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BUSINESS STATISTICS›PROBLEM SOLVING›UNIT 06

Unit 06 — How to Solve Anything

Three topics, all built on the one substitution from Unit 5. Topic 3 is Q3 (b) — 5 marks in both papers, and the most mechanical question on the whole paper once you know which of the two formulas to reach for.

TOPICS: 3 TEXTBOOK: LEVIN §6.5, pp. 302–311 & §7.8, pp. 351–359 EXAM WEIGHT: Q3 (b) — 5 MARKS  ·  HALF OF Q4 (a) PAIRS WITH: Unit 06 notes
◆ The three formulas this unit is made of
Finite Population Multiplier
\[ \text{FPM} = \sqrt{\frac{N - n}{N - 1}} \qquad \text{(Apply when } n/N > 0.05 \text{ without replacement)} \]
Sampling Proportion & SE
\[ \hat{p} = \frac{x}{n}, \qquad \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}, \qquad z = \frac{\hat{p} - p}{\sigma_{\hat{p}}} \]
Sample Size Formulas
\[ n = \left(\frac{z \cdot \sigma}{E}\right)^2 \quad (\text{Mean}), \qquad n = \frac{z^2 \cdot p \cdot q}{E^2} \quad (\text{Proportion, } p=0.5 \text{ if unknown}) \]

Q3 (b) is formula 3, every time. The Final asked it for a mean; the Re-Exam asked it for a proportion with no prior estimate. Both worked here →

Topic 1 of 3 · Levin §6.5 · pp. 306–308 · 7 textbook + 5 exam-shaped

The Finite Population Correction

One extra factor, applied in one specific circumstance. When your sample is a large fraction of a finite population, \(\sigma/\sqrt{n}\) overstates the uncertainty — you already know a lot about the population — and the correction shrinks it. The examinable skill is knowing when to apply it, not how.

How to know a question belongs to this topic
The single tell
The question gives you \(N\) — a population size — as well as \(n\). Unit 5 questions never do. If \(N\) appears, check \(n/N\).
The rule
Compute \(n/N\). If it exceeds 0.05 (5%), apply the correction. Below that, the factor is so close to 1 that it is conventionally ignored — and Levin's own exercises are built to sit on either side of that line.
Give-away words
a population of 75 items N = 80 a chain of 145 supermarkets sampled without replacement 30 airliners, 6 checked
"Without replacement" is the other trigger
Sampling with replacement makes the population effectively infinite — no correction. Sampling without replacement from a finite population is what creates the need for it. Levin 6-36 asks for both, side by side.
Which direction does it move things?
The multiplier \(\sqrt{(N-n)/(N-1)}\) is always less than 1, so it reduces the standard error and increases precision. If your corrected standard error came out larger, you have inverted the fraction.
Don't confuse it with
If the question…then…
gives no \(N\) at allUnit 5 — plain \(\sigma/\sqrt{n}\)
gives \(N\) but \(n/N \le 0.05\)Unit 5 — say you checked and it is negligible
gives \(N\) and \(n/N > 0.05\)Here
asks how large \(n\) should beTopic 3 — sample size
Master Method — Topic 1
Formula bank
Standard error with
the correction
\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\times\sqrt{\frac{N-n}{N-1}} \]
The finite population
multiplier itself
\[ \text{fpm} = \sqrt{\frac{N-n}{N-1}} \qquad (\le 1 \text{ always}) \]
When to use it
\[ \frac{n}{N} > 0.05 \quad\text{and sampling is \textbf{without replacement}} \]
\(N\) population size
\(n\) sample size
\(\sigma\) population standard deviation
\(\sigma_{\bar{x}}\) standard error of the mean
The routine
  1. Write \(N\), \(n\), \(\mu\), \(\sigma\). If a variance was given, root it now.
  2. Compute \(n/N\) and compare with 0.05. Write the comparison down — it is a mark, and it justifies everything that follows.
  3. Compute \(\sigma/\sqrt{n}\) first, on its own.
  4. Compute the multiplier \(\sqrt{(N-n)/(N-1)}\) separately — note it is \(N-1\) on the bottom, not \(N\).
  5. Multiply, then proceed exactly as in Unit 5: standardise, sketch, read, combine.
Traps
  • Using \(N\) instead of \(N-1\) in the denominator. Small effect, easy mark lost.
  • Forgetting the square root over the whole fraction.
  • Applying it when \(n/N\) is tiny. Harmless numerically but it shows you did not check.
  • Not applying it when \(n/N\) is large. This overstates the standard error, sometimes badly — at \(n/N = 0.43\) the correction cuts it by nearly a quarter.
  • Applying it to sampling with replacement. Then the population is effectively infinite and no correction belongs.
Textbook bank §6.5, pp. 306–308 · the finite-population exercises, solved
7 exercises Levin §6.5 The relationship between sample size and standard error pp. 302–308 The sample-size exercises from this section are in Topic 3.
6-36With and without replacementLevin Ex. 6-36 · p. 306

Q. From a population of 75 items with mean 364 and variance 18, 32 items are selected at random without replacement. (a) Find the standard error of the mean. (b) Find \(P(363<\bar{x}<366)\). (c) What would (a) be if sampling were with replacement?

