BUSINESS STATISTICS›UNIT 06›SAMPLING THEORY II

Unit 06 — Sampling Theory II

Standardizing the sample mean, importance of sample size, concept of diminishing returns in sampling, sampling distributions for the statistic proportion of success — the concept of standard error, standardizing the proportion of success, importance of sample size.

SESSIONS: 2 TEXTBOOK: LEVIN §6.7, pp. 302–307 & §7.8, pp. 351–356 EXAM WEIGHT: Q3 (b) + HALF OF Q5 (b) — 7 MARKS Full PPT: Proportion 13–23 Sample Size deck 43 Class notes 6–8

PRACTISE THIS UNIT

Every question shape in this unit is drilled on Unit 06 — How to Solve Anything: three topics, each with a "how to recognise this topic" card, a master method, and every exercise Levin sets solved with its page number. Topic 3 is Q3 (b) — 5 marks, both the mean and the proportion version.

◆ What the papers ask from this unit

Q3 (b), 5 marks, both papers — sample size determination, and the two versions are the two halves of the topic:

Final: σ = 32 mg/dL, 99% confidence, E = 5 → sample size for a mean. Answer n = 272 Key p. 3.
Re-Exam: 95% confidence, E = 0.04, no prior estimate of p → sample size for a proportion, using p = 0.5. Answer n = 601 Key p. 3.

Learn both formulas and the p = 0.5 convention and this slot is free marks. Both solved →

⚠ A syllabus / deck boundary you should know about

The deck for this material is titled "Unit 6 (Part II) — Determination of Sample Size" slide 1, but the syllabus lists "Determination of Sample Size" under Unit 7 and gives Unit 6 only "Importance of Sample size" Syllabus p. 2. It is taught as Unit 6, so it is covered here. If your mid-term is strictly Units 1–6 by the syllabus, treat Concept 5 as a bonus that costs you ten minutes and might be worth 5 marks.

1. Standardizing the Sample Mean Core syllabus concept

1The formula, and the one thing that changes from Unit 4

Individual Observation

\[ z = \frac{x - \mu}{\sigma} \]

Sample Mean

\[ z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}} = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \]

The whole difference is the denominator: σ/√n is always smaller than σ, so sample means cluster more tightly around the centre.

Your class notes write it with the subscripts in place, twice Class notes p. 2, p. 4.

2Worked example — the class's interval problem

μ = 12, σ = 2, n = 10. Find P(14 ≤ x̄ ≤ 16). Class notes p. 2

Step 1 — Standard error σ_x̄ = σ/√n = 2/√10 = 2/3.162 = 0.632
Step 2 — Standardise both ends z₁ = (14 − 12)/0.632 = 3.16 z₂ = (16 − 12)/0.632 = 6.32
Step 3 — Read the table P(3.16 ≤ Z ≤ 6.32) = A(6.32) − A(3.16) ≈ 0.5000 − 0.4992 = 0.0008
Both z's are far out in the right tail, so the probability is tiny. Note what the small sample did: with σ = 2 an individual value of 14 is only 1 SD above the mean and unremarkable, but a mean of ten equal to 14 is 3.16 standard errors out and very nearly impossible. That contrast is the point of the whole unit.

Sources

Primary Class notes — the z formula for x̄ and the worked interval pp. 2–3

Textbook Levin §6.6 Sampling distributions in more detail pp. 289–301

2. The Sampling Distribution of a Proportion Core syllabus concept

1Understand the Concept

The deck's framing slide 13: the sampling distribution of a proportion is "used when the issue of interest is categorical in nature — classified as occurrences or non-occurrences. We estimate the proportion of occurrences… since complete information about the population is not available, we use the sample proportion to estimate the true proportion."

Your class notes compress it usefully Class notes p. 6: "Like binomial, only 2 options — M/F, Pass/Fail, etc. CLT applies here. If N is given, the σ_x̄ ka change applies here [too]."

The logic underneath: let x = the number of occurrences in n trials. Then x follows a binomial distribution with probability p. If n is large, that binomial can be approximated by a normal distribution slide 15 — which is why you can put a z on a proportion at all.

