Statistics for Business — Week 7

Sampling Distributions

How do we draw conclusions about entire populations when we can only measure a small sample? This lecture answers that question — rigorously.

Standard
Error

Central
Limit Theorem

Proportion
Distributions

Finite Pop.
Correction

1.1

1.1 Why Do We Need Sampling Distributions?

In practice, we cannot measure an entire population. Instead, we take a sample and compute a statistic (e.g., the sample mean $\bar{X}$) to estimate the parameter (e.g., population mean $\mu$).

The Problem: Different random samples from the same population give different sample means. How reliable is any one sample mean?

A sampling distribution describes how a sample statistic varies across all possible samples of the same size.

Each sample gives a different x̄; their distribution is the sampling distribution

1.2

1.2 Standard Error of the Mean

The standard error measures how much the sample mean $\bar{X}$ typically varies from sample to sample.

Standard Error of the Mean

$$\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$$

where $\sigma$ = population standard deviation, $n$ = sample size.

Key insight: Bigger samples → smaller standard error → more reliable estimates. The denominator $\sqrt{n}$ tells us improvement slows down as $n$ grows.

Effect of sample size

Doubling sample size halves the standard error only when going from n=1 to n=4

1.3

1.3 The Mean is Unbiased

Consider a tiny population — 4 components, with fault counts: A=5, B=3, C=6, D=2.

Population mean: $\mu = \frac{5+3+6+2}{4} = 4$

Take all samples of size $n=2$. There are $\binom{4}{2}=6$ possible samples:

Sample	Values	$\bar{x}$
A, B	5, 3	4.0
A, C	5, 6	5.5
A, D	5, 2	3.5
B, C	3, 6	4.5
B, D	3, 2	2.5
C, D	6, 2	4.0
Average of all $\bar{x}$		4.0

Conclusion: The average of all possible sample means equals the population mean $\mu$. This property is called unbiasedness.

$$E(\bar{X}) = \mu$$

Individual sample means scatter around $\mu$, but they do not systematically over- or under-estimate it.

Accounting context

If an auditor randomly samples invoices, the average error rate across all possible samples will equal the true population error rate — the audit process is unbiased.

Section 2

The Central Limit Theorem

2.1

2.1 The Central Limit Theorem (CLT)

Central Limit Theorem:
Regardless of the shape of the population distribution, the sampling distribution of $\bar{X}$ approaches a normal distribution as the sample size $n$ increases.

Sampling distribution of the mean

$$\bar{X} \sim N\!\left(\mu,\; \frac{\sigma^2}{n}\right)$$

Rule of thumb: $n \geq 30$ is usually sufficient (more if the population is highly skewed).

Why does this matter? It lets us compute probabilities about $\bar{X}$ even when we don't know the population's shape.

CLT in action: the distribution of x̄ normalises regardless of parent shape

2.2

2.2 Computing Probabilities for $\bar{X}$

Once we know $\bar{X}$ is normally distributed, we standardise to use the Z-table:

$$Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}}$$

Step-by-step process

1Identify $\mu$, $\sigma$, and $n$

2Compute $\sigma_{\bar{X}} = \sigma/\sqrt{n}$

3Convert $\bar{X}$ to $Z$

4Find area using Z-table

5Interpret in context

Rearranging for intervals:
To find the range that contains a fixed proportion of sample means:
$$\bar{X} = \mu \pm Z \cdot \frac{\sigma}{\sqrt{n}}$$ For 95%: use $Z = \pm 1.96$

Note: If the population is already normal, this works for any sample size. The CLT is only needed for non-normal populations.

2.3

2.3 Worked Example — Office Task Times

Example 7-1

Task completion times: $\mu = 8$ min, $\sigma = 2$ min. Random samples of $n = 40$ tasks.

(a) P(x̄ > 9 min)?