Why: the perfect first question — it asks for the corrected and uncorrected standard error side by side, so the size of the correction is visible.
Why we do itWhat to write
Variance → SD, then test the 5% rule.
\[ \sigma=\sqrt{18}=4.2426 \] \[ \frac{n}{N}=\frac{32}{75}=0.427 > 0.05 \;\Rightarrow\; \text{correction required} \]
(a) Uncorrected part first, then the multiplier.
\[ \frac{\sigma}{\sqrt{n}}=\frac{4.2426}{\sqrt{32}}=0.7500 \] \[ \sqrt{\frac{75-32}{74}}=\sqrt{0.5811}=0.7623 \] \[ \sigma_{\bar{x}}=0.7500\times0.7623=\mathbf{0.5717} \]
(b) Standardise with the corrected value. 363 is below the mean, 366 above ⇒ add.
\[ z_1=\frac{363-364}{0.5717}=-1.75 \qquad z_2=\frac{366-364}{0.5717}=3.50 \] \[ P=0.4599+0.4998=\mathbf{0.9597} \]
(c) With replacement the population behaves as infinite — drop the multiplier.
\[ \sigma_{\bar{x}}=\frac{4.2426}{\sqrt{32}}=\mathbf{0.7500} \] The correction cut the standard error by 24% — sampling 43% of the population buys real precision.
6-37Corrected intervalLevin Ex. 6-37 · p. 306

Q. \(N = 80\), mean 22, standard deviation 3.2. What is the probability that a sample of 25 has a mean between 21 and 23.5?

Check the rule.
\[ \frac{25}{80}=0.3125>0.05 \;\Rightarrow\; \text{correct} \]
Standard error.
\[ \sigma_{\bar{x}}=\frac{3.2}{5}\times\sqrt{\frac{55}{79}}=0.64\times0.8344=\mathbf{0.5340} \]
Opposite sides of the mean ⇒ add.
\[ z_1=\frac{21-22}{0.5340}=-1.87 \qquad z_2=\frac{23.5-22}{0.5340}=2.81 \] \[ P=0.4693+0.4975=\mathbf{0.9668} \]
6-38Three sample sizesLevin Ex. 6-38 · p. 306

Q. For \(N = 80\), mean 8.2, standard deviation 2.1, find the standard error for (a) \(n=16\), (b) \(n=25\), (c) \(n=49\).

Why: three corrections on one population — it shows how the multiplier bites harder as \(n\) grows toward \(N\).
nn/Nσ/√n√((N−n)/(N−1))σx̄
160.2000.52500.90010.4725
250.3130.42000.83440.3504
490.6130.30000.62640.1879
The pattern: at \(n=16\) the correction shaves 10% off; at \(n=49\) it removes 37%. As \(n\) approaches \(N\), the multiplier approaches 0 — and at \(n=N\) the standard error is exactly zero, because you have measured everyone and \(\bar{x}=\mu\) with certainty.
6-40Total in disguiseLevin Ex. 6-40 · p. 307

Q. Wrecks generate an average of ₹2,25,000 revenue with standard deviation ₹39,000. If exploration expenses of ₹21,00,000 are not recouped from the first nine wrecks, the financier cancels. What is the probability that exploration continues past nine wrecks?

Why: the threshold is on a total (₹21 lakh from nine wrecks) but the parameters are per wreck. Convert the total into a mean per wreck and the question becomes an ordinary \(\bar{x}\) problem.
Why we do itWhat to write
Turn the total threshold into a per-wreck mean — this is the whole trick.
\[ \bar{x}_{\text{needed}}=\frac{21{,}00{,}000}{9}=₹2{,}33{,}333 \]
Standard error. No \(N\) is given, so no correction.
\[ \sigma_{\bar{x}}=\frac{39{,}000}{\sqrt{9}}=₹13{,}000 \]
Standardise.
\[ z=\frac{2{,}33{,}333-2{,}25{,}000}{13{,}000}=0.64 \]
Upper tail — exploration continues if the mean is high enough.
\[ P=0.5-A(0.64)=0.5-0.2389=\mathbf{0.2611} \] Only about a 26% chance the exploration continues past nine wrecks.
6-42Correction + decisionLevin Ex. 6-42 · p. 307

Q. A de-icer is installed on 30 airliners; ice removed is normal with mean 96% and standard deviation 7. The FAA spot-checks six aircraft and approves if at least 98% is removed on average. What is the probability of approval?