Standard error of proportion

\[ \sigma_{\hat{p}} = \sqrt{\frac{p q}{n}} \qquad (\mu_{\hat{p}} = p, \; q = 1-p) \]

Standardising proportion

\[ z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} = \frac{\hat{p} - p}{\sigma_{\hat{p}}} \]

Deck slides 14–15; class notes with σ_p̂ circled and labelled "standard error of proportion", and the z formula written out Class notes p. 7.

⚠ Check the normal approximation before you use it

The deck's solution does this as an explicit Step 1 slide 18: compute n·p and n·(1−p) and confirm both are above 5. Only then may you "use the binomial approximation to normal distribution". Write that check out — the official Q4 (b) key does the same thing with np₀ = 150 and n(1−p₀) = 450 Key p. 4. It is a free line of method marks.

2Worked example — the deck's customer-gender problem

A company's customers are 43% women and 57% men. In a sample of 50 customers, how likely is it that 46% or more are women? slides 17–18

Step 1 — Check the sample is large enough n·p = 50 × 0.43 = 21.5 n·(1−p) = 50 × 0.57 = 28.5
Both above 5, so the normal approximation is valid.
Step 2 — Standard error σ_p̂ = √[p(1−p)/n] = √[0.43 × 0.57 / 50] = √0.004902 = 0.07
Step 3 — Standardise z = (p̂ − p)/σ_p̂ = (0.46 − 0.43)/0.07 = 0.43
Step 4 — Read the table P(Z ≥ 0.43) = 0.5 − A(0.43) = 0.5 − 0.1664 = 0.3336 = 33.36%
Deck's answer slide 18.

3Worked example — your class's SVKM problem

The proportion of female students is p = 0.40. In a sample of n = 30, what is P(p̂ > 0.5)? Class notes pp. 6–8

Step 1 — Identify p and q, and state μ_p̂ p = 0.40 (proportion of interest — female) q = 0.60 μ_p̂ = p = 0.40
Your notes label q explicitly as "male wala (not interest)" — always say in words which category you called a success p. 7.
Step 2 — Standard error of the proportion σ_p̂ = √(pq/n) = √(0.40 × 0.60 / 30) = √0.008 = 0.089
Step 3 — Draw both curves, then standardise
Your notes draw the p̂ curve centred at 0.4 with 0.5 marked, then the z curve with 1.12 marked p. 7.
z = (0.5 − 0.4)/0.089 = 1.12
Step 4 — Read the table P(Z > 1.12) = 0.5 − A(1.12) = 0.5 − 0.3686 = 0.1314 = 13.14%
Class notes p. 8

ENSampling distributions of the sample proportion (CLT) Paul Miller ENCentral limit theorem for a sample proportion Katie Ann Jager

Practice H1–H4 — sampling distribution of a proportion →

Sources

Primary Sampling Distribution deck — proportion, binomial link, worked example slides 13–18

Class SVKM problem worked in full, with both curves drawn pp. 6–8

Textbook Levin §6.6 — sampling distribution of the proportion pp. 286–301

3. Diminishing Returns in Sampling Core syllabus concept

1Understand the Concept

Your class notes pose it as a question and then draw the answer Class notes p. 2: "For 50 crore people's average height, why only use 1K samples?" Beneath it is a curve of error E against sample size n, falling steeply and then flattening, with the annotation "yaha pe 1K samples — now this will stay same, i.e. parallel".

The mathematics behind that curve:

Standard error & error bound

\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \qquad E = z \cdot \frac{\sigma}{\sqrt{n}} \]

Diminishing returns rule

\[ \text{Error } \propto \frac{1}{\sqrt{n}} \implies n \times 4 \implies \text{Error} \div 2; \quad n \times 9 \implies \text{Error} \div 3 \]

So to halve your error you must quadruple the cost of the survey. Past a point the extra precision is not worth buying — which is exactly why a national poll of 1.4 billion people uses a sample of a few thousand and stops. The deck's own opening example: the Federal Reserve sampled 2,700 of 14,117 institutions for a 95% confidence interval with ±5% accuracy slide 2.