1Std error: $\sigma_{\bar{X}} = \frac{2}{\sqrt{40}} = 0.3162$

2Z-value: $Z = \frac{9-8}{0.3162} = 3.16$

3$P(\bar{X}>9) = P(Z>3.16) = 0.5 - 0.4992 = \mathbf{0.0008}$

There is only a 0.08% chance the sample mean exceeds 9 minutes — extremely unlikely!

(b) P(7.2 < x̄ < 8.5)?

1$Z_1 = \frac{7.2 - 8}{0.3162} = -2.53$

2$Z_2 = \frac{8.5 - 8}{0.3162} = +1.58$

3$P(-2.53 < Z < 1.58)$
$= 0.4943 + 0.4429 = \mathbf{0.9372}$

93.72% of all sample means (n=40) will fall between 7.2 and 8.5 minutes.

2.4

2.4 Effect of Sample Size — Example 7-1 (c) & (d)

(c) What if n = 20?

n = 20 is below the CLT threshold of 30. To proceed, we would need to assume the population is normal (not just approximately symmetric).

If the population is normal, the sample mean is still normally distributed for any n — CLT is not required in that case.

If population normality cannot be assumed and n < 30, we cannot reliably use the normal distribution for $\bar{X}$.

(d) Which is most likely? ($\mu=8, \sigma=2$)

Event	Z	Prob.
$\bar{x} < 7.5$, n=30	$\frac{7.5-8}{2/\sqrt{30}}=-1.37$	0.0853
$\bar{x} < 7.5$, n=50	$\frac{7.5-8}{2/\sqrt{50}}=-1.77$	0.0384
Individual $X < 2$	$\frac{2-8}{2}=-3.0$	0.0013

Answer

$\bar{x} < 7.5$ with n=30 is most likely (8.53%). As n increases, extreme sample means become less probable.

2.Q

Knowledge Check — Sampling Distribution of the Mean

Q1: A population has σ = 12. If you increase sample size from n=9 to n=36, what happens to the standard error?

σ/√9 = 12/3 = 4; σ/√36 = 12/6 = 2. Quadrupling n halves the standard error (because of the square root).

Q2: You sample n=25 items from a heavily right-skewed population. Can you use the CLT?

The n≥30 rule is a guideline. For heavily skewed populations, a larger sample may be needed before the sampling distribution becomes approximately normal.

Section 3

Sampling Distribution of the Proportion

3.1

3.1 Sampling Distribution of the Proportion

Sometimes we care about a proportion rather than a mean — e.g., "What fraction of accounts are overdue?"

Sample proportion: $$\hat{p} = \frac{\text{number with characteristic}}{n}$$ This estimates the population proportion $\pi$.

Standard Error of the Proportion

$$\sigma_{\hat{p}} = \sqrt{\frac{\pi(1-\pi)}{n}}$$

When $n\pi \geq 5$ and $n(1-\pi) \geq 5$, we can approximate with the normal distribution:

$$Z = \frac{\hat{p} - \pi}{\sqrt{\pi(1-\pi)/n}}$$

Visualising the condition

Both the number of "successes" and "failures" in the sample must be at least 5 for the normal approximation to be valid.

3.2

3.2 Worked Example — Child Care Survey

Example 7-2

40% of two-parent families use child care ($\pi = 0.40$). Random sample of $n=100$ families.

(a) P(0.40 ≤ p̂ ≤ 0.50)?

Check: $100(0.4)=40\geq5$ ✓

1$\sigma_{\hat{p}} = \sqrt{\frac{0.4 \times 0.6}{100}} = 0.049$

2$Z_1 = \frac{0.40-0.40}{0.049}=0$

3$Z_2 = \frac{0.50-0.40}{0.049}=2.04$

4$P = 0 + 0.4793 = \mathbf{0.4793}$

(b) P(p̂ > 0.45)?

1$Z = \frac{0.45-0.40}{0.049}=1.02$

2$P = 0.5 - 0.3461 = \mathbf{0.1539}$

About 15% of samples of 100 will show over 45% usage.

(c) 95% interval?

