chance μ is in this interval." The correct wording is about the long-run proportion of intervals (built this way) that would contain μ. Examiners specifically test this misconception.
Overstating hypothesis test conclusions
Never write "H₀ is true" or "we have proven the claim." Always use language like "insufficient/sufficient evidence to reject H₀" — hypothesis tests deal in evidence, not proof.
Forgetting to explain the CLT when asked
"State the importance of the CLT" questions want the specific phrase: it allows us to assume the sample mean is approximately normally distributed, enabling use of the normal/z-distribution — even when the original population isn't normal.
Using the wrong sample's variance
When multiple variance estimates are available (e.g. from different sample sizes), always use the one from the largest sample — it's the most reliable.
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