The exam sentence: "Because the standard error falls with √n rather than n, each additional observation buys less precision than the one before. Beyond a certain sample size the reduction in error no longer justifies the additional cost — this is the law of diminishing returns in sampling."

2Precision and confidence — the trade-off

The deck defines both, and then names the tension between them slides 22–24:

	Precision	Confidence
Definition	The closeness of our estimate to the true population characteristic.	The certainty that our estimate really holds true for the population.
Measured by	The standard error of the sampling distribution of the mean.	The confidence level — 90%, 95%, 99%.
Improve it by	Narrowing the interval (smaller E).	Widening the interval (larger z).

The deck's conclusion: "Other things being equal, the narrower the range, the lower the confidence. Hence a trade-off between precision and confidence is a must" slide 23.

Your class notes make the same point with a concrete pair Class notes pp. 1–2:
μ = 5'6″ ± 10″ = (4'8″ to 6'4″) → 95% confidence
μ = 5'6″ ± 3′ = (2'6″ to 8'6″) → 99% confidence
"If CL increases, z also increases, which causes MOE to increase as well." The second interval is almost certainly right and almost completely useless. That is the trade-off in one example.

The margin-of-error formula your notes open with, and which drives all of this:

Margin of Error (M.E.)

\[ \text{M.E.} = E = z \cdot \frac{\sigma}{\sqrt{n}} \qquad \mu = \bar{x} \pm E \]

Class notes p. 1, with "3σ/√n" circled and labelled M.E.

ENMargin of error & sample size for a confidence interval MarinStatsLectures ENHow sample size and margin of error relate statisticsfun

Sources

Primary Sample Size deck — precision, confidence, the trade-off, diminishing returns slides 22–30

Class Margin of error, the two confidence bands, the E-vs-n curve pp. 1–2

Textbook Levin §6.7 Sample size and standard error pp. 302–307

4. What Actually Decides Your Sample Size Supporting

1The three factors

The deck names exactly three slide 31, and they are the three symbols in the formula:

Factor	Deck's wording	Symbol	Effect on n
Variance (heterogeneity)	"A heterogeneous population has more variance (a larger standard deviation) which will require a larger sample. A homogeneous population has less variance which permits a smaller sample."	σ	σ ↑ → n ↑ (as σ²)
Magnitude of error	"How precise must the estimate be?"	E	E ↓ → n ↑ (as 1/E²)
Confidence level	"How much error will be tolerated?"	z	z ↑ → n ↑ (as z²)

Population size is not on the list. That surprises people, and it is worth a sentence in an answer: for large populations, n depends on σ, E and z — not on N. N only enters through the finite population correction when the sample is a large share of the population.

2Roscoe's rules of thumb

Useful when a question gives no σ at all and asks for judgement slide 43:

Sample sizes larger than 30 and less than 500 are appropriate for most research.
Where samples are broken into subsamples (males/females, juniors/seniors), a minimum of 30 per category is necessary.
In multivariate research (including multiple regression), the sample should be several times — preferably ten times or more — as large as the number of variables.
For simple experimental research with tight controls (matched pairs), successful research is possible with samples as small as 10 to 20.

3Sequential sampling

The chicken-and-egg problem: the formula needs σ, but you only learn σ by sampling. The deck's fix slide 32: sequential sampling — "conducting a pilot study to estimate the population parameters so that another, larger sample of the appropriate sample size may be drawn." If a question gives you no σ and no pilot, say you would run one.

Sources

Primary Sample Size deck — factors of concern, sequential sampling, Roscoe's thumb rules slides 31–43

5. Determining the Sample Size Q3 (b), both papers

1The two formulas — derive them once, then just use them

Both come from setting the margin of error equal to the tolerance E and solving for n.

Sample size for a Mean

\[ n = \left(\frac{z \cdot \sigma}{E}\right)^2 \]

Sample size for a Proportion

\[ n = \frac{z^2 p q}{E^2} \qquad \text{(Use } p = 0.5 \text{ if no prior estimate)} \]

Deck: means slide 32, proportions slides 36–37.