95% of sample proportions fall within $\pm1.96$ standard errors of $\pi$:

$$\pi \pm 1.96 \cdot \sigma_{\hat{p}}$$ $$= 0.40 \pm 1.96(0.049)$$

Between 30.4% and 49.6%

3.Q

Knowledge Check — Proportion

Q3: A survey finds that 10% of invoices contain errors (π = 0.10). You sample n = 40 invoices. Can you use the normal approximation?

Both conditions must hold. nπ = 4 < 5 fails the first check. You would need a larger sample (n ≥ 50) to use the normal approximation here.

Q4: For a proportion with π = 0.5, which sample size gives the LARGEST standard error?

σ = √(0.5×0.5/n) = 0.5/√n. The smallest n gives the largest standard error: 0.5/√25 = 0.10, vs 0.5/√100 = 0.05.

Section 4

Normal Approximations to Binomial & Poisson

4.1

4.1 Normal Approximation to the Binomial

The binomial distribution $X \sim B(n, \pi)$ can be approximated by the normal when:

$$n\pi \geq 5 \quad\text{and}\quad n(1-\pi) \geq 5$$

Then use: $\mu = n\pi$ and $\sigma = \sqrt{n\pi(1-\pi)}$

And standardise as: $Z = \dfrac{X_a - n\pi}{\sqrt{n\pi(1-\pi)}}$

$X_a$ is the continuity-corrected value — necessary because we are approximating a discrete distribution with a continuous one.

The Continuity Correction

To include the full bar for X≥3, use X_a = 2.5 in the normal approximation

4.2

4.2 Continuity Correction — Rules

When converting a discrete event to a continuous normal probability, shift by 0.5:

Discrete (Binomial)	Continuous (Normal) — use X_a	Reason
$P(X \geq k)$	$P(X \geq k - 0.5)$	Include the full bar at k
$P(X > k)$	$P(X > k + 0.5)$	Exclude the bar at k
$P(X \leq k)$	$P(X \leq k + 0.5)$	Include the full bar at k
$P(X < k)$	$P(X < k - 0.5)$	Exclude the bar at k
$P(X = k)$	$P(k - 0.5 \leq X \leq k + 0.5)$	Capture the whole bar width

Memory aid: The correction always moves the boundary by 0.5 towards the values you want to include. Inclusive inequality (≥, ≤) → extend outward by 0.5.

4.3

4.3 Worked Example — Salary Package Choice

Example 7-3

Staff choose equally among 3 packages (A, B, C). P(choose C) = 1/3. (a) n=6: exact. (b) n=20: approximate.

(a) n=6, P(X ≥ 3) — use Binomial exactly

Check: $n\pi = 6 \times \frac{1}{3} = 2 < 5$ → cannot use normal.

Use binomial formula $\binom{n}{x}\pi^x(1-\pi)^{n-x}$:

P(X=3) = C(6,3)(1/3)³(2/3)³ = 0.2195
P(X=4) = C(6,4)(1/3)⁴(2/3)² = 0.0823
P(X=5) = C(6,5)(1/3)⁵(2/3)¹ = 0.0165
P(X=6) = C(6,6)(1/3)⁶(2/3)⁰ = 0.0014
─────────────────────────────────────
P(X≥3) = 0.3196

(b) n=20, P(X ≥ 3) — normal approximation

Check: $n\pi = 20/3 = 6.67 \geq 5$ ✓

1$\mu = 20 \times \frac{1}{3} = 6.67$

2$\sigma = \sqrt{20 \times \frac{1}{3} \times \frac{2}{3}} = 2.108$

3Continuity correction: $X_a = 2.5$ (for $X \geq 3$)

4$Z = \frac{2.5 - 6.67}{2.108} = -1.98$

5$P(X \geq 3) = P(Z \geq -1.98) = 0.5 + 0.4761 = \mathbf{0.9761}$

4.4

4.4 Normal Approximation to the Poisson

For a Poisson distribution $X \sim \text{Pois}(\lambda)$, when $\lambda \geq 5$ we can approximate:

$$\mu = \lambda, \quad \sigma = \sqrt{\lambda}$$ $$Z = \frac{X_a - \lambda}{\sqrt{\lambda}}$$

Apply the same continuity correction rules as for the binomial.