\(N=30\) is the whole fleet with the system, \(n=6\) — check the rule.
\[ \frac{6}{30}=0.20>0.05 \;\Rightarrow\; \text{correct} \]
Standard error.
\[ \sigma_{\bar{x}}=\frac{7}{\sqrt{6}}\times\sqrt{\frac{24}{29}}=2.8577\times0.9097=\mathbf{2.5997} \]
Upper tail at 98.
\[ z=\frac{98-96}{2.5997}=0.77 \] \[ P=0.5-A(0.77)=0.5-0.2794=\mathbf{0.2206} \] Only a 22% chance of approval — the system's mean of 96% is simply below the 98% the FAA wants, and no amount of luck in the spot-check reliably fixes that.
6-43Symmetric intervalLevin Ex. 6-43 · p. 307

Q. A chain of 145 supermarkets; 36 stores' records are examined. Store profits are normal with standard deviation $1,200. What is the probability that the sample mean falls within $200 of the true mean?

Check the rule — 145 stores is a genuinely finite population.
\[ \frac{36}{145}=0.248>0.05 \;\Rightarrow\; \text{correct} \]
Standard error.
\[ \sigma_{\bar{x}}=\frac{1200}{6}\times\sqrt{\frac{109}{144}}=200\times0.8700=\mathbf{174.01} \]
"Within $200" is symmetric about \(\mu\) ⇒ \(2A(z)\).
\[ z=\pm\frac{200}{174.01}=\pm1.15 \] \[ P=2A(1.15)=2(0.3749)=\mathbf{0.7498} \] About a 75% chance the sample mean lands within $200 of the true average store profit.
Without the correction the standard error would be 200, giving \(z=\pm1.00\) and \(P=0.6826\) — nearly 7 percentage points lower. Sampling a quarter of the chain buys real extra confidence, and omitting the correction throws it away.
SC 6-7Correction + narrow bandLevin SC 6-7 · p. 308

Q. \(N = 125\), \(\mu = 105\), \(\sigma = 17\), \(n = 64\). Find (a) the standard error, (b) \(P(107.5<\bar{x}<109)\).

(a) \(n/N = 0.512\) — a very large fraction, so the correction matters a lot.
\[ \sigma_{\bar{x}}=\frac{17}{8}\times\sqrt{\frac{61}{124}}=2.125\times0.7014=\mathbf{1.4904} \]
(b) Both limits are above the mean ⇒ same side ⇒ subtract.
\[ z_1=\frac{107.5-105}{1.4904}=1.68 \qquad z_2=\frac{109-105}{1.4904}=2.68 \] \[ P=A(2.68)-A(1.68)=0.4963-0.4535=\mathbf{0.0428} \]
Matches Levin's printed answers (1.4904 and 0.0428) exactly. Note the subtraction — a band entirely on one side of the mean is the case people most often get wrong.
Five more, shaped like Q4 (a) the paper's Q4 (a) needs exactly this correction inside a confidence interval
Always write the \(n/N\) comparison as a line of working.
X1Apply or not?

Q. For each, say whether the finite population correction is needed: (a) \(N=5{,}000\), \(n=100\); (b) \(N=200\), \(n=40\); (c) \(N=1{,}000\), \(n=45\); (d) \(N=60\), \(n=30\).

Solution
Why: one division each, compared with 0.05.
Compute \(n/N\).
(a) \(100/5000=0.020\)   (b) \(40/200=0.200\)   (c) \(45/1000=0.045\)   (d) \(30/60=0.500\)
Compare with 0.05.
(a) No (2%)  ·  (b) Yes (20%)  ·  (c) No (4.5%, just under)  ·  (d) Yes (50%, and the multiplier is \(\sqrt{30/59}=0.713\) — a 29% reduction)
X2Corrected standard error

Q. A company has 500 employees; 60 are sampled. Salaries have \(\sigma = ₹8{,}000\). Find the standard error with and without the correction, and state which to report.

Solution
Check the rule.
\[ \frac{60}{500}=0.12>0.05 \;\Rightarrow\; \text{correct} \]
Both versions.
Uncorrected: \(\dfrac{8000}{\sqrt{60}}=₹1032.80\)
Multiplier: \(\sqrt{\dfrac{440}{499}}=0.9390\)
Corrected: \(1032.80\times0.9390=\mathbf{₹969.82}\)

Report the corrected figure — sampling 12% of a finite workforce genuinely reduces uncertainty, and the uncorrected value overstates the error by about 6%.
X3Probability with correction

Q. A college has 400 students; 100 are sampled. Study hours per week have \(\mu = 14\), \(\sigma = 5\). Find \(P(\bar{x} > 15)\).

Solution
\(n/N = 0.25 > 0.05\).
\[ \sigma_{\bar{x}}=\frac{5}{10}\times\sqrt{\frac{300}{399}}=0.5\times0.8671=\mathbf{0.4336} \]
Upper tail.
\[ z=\frac{15-14}{0.4336}=2.31 \] \[ P=0.5-A(2.31)=0.5-0.4896=\mathbf{0.0104} \]
X4Direction check

Q. A student computes a corrected standard error larger than the uncorrected one. What has gone wrong, and what is the largest value the multiplier can take?

Solution
AnswerThey have inverted the fraction — written \(\sqrt{(N-1)/(N-n)}\) instead of \(\sqrt{(N-n)/(N-1)}\).