⚠ Always round UP

n is a minimum. Rounding 600.25 down to 600 gives a margin of error slightly larger than the one you promised. Both official keys round up — 271.9 → 272 Key p. 3, 600.25 → 601 Key p. 3. The deck does the same, twice: 600.25 → 601 and 450.x → 451 slides 38–39.

2When no prior estimate of p is available — the p = 0.5 rule

The proportion formula needs p, but p is what you are trying to find out. The deck's answer slide 38: "Because there is no estimate of the proportion given, we use p = 0.50 for a conservative estimate."

The reason it is conservative: p·q = p(1−p) is largest when p = 0.5, where it equals 0.25. Any other p gives a smaller product and hence a smaller n. Choosing 0.5 therefore guarantees your sample is big enough whatever the true p turns out to be. The Re-Exam key calls this "maximum variability" Key p. 3.

3Worked example — sample size for a mean (the Final's Q3 b)

σ = 32 mg/dL. How large a sample to be 99% confident that the sample mean cholesterol is within E = 5 mg/dL?

Step 1 — Pull out the three inputs σ = 32 E = 5 confidence 99% ⟹ z = 2.576
Step 2 — Substitute n = (z·σ / E)² = (2.576 × 32 / 5)² = (82.432 / 5)² = (16.4864)²
Step 3 — Evaluate and round UP n = 271.8 ≈ 272 people
The official key writes 271.9 → 272 Key p. 3.

4Worked example — sample size for a proportion (the Re-Exam's Q3 b)

95% confidence, margin of error ±0.04, no prior estimate of p.

Step 1 — Inputs, and the p = 0.5 decision z = 1.96 E = 0.04 no prior estimate ⟹ p = q = 0.5
Write the reason down: "no prior estimate is available, so p = 0.5 is used for maximum variability." That sentence is worth a mark on its own.
Step 2 — Substitute n = z²·p·q / E² = (1.96)²(0.5)(0.5) / (0.04)² = 3.8416 × 0.25 / 0.0016 = 0.9604 / 0.0016
Step 3 — Evaluate and round UP n = 600.25 ≈ 601 consumers
Step 4 — Interpret
"To estimate the proportion of consumers who prefer online shopping with 95% confidence and a margin of error of ±0.04, the marketing firm should select a sample of at least 601 people" Key p. 3.

5What a prior estimate buys you

The deck runs the same 95%/E=0.04 problem twice to show the difference slides 38–39:

Prior estimate of p	p·q	n required
None → use p = 0.50	0.2500	601
p = 0.25 known	0.1875	451
p = 0.05 known (deck's retail example)	0.0475	73

The deck's business framing slide 41: a retailer estimating its website conversion rate at 95% confidence and 5% margin of error needs 385 customers if it knows nothing, but only 73 if previous studies suggest a 5% conversion rate. Knowing something in advance is worth 312 surveys.

⚠ Common mistakes in this slot

Using the mean formula for a proportion. If the question is about a percentage or proportion of people, use z²pq/E². If it is about an average of a measured quantity, use (zσ/E)².
Mixing units for E. "±4%" means E = 0.04 in the proportion formula, not 4.
Grabbing the wrong z. 99% is 2.576, not 1.96. The Final's Q3 (b) is 99%.
Forgetting to square. Both formulas square the whole bracket or the z.
Rounding down. Always up.

ENCalculating sample size given the confidence level & margin of error The Organic Chemistry Tutor ENSample size n to estimate a population mean Joshua Emmanuel ENSample size from a given margin of error — proportions Jennifer Ward

Practice I1–I6 — sample size for means and proportions →

Sources

Primary Sample Size deck — both formulas, the p = 0.5 rule, four worked examples, the business problem slides 32–42

Textbook Levin §7.8 Determining the sample size in estimation pp. 351–356

Exam Final Q3 (b) p. 2 · key p. 3 · Re-Exam Q3 (b) p. 2 · key p. 3