The Poisson models counts of events in a fixed interval — e.g., customer arrivals per minute, defects per batch, claims per month.

Example 7-4 — Coffee Arrivals

Customers arrive at 5/min ($\lambda=5$). P(X ≤ 3) in any minute?

Exact Poisson:
$P(X \leq 3) = e^{-5}\left(1 + 5 + \frac{25}{2} + \frac{125}{6}\right) = 0.2650$

Normal approximation (since $\lambda = 5 \geq 5$):
$\mu=5,\; \sigma=\sqrt{5}=2.236$
Continuity correction: $X_a = 3.5$
$Z = \frac{3.5 - 5}{2.236} = -0.67$
$P(Z \leq -0.67) = 0.5 - 0.2486 = \mathbf{0.2514}$

4.Q

Knowledge Check — Normal Approximations

Q5: You want P(X ≥ 8) using the normal approximation to a binomial. What is the continuity-corrected value X_a?

For P(X ≥ 8), we want to include the full bar at X=8, so we shift the boundary to the left: X_a = 8 − 0.5 = 7.5.

Q6: A Poisson random variable has λ = 3. Should you use the normal approximation?

The standard rule is λ ≥ 5. With λ = 3 the Poisson distribution is still noticeably skewed and the normal approximation performs poorly.

Section 5

Finite Population Correction Factor

5.1

5.1 When the Population is Not "Infinite"

So far we assumed the population is very large relative to our sample. In practice, if we sample a significant fraction of the population, our sample is more informative than the formula suggests.

Apply the Finite Population Correction (FPC) when sampling without replacement and: $$\frac{n}{N} > 0.05$$ (sample is more than 5% of the population)

FPC Factor

$$\text{FPC} = \sqrt{\frac{N - n}{N - 1}}$$

Corrected Standard Error

$$\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} \times \sqrt{\frac{N-n}{N-1}}$$

Intuition

5.2

5.2 Worked Example — Smoking Breaks Audit

Example 7-5

Population: N=500 smokers. Smoking time: μ=25 min, σ=8 min (normally distributed). Sample n=50 without replacement. Find P(x̄ > 26).

1Check ratio: $\frac{n}{N} = \frac{50}{500} = 0.10 > 0.05$ → FPC required

2Compute FPC: $$\text{FPC} = \sqrt{\frac{500-50}{500-1}} = \sqrt{\frac{450}{499}} = 0.9506$$

3Corrected standard error: $$\sigma_{\bar{X}} = \frac{8}{\sqrt{50}} \times 0.9506 = 1.1314 \times 0.9506 = 1.0754$$

4Z-value: $$Z = \frac{26-25}{1.0754} = 0.93$$

5Probability: $$P(\bar{X}>26) = P(Z>0.93) = 0.5 - 0.3238 = \mathbf{0.1762}$$

17.62% of samples of 50 will have an average smoking time above 26 minutes.

Without FPC: σ = 8/√50 = 1.131, Z = 0.885, P = 0.188 — slightly overestimates uncertainty.

5.Q

Knowledge Check — Finite Population Correction

Q7: When sampling WITHOUT replacement, when must you apply the FPC?

The FPC corrects for the fact that sampling without replacement from a small population is more informative than sampling with replacement. The 5% rule is the standard threshold.

Q8: Applying the FPC always _______ the standard error compared to not applying it.

FPC = √((N−n)/(N−1)) is always less than 1 (when n > 1). Multiplying the standard error by a number less than 1 decreases it — reflecting that we have more information when sampling a large fraction of the population.