Since \(n \ge 1\), the numerator \(N-n\) is always less than or equal to \(N-1\), so the multiplier is always \(\le 1\) and the corrected standard error can never exceed the uncorrected one. Its maximum is 1, reached only when \(n = 1\); its minimum is 0, at \(n = N\) — a census, where there is no sampling error at all.
Use this as a check on every answer: if the correction made the standard error bigger, the fraction is upside down.
X5Q4 (a) half

Q. A hospital has 250 beds; occupancy is recorded for 50 days. Daily occupancy has \(\sigma = 12\) beds. Compute the standard error of the mean, and explain in one sentence why the correction applies here.

Solution
Why: a trap — read carefully what \(N\) actually is.
Identify the population being sampled. It is days, not beds.
The 250 beds are not the population — the sampling unit is a day, and days are effectively unlimited.
So no correction applies.
\[ \sigma_{\bar{x}}=\frac{12}{\sqrt{50}}=\mathbf{1.697 \text{ beds}} \] No correction. The finite population multiplier applies only when the sampling frame itself is finite and heavily sampled. Here 50 days is a sample from an unbounded stream of days, so \(n/N \approx 0\).
Topic 2 of 3 · deck slides 13–18 · Levin §7.4, pp. 337–338 · 12 questions

The Sampling Distribution of a Proportion

Everything from Unit 5, with \(p\) in place of \(\mu\). The only new thing to learn is where the standard error comes from: for a proportion you do not need a separate \(\sigma\), because \(p\) determines the spread by itself — \(\sigma_{\hat{p}} = \sqrt{pq/n}\).

How to know a question belongs to this topic
The single tell
The variable is a percentage or a fraction of a group — "40% of subscribers are aware", "the proportion defective", "the percentage in favour". There is no \(\sigma\) anywhere in the question, and none is needed.
Give-away words
proportion percentage of customers who… p̂ what fraction of the sample 40% of subscribers are aware in favour / defective / aware
The whole translation
Mean: \(\mu \to \bar{x}\), spread \(\sigma/\sqrt{n}\).
Proportion: \(p \to \hat{p}\), spread \(\sqrt{pq/n}\).
Same normal curve, same table, same five readings. Only the standard error formula changes.
Why no σ is given
A proportion comes from a binomial, and for a binomial the spread is fixed by \(p\) alone (\(\sigma=\sqrt{npq}\)). Divide by \(n\) to turn a count into a proportion and you get \(\sqrt{pq/n}\). That is why \(p\) is the only parameter you need.
Don't confuse it with
If the question…then…
gives \(\mu\) and \(\sigma\) of a measured quantityUnit 5 — \(\sigma/\sqrt{n}\)
gives a percentage and \(n\)Here — \(\sqrt{pq/n}\)
asks for the number of successes, not the fractionUnit 4 — binomial, or its normal approximation
asks how large \(n\) must be for a given precisionTopic 3 — sample size for a proportion
Master Method — Topic 2
Formula bank
Mean of the sampling
distribution of p̂
\[ \mu_{\hat{p}} = p \]
Standard error
of the proportion
\[ \sigma_{\hat{p}} = \sqrt{\frac{pq}{n}} \qquad q = 1-p \]
Standardising
\[ z = \frac{\hat{p}-p}{\sqrt{pq/n}} \]
Estimated standard error
(p unknown)
\[ \hat{\sigma}_{\hat{p}} = \sqrt{\frac{\hat{p}\hat{q}}{n}} \qquad\text{use the sample's own proportion} \]
With a finite
population
\[ \sigma_{\hat{p}} = \sqrt{\frac{pq}{n}}\times\sqrt{\frac{N-n}{N-1}} \]

The maximum-spread fact, and it matters in Topic 3: \(pq\) is largest when \(p = 0.5\), where \(pq = 0.25\). So a proportion near 50% is the hardest to estimate, and \(p=0.5\) is the safe assumption when you know nothing.

The routine
  1. Write \(p\), \(q = 1-p\), \(n\). Convert any percentage to a decimal first.
  2. Check normality: \(np \ge 5\) and \(nq \ge 5\) — the same condition as the normal approximation to the binomial, because that is exactly what this is.
  3. Compute \(\sigma_{\hat{p}} = \sqrt{pq/n}\) on its own line.
  4. Standardise, sketch, read, combine — identical to Units 4 and 5.
  5. Answer as a percentage if the question was posed in percentages.
Traps
  • Using \(\sqrt{pq}/n\) instead of \(\sqrt{pq/n}\). The \(n\) is inside the root.
  • Forgetting to convert percentages. \(p = 40\) instead of 0.40 makes \(pq\) negative.
  • Mixing counts and proportions. "How many of the 200" needs a count; \(\hat{p}\) is a fraction.
  • Skipping the \(np\ge5\), \(nq\ge5\) check — with a very small \(p\) the normal curve is a poor fit and you should use the binomial directly.
Twelve questions Levin sets the proportion's exercises in Ch. 7 (out of mid-term scope), so these are built to the deck's treatment
12 questions Formulas from Levin §7.4 pp. 337–338 · taught from the deck slides 13–18
Write \(\sqrt{pq/n}\) as your first line every time.
Q1Standard error

Q. 40% of a population supports a policy. Find the standard error of the sample proportion for \(n = 100\), and check that the normal approximation is valid.

