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Estimation, Confidence Intervals and Tests using the Normal Distribution

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Summary — What This Chapter Covers
  • 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.
1. Statistics, Estimators & Bias

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, ... Xₙ taken independently from the population. Each Xᵢ is itself a random variable and follows the same distribution as the population variable X.
  • Statisticany function of the sample data that does NOT contain an unknown parameter. For example, the sample mean is 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 difference E(T) − θ.
This is why we divide by 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 .

The formulas you need

Unbiased mean:   x̄ = Σx / n

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."
Worked Example

The table shows the lowest temperature of the day, x (°C), recorded on 25 random days:

x3, 4, 5, 6, 7, 8
Frequency5, 4, 3, 5, 4, 4
Step 1: Find Σx and Σx²Σx = (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 meanx̄ = 136 / 25 = 5.44
Step 3: Unbiased variances² = (816 − 25×5.44²) / 24 = 238/75 = 3.17 (3 s.f.)
Practice Question 1

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.

If you're given a sample's mean x̄ and variance s² (with sample size n), you can always recover:

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.
Worked Example (continuing Example 1)

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 2Σy = 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 n25 + 20 = 45
New unbiased meanw̄ = 260.4 / 45 = 5.79
New unbiased variances² = (1667.098 − 45×5.79²) / 44 = 3.60
Practice Question 2

Sample A: n = 10, mean = 12, variance = 5. Sample B: n = 15, mean = 14, variance = 6. Find the combined unbiased mean and variance.

2. Standard Error of the Mean

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.

Standard error of the mean = σ / √n where σ is the population standard deviation (or its best available unbiased estimate) and n is the sample size.
If you have several samples with different variances, always use the estimate from the largest sample size — it's considered the most accurate/reliable estimate of σ².
Worked Example

Using the combined sample from before (n = 45, s² = 3.60), find the standard error of the mean.

Use the largest, most accurate variances² = 3.60
Standard error = √s² / √n√3.60 / √45 = 0.283 (3 s.f.)
Practice Question

A sample of size 64 has an unbiased variance estimate of 12.96. Find the standard error of the mean.

3. Confidence Intervals

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.

A 95% confidence interval does NOT mean "there's a 95% probability the true mean is in this specific interval." Once you've calculated it, the true mean either is or isn't in there — no probability about it anymore. The correct interpretation is: if you repeated this sampling process many times and built a new confidence interval each time, 95% of those intervals would contain the true mean. Examiners love catching students on this.

The formula

CI = x̄ ± z × (σ / √n) x̄ = sample mean, z = critical value from the standard normal table, σ = population std. dev., n = sample size.

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:

Width = 2 × z × (σ / √n)
Worked Example

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 μz = 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% CIWidth = 2 × 2.5758 × (4/√9) = 6.8688
(c) Smallest n so 99% CI width < 55 > 2 × 2.5758 × (4/√n)
√n > 20.6064/5 = 4.121
n > 16.98 → round UP → n = 17
For "minimum sample size" questions, always round UP — even if you get n > 16.02, you still need n = 17. A smaller sample wouldn't satisfy the inequality. This trips up a huge number of students who round to the nearest whole number instead of rounding up.
Practice Question

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.

4. The Central Limit Theorem

This is the quiet hero that makes everything else in this chapter legal. It says:

No matter what shape the original population's distribution is — skewed, uniform, weird, anything — as the sample size n gets larger, the distribution of the sample mean X̄ approaches a normal distribution.
If X₁, X₂, ... Xₙ is a random sample from a population with mean μ and variance σ², then:

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.

Exam questions frequently ask: "why is the Central Limit Theorem important in this test/interval?" The expected answer is almost always some version of: "it allows us to assume the sample mean is (approximately) normally distributed, which lets us use the normal distribution / z-tables to calculate the interval or test statistic." Learn this phrasing — it's basically a guaranteed mark if you know to say it.
Practice Question

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.

5. Hypothesis Testing for the Mean

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

  1. 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)
  2. Calculate the test statistic Z using the formula below.
  3. Find the critical value(s) from the z-table for your chosen significance level, and identify the critical region.
  4. 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.
Z = ( X̄ − μ ) / ( σ / √n ) X̄ = sample mean, μ = the value stated in H₀, σ = population standard deviation, n = sample size.
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
Worked Example

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: HypothesesH₀: μ = 250    H₁: μ < 250 (one-tailed — "less than" is the suspicion)
Step 2: Test statisticZ = (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.
ConclusionInsufficient evidence to reject H₀. There is not enough evidence to support Jason's claim that the volume is below 250mL.
Always write your conclusion in context and never say "H₀ is true" — you can only say there's "insufficient evidence to reject H₀" or "sufficient evidence to reject H₀ in favour of H₁." Hypothesis tests never prove anything; they only weigh evidence.
Practice Question

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.

What to Memorise
Unbiased mean
x̄ = Σx / n
Unbiased variance
s² = (Σx² − n·x̄²) / (n − 1)
Recovering Σx, Σx²
Σx = n·x̄  |  Σx² = s²(n−1) + n·x̄²
Standard error of the mean
σ / √n — decreases as n increases
Confidence interval
CI = x̄ ± z × (σ / √n)
CI width
2 × z × (σ / √n)
Common z-values
90% → 1.6449  |  95% → 1.96  |  99% → 2.5758
Central Limit Theorem
X̄ ~ N(μ, σ²/n) for large n, regardless of the population's original distribution
Test statistic for μ
Z = (X̄ − μ) / (σ / √n)
One-tailed vs two-tailed
"less than"/"greater than" → one-tailed. "different from"/"changed" → two-tailed
Concepts Checklist
Exam Tips & Common Mistakes

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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