Summary

Key formulae and decision rules

6.1

6.1 Summary — Formulae and When to Use Them

Situation	Formula	Condition to check
Sample mean, large / normal population	$Z = \dfrac{\bar{X}-\mu}{\sigma/\sqrt{n}}$	Population normal, or $n \geq 30$ (CLT)
Sample mean, finite population	$Z = \dfrac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}\sqrt{\frac{N-n}{N-1}}}$	$n/N > 0.05$ and sampling without replacement
Sample proportion	$Z = \dfrac{\hat{p}-\pi}{\sqrt{\pi(1-\pi)/n}}$	$n\pi \geq 5$ and $n(1-\pi) \geq 5$
Binomial → Normal approx	$Z = \dfrac{X_a - n\pi}{\sqrt{n\pi(1-\pi)}}$	$n\pi \geq 5$ and $n(1-\pi) \geq 5$; apply continuity correction
Poisson → Normal approx	$Z = \dfrac{X_a - \lambda}{\sqrt{\lambda}}$	$\lambda \geq 5$; apply continuity correction

Decision order: (1) Is population normal or n≥30? → use Z for means. (2) Is n/N > 5%? → add FPC. (3) Proportion? → check $n\pi \geq 5$. (4) Discrete? → add continuity correction.

6.2

6.2 Brain Teasers — Think It Through

Q1: What is the probability distribution of $\bar{X}$ for a random sample of size 30?
→ Normal with mean $\mu$ and variance $\sigma^2/30$ (by CLT).

Q2: We studied the sampling distribution of the mean. Is there a sampling distribution of the variance too?
→ Yes! It follows a Chi-square distribution $\left(\chi^2\right)$ — not normal. This becomes important when constructing confidence intervals for $\sigma^2$.

Q3: Why is the Central Limit Theorem so important?
→ Because it lets us make inferences about populations whose distribution is unknown. Most real-world populations are not normally distributed — but their sample means are, when n is large enough.

Event	Z	Prob.
\(\bar{x} < 7.5\), n=30	\(\frac{7.5-8}{2/\sqrt{30}}=-1.37\)	0.0853
\(\bar{x} < 7.5\), n=50	\(\frac{7.5-8}{2/\sqrt{50}}=-1.77\)	0.0384
Individual \(X < 2\)	\(\frac{2-8}{2}=-3.0\)	0.0013

Discrete (Binomial)	Continuous (Normal) — use X_a	Reason
\(P(X \geq k)\)	\(P(X \geq k - 0.5)\)	Include the full bar at k
\(P(X > k)\)	\(P(X > k + 0.5)\)	Exclude the bar at k
\(P(X \leq k)\)	\(P(X \leq k + 0.5)\)	Include the full bar at k
\(P(X < k)\)	\(P(X < k - 0.5)\)	Exclude the bar at k
\(P(X = k)\)	\(P(k - 0.5 \leq X \leq k + 0.5)\)	Capture the whole bar width

Situation	Formula	Condition to check
Sample mean, large / normal population	\(Z = \dfrac{\bar{X}-\mu}{\sigma/\sqrt{n}}\)	Population normal, or \(n \geq 30\) (CLT)
Sample mean, finite population	\(Z = \dfrac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}\sqrt{\frac{N-n}{N-1}}}\)	\(n/N > 0.05\) and sampling without replacement
Sample proportion	\(Z = \dfrac{\hat{p}-\pi}{\sqrt{\pi(1-\pi)/n}}\)	\(n\pi \geq 5\) and \(n(1-\pi) \geq 5\)
Binomial → Normal approx	\(Z = \dfrac{X_a - n\pi}{\sqrt{n\pi(1-\pi)}}\)	\(n\pi \geq 5\) and \(n(1-\pi) \geq 5\); apply continuity correction
Poisson → Normal approx	\(Z = \dfrac{X_a - \lambda}{\sqrt{\lambda}}\)	\(\lambda \geq 5\); apply continuity correction

Contents — Week 7: Sampling Distributions