Solution
Validity check first.
\(np = 40 \ge 5\) ✓   \(nq = 60 \ge 5\) ✓ — normal approximation valid.
Standard error.
\[ \sigma_{\hat{p}}=\sqrt{\frac{(0.40)(0.60)}{100}}=\sqrt{0.0024}=\mathbf{0.0490} \] i.e. about 4.9 percentage points.
Q2Effect of n

Q. With \(p = 0.60\), find \(\sigma_{\hat{p}}\) for \(n = 200\). How much larger a sample would halve it?

Solution
Substitute.
\[ \sigma_{\hat{p}}=\sqrt{\frac{(0.6)(0.4)}{200}}=\sqrt{0.0012}=\mathbf{0.0346} \]
\(n\) enters through \(\sqrt{n}\), so quadruple it.
\(n = 800\) gives \(\sqrt{0.24/800}=0.0173\) — exactly half. Quadrupling \(n\) halves the standard error, the same diminishing return as for the mean.
Q3Upper tail

Q. 30% of customers use online banking. In a random sample of 150, what is the probability that more than 35% do?

Solution
Check and compute the standard error.
\(np=45\) ✓ \(nq=105\) ✓ \[ \sigma_{\hat{p}}=\sqrt{\frac{(0.30)(0.70)}{150}}=\sqrt{0.0014}=0.0374 \]
Standardise and read the upper tail.
\[ z=\frac{0.35-0.30}{0.0374}=1.34 \] \[ P=0.5-A(1.34)=0.5-0.4099=\mathbf{0.0901} \] About a 9% chance.
Q4Symmetric interval

Q. 20% of items in a large batch are seconds. For a sample of 400, find the probability that the sample proportion lies between 17% and 23%.

Solution
Standard error.
\[ \sigma_{\hat{p}}=\sqrt{\frac{(0.20)(0.80)}{400}}=\sqrt{0.0004}=0.0200 \]
The interval is symmetric about \(p\) ⇒ one \(z\), doubled area.
\[ z=\pm\frac{0.03}{0.02}=\pm1.50 \] \[ P=2A(1.50)=2(0.4332)=\mathbf{0.8664} \]
Q5Maximum spread

Q. For a fixed \(n = 100\), compare \(\sigma_{\hat{p}}\) at \(p = 0.10\), \(p = 0.50\) and \(p = 0.90\). What do you conclude?

Solution
Three substitutions.
\(p=0.10\): \(\sqrt{0.09/100}=\mathbf{0.0300}\)
\(p=0.50\): \(\sqrt{0.25/100}=\mathbf{0.0500}\)
\(p=0.90\): \(\sqrt{0.09/100}=\mathbf{0.0300}\)
Read the pattern.
The standard error is largest at \(p = 0.5\) and symmetric about it — 0.10 and 0.90 give the same value. So a proportion near half is the hardest to pin down, and one near 0 or 1 is easiest. This is why \(p = 0.5\) is the conservative assumption in sample-size work — it guarantees a large enough sample whatever the truth turns out to be.
Q6Lower tail

Q. A machine produces 8% defectives. In a sample of 300, find the probability that fewer than 5% are defective.

Solution
Check and compute.
\(np=24\) ✓ \(nq=276\) ✓ \[ \sigma_{\hat{p}}=\sqrt{\frac{(0.08)(0.92)}{300}}=\sqrt{0.0002453}=0.01566 \]
Lower tail.
\[ z=\frac{0.05-0.08}{0.01566}=-1.92 \] \[ P=0.5-A(1.92)=0.5-0.4726=\mathbf{0.0274} \]
Q7Judge a claim

Q. A firm claims 75% customer satisfaction. A survey of 250 customers finds 68% satisfied. Is the claim credible?

Solution
Standard error under the claim.
\[ \sigma_{\hat{p}}=\sqrt{\frac{(0.75)(0.25)}{250}}=\sqrt{0.00075}=0.0274 \]
Measure the gap in standard errors.
\[ z=\frac{0.68-0.75}{0.0274}=-2.56 \] \[ P(\hat{p}\le0.68)=0.5-A(2.56)=0.5-0.4948=\mathbf{0.0052} \]
Judge it.
If the 75% claim were true, a survey result this low would occur about once in 200 surveys. That is strong evidence against the claim — the true satisfaction rate is probably below 75%.
Q8Count vs proportion

Q. 25% of 400 sampled voters are expected to favour a candidate. Find (a) the probability that more than 110 favour them, (b) the probability that more than 27.5% do. Comment.

Solution
Why: (a) is a count and (b) is a proportion — but \(110/400 = 0.275\), so they are the same event asked two ways.
(a) As a count — binomial, normal approximation with the continuity correction.
\[ \mu=100,\ \sigma=\sqrt{400(0.25)(0.75)}=8.660 \] \[ z=\frac{110.5-100}{8.660}=1.21 \;\Rightarrow\; P=0.5-0.3869=\mathbf{0.1131} \]
(b) As a proportion — no continuity correction.
\[ \sigma_{\hat{p}}=\sqrt{\frac{(0.25)(0.75)}{400}}=0.02165 \] \[ z=\frac{0.275-0.25}{0.02165}=1.15 \;\Rightarrow\; P=0.5-0.3749=\mathbf{0.1251} \]
Comment on the small difference.
They describe the same event and differ only because the count version applies the ±0.5 continuity correction and the proportion version does not. Both are acceptable; the count version is slightly more accurate. Note \(\sigma_{\hat{p}} = \sigma/n = 8.660/400\) — the two standard errors are the same quantity in different units.
Q9Finite population

Q. A club has 500 members, 60% of whom are expected to renew. 100 are surveyed. Find the standard error of the sample proportion.

Solution
Check \(n/N\).
\[ \frac{100}{500}=0.20>0.05 \;\Rightarrow\; \text{correction applies} \]
Both factors.
\[ \sqrt{\frac{(0.6)(0.4)}{100}}=0.04899 \qquad \sqrt{\frac{400}{499}}=0.8953 \] \[ \sigma_{\hat{p}}=0.04899\times0.8953=\mathbf{0.0439} \]
Q10p unknown

Q. A sample of 500 households finds 165 own two cars. Estimate the population proportion and its standard error.

Solution
Why: \(p\) is unknown, so use the sample's own \(\hat{p}\) in the formula — this is the estimated standard error.
Point estimate.
\[ \hat{p}=\frac{165}{500}=\mathbf{0.33} \]
Estimated standard error using \(\hat{p}\hat{q}\).
\[ \hat{\sigma}_{\hat{p}}=\sqrt{\frac{(0.33)(0.67)}{500}}=\sqrt{0.0004422}=\mathbf{0.0210} \] So about 33% ± 2.1 percentage points per standard error.
Q11Validity

Q. A rare defect occurs in 1% of items. A sample of 200 is taken. Can you use the normal curve to find \(P(\hat{p}>0.02)\)? What should you do?

Solution
Run the check.
\[ np = 200(0.01) = 2 \;\;\mathbf{< 5} \quad ✗ \] \(nq = 198\) passes, but both are required — the normal approximation is not valid.
Name the right tool.
Treat it as a count: \(\hat{p}>0.02\) means more than 4 defectives out of 200. Since \(n\ge20\) and \(p\le0.05\), use the Poisson approximation with \(\lambda = np = 2\): \[ P(x\ge5)=1-P(x\le4)=1-0.9473=\mathbf{0.0527} \] The normal route would have given 0.0778 — badly wrong, because the distribution is strongly right-skewed at \(p=0.01\).
Q12Full comparison

Q. Write the mean and standard error of the sampling distribution for (a) a sample mean and (b) a sample proportion, and state the one condition each needs before the normal curve may be used.

Solution
Answer (a) Sample mean: \(\mu_{\bar{x}} = \mu\),   \(\sigma_{\bar{x}} = \sigma/\sqrt{n}\).
Condition: the population is normal, or \(n \ge 30\) so the Central Limit Theorem applies.

(b) Sample proportion: \(\mu_{\hat{p}} = p\),   \(\sigma_{\hat{p}} = \sqrt{pq/n}\).
Condition: \(np \ge 5\) and \(nq \ge 5\).

Both get multiplied by \(\sqrt{(N-n)/(N-1)}\) when sampling more than 5% of a finite population, and both are then standardised with the same \(z = (\text{statistic} - \text{parameter})/\text{standard error}\) and read from the same table.
Topic 3 of 3 · Levin §6.5 & §7.8 · pp. 307, 356–357 · Q3 (b) — 5 marks in both papers

Determining the Sample Size

The most mechanical 5 marks on the paper. Two formulas, one decision between them, and a rounding rule. The Final asked it for a mean; the Re-Exam asked it for a proportion with no prior estimate — so learn both, and learn to tell them apart in one read.

How to know a question belongs to this topic
The single tell
The question asks "how large a sample…" and gives you a confidence level and a tolerable error. The unknown is \(n\); everything else is supplied.
Give-away words
how large a sample is needed what size sample within 50 points within 3% of 95 percent confident margin of error to guarantee this accuracy
Which of the two formulas?
Is a \(\sigma\) given? → it is a MEAN question → \(n = (z\sigma/E)^2\).
Is the answer a percentage or proportion? → it is a PROPORTION question → \(n = z^2pq/E^2\).
A σ in the question is the single cleanest discriminator. Proportion questions never give one.
"No strong feelings about the proportion"
Use \(p = q = 0.5\). It maximises \(pq\) at 0.25, so it gives the largest — safest — sample size. The Re-Exam's Q3 (b) is exactly this case, and saying why you chose 0.5 is worth a mark.
Always round UP
\(n = 384.16\) becomes 385, never 384. Rounding down breaks the precision you just promised. Write "so \(n \ge 385\)".
Watch for the width trap
If the question gives the width of an interval, \(E\) is half of it. "A confidence interval of width 8" means \(E = 4\).
Master Method — Topic 3
Formula bank
Sample size
for a MEAN
\[ n = \left(\frac{z\,\sigma}{E}\right)^{2} \]
Sample size
for a PROPORTION
\[ n = \frac{z^{2}\,p\,q}{E^{2}} \qquad\text{use } p=q=0.5 \text{ if unknown} \]
With a finite
population
\[ n = \frac{n_0}{1+\dfrac{n_0-1}{N}} \qquad n_0 = \text{the uncorrected size} \]
σ from a range
(when σ is not given)
\[ \sigma \approx \frac{\text{range}}{6} \qquad\text{(the range spans about } \pm3\sigma) \]
\(E\) tolerable error — the "within …" figure
\(z\) from the confidence level
\(\sigma\) population standard deviation
\(p,q\) the proportion and its complement
The z values — know these cold
ConfidenceA(z) to look upzCommon use
90%0.45001.64 (or 1.645)lenient
95%0.47501.96the default
96%0.48002.05occasionally set
98%0.49002.33strict
99%0.49502.58 (or 2.576)very strict

To find \(z\) for any confidence level \(C\): look up \(A(z) = C/2\) in the body of the table and read off \(z\). That works for the odd ones like 96% that nobody memorises.

The routine — five lines
  1. Decide: mean or proportion? Look for a \(\sigma\). Write which one you are doing.
  2. Write \(z\) from the confidence level, and say where it came from.
  3. Write \(E\) — and halve it first if the question gave an interval width.
  4. Substitute into the matching formula, showing the arithmetic.
  5. Round UP and state the answer as "a sample of at least \(n\)".
How the exam asks it
PaperThe question
FinalQ3 (b), 5 marks — for a mean: σ = 32 mg/dL, 99% confidence, E = 5 mg/dL QP p. 2 · worked here
Re-ExamQ3 (b), 5 marks — for a proportion with no prior estimate, so \(p = 0.5\) QP p. 2 · worked here
Traps
  • Rounding down. Always up, always.
  • Using the interval width as \(E\). \(E\) is the half-width.
  • Forgetting to square. The whole bracket is squared.
  • Using the mean formula on a proportion question because you invented a σ. If the answer is a percentage, use \(z^2pq/E^2\).
  • Not justifying \(p = 0.5\). Say "no prior estimate, so take \(p = 0.5\), which maximises \(pq\) and gives the most conservative sample size."
  • Mixing percentage and decimal in \(E\). "Within 3%" means \(E = 0.03\), not 3.
Textbook bank — every Levin sample-size exercise §6.5 and §7.8 · all 16, solved
16 exercises Levin §6.5 p. 307 · §7.8 Determining the sample size in estimation pp. 351–357
7-45MeanLevin Ex. 7-45 · p. 356

Q. Population standard deviation 78. Find the sample size needed to estimate the true mean within 50 points at 95% confidence.

Why we do itWhat to write
A σ is given ⇒ mean formula. Read \(z\) and \(E\) off the question.
\(\sigma=78\), \(E=50\), 95% ⇒ \(z=1.96\)
Substitute and round up.
\[ n=\left(\frac{1.96\times78}{50}\right)^2=(3.0576)^2=9.35 \] \[ n=\mathbf{10} \]
7-46Proportion, p knownLevin Ex. 7-46 · p. 356

Q. The proportion is thought to be around 0.7. Find the sample size to estimate it within 0.02 at 90% confidence.

A proportion is given ⇒ proportion formula. No σ appears anywhere.
\(p=0.7,\ q=0.3,\ E=0.02\), 90% ⇒ \(z=1.64\)
Substitute and round up.
\[ n=\frac{(1.64)^2(0.7)(0.3)}{(0.02)^2}=\frac{2.6896\times0.21}{0.0004}=1412.04 \] \[ n=\mathbf{1413} \]
Why the prior estimate matters: had you assumed \(p=0.5\), you would get \(n = 1681\) — 268 extra interviews. Knowing that \(p\) is near 0.7 saves real money, because \(pq = 0.21\) is well below the maximum 0.25.
7-47Mean, 99%Levin Ex. 7-47 · p. 356

Q. Standard deviation 8.6. What size sample estimates the mean within 0.5 with 99% confidence?

Mean formula; 99% ⇒ \(A(z)=0.495\) ⇒ \(z=2.58\).
\(\sigma=8.6,\ E=0.5,\ z=2.58\)
Substitute and round up.
\[ n=\left(\frac{2.58\times8.6}{0.5}\right)^2=(44.376)^2=1969.23 \] \[ n=\mathbf{1970} \] Using \(z=2.576\) gives 1963.13 ⇒ 1964. Either is fine if you state your \(z\).
The cost of precision: demanding \(E=0.5\) on data with \(\sigma=8.6\) — a tolerance one-seventeenth of the spread — needs nearly 2,000 observations. Halving \(E\) again would need four times as many. This is the diminishing-returns law seen from the other end.
7-48Three values of pLevin Ex. 7-48 · p. 356

Q. Find the sample size to estimate a proportion within 0.05 at 95% confidence, assuming no strong feelings about the proportion. How does it change if about 75% favour the proposal? If only about 25% do?

Why: the perfect demonstration of why \(p = 0.5\) is the safe default — and of the symmetry of \(pq\) about 0.5.
Why we do itWhat to write
No information ⇒ \(p = 0.5\), which maximises \(pq\).
\[ n=\frac{(1.96)^2(0.5)(0.5)}{(0.05)^2}=\frac{0.9604}{0.0025}=384.16 \;\Rightarrow\; \mathbf{385} \]
\(p = 0.75\) ⇒ \(pq = 0.1875\), smaller than 0.25.
\[ n=\frac{3.8416\times0.1875}{0.0025}=288.12 \;\Rightarrow\; \mathbf{289} \]
\(p = 0.25\) ⇒ \(pq = 0.1875\) — identical, because \(pq\) is symmetric about 0.5.
\[ n=288.12 \;\Rightarrow\; \mathbf{289} \] Exactly the same as \(p=0.75\).
The conclusion the question is driving at.
Knowing \(p\) is away from 0.5 reduces the required sample — here by 96 respondents, a 25% saving. And \(p=0.5\) always gives the largest \(n\), which is why it is the safe assumption when you know nothing: you can never be undersampled by using it.
7-49σ from a rangeLevin Ex. 7-49 · p. 356

Q. Life-spans differ by no more than 550 weeks between the upper and lower limits. At 98% confidence, how large a sample finds the average life-span within 30 weeks?

Why: no \(\sigma\) is given — only a range. Since a normal distribution spans about \(\pm3\sigma\), the range covers roughly \(6\sigma\), so \(\sigma \approx \text{range}/6\). That estimate is the whole first half of the question.
Why we do itWhat to write
Estimate σ from the range using the empirical rule.
\[ \sigma \approx \frac{\text{range}}{6}=\frac{550}{6}=91.67 \text{ weeks} \]
98% ⇒ \(A(z)=0.49\) ⇒ \(z=2.33\).
\(z=2.33,\ E=30\)
Substitute and round up.
\[ n=\left(\frac{2.33\times91.67}{30}\right)^2=(7.1197)^2=50.69 \] \[ n=\mathbf{51} \]
State the assumption. The \(\text{range}/6\) rule assumes the life-spans are roughly normal. Say so — if the distribution were heavily skewed the estimate of σ would be poor and \(n\) would be wrong.
7-50MeanLevin Ex. 7-50 · p. 356

Q. Garbage-bag breaking weight has standard deviation 1.2 lb. How many bags must be tested to be 95% confident the sample average is within 0.5 lb of the true average?

Mean formula, \(z=1.96\), \(E=0.5\).
\[ n=\left(\frac{1.96\times1.2}{0.5}\right)^2=(4.704)^2=22.13 \] \[ n=\mathbf{23 \text{ bags}} \]
7-51"Regardless of the true percentage"Levin Ex. 7-51 · p. 356

Q. The university needs to be 90% confident that the percentage favouring a tuition rise has been estimated within 2% of the true value. How large a sample guarantees this accuracy regardless of the true percentage?

Why: "regardless of the true percentage" is the instruction to use \(p = 0.5\) — it is the only value that guarantees the accuracy whatever \(p\) turns out to be.
Justify \(p = 0.5\) explicitly — that justification is a mark.
To guarantee accuracy for any \(p\), use the \(p\) that maximises \(pq\): \(p=q=0.5\), \(pq=0.25\).
90% ⇒ \(z = 1.64\); \(E = 0.02\).
\[ n=\frac{(1.64)^2(0.25)}{(0.02)^2}=\frac{0.6724}{0.0004}=1681 \] \[ n=\mathbf{1681} \]
7-52Width, not half-widthLevin Ex. 7-52 · p. 356

Q. Daily customer numbers have standard deviation 15. What sample size gives a 96% confidence interval for the true mean with a width of only eight customers?

Why: two traps in one question — the interval width must be halved to get \(E\), and 96% is a confidence level nobody memorises, so you must read \(z\) out of the table.
Why we do itWhat to write
Halve the width. \(E\) is the distance from the centre to either end.
\[ E=\frac{8}{2}=4 \text{ customers} \]
Get \(z\) for 96% by looking up \(A(z)=0.48\) in the body of the table.
\[ A(z)=\frac{0.96}{2}=0.48 \;\Rightarrow\; z=2.05 \]
Substitute and round up.
\[ n=\left(\frac{2.05\times15}{4}\right)^2=(7.6875)^2=59.10 \] \[ n=\mathbf{60 \text{ days}} \]
If you had used \(E = 8\) you would get \(n = 15\) — a quarter of the right answer, and an interval twice as wide as asked for. Read the word "width" carefully.
6-39Two partsLevin Ex. 6-39 · p. 307

Q. Tread life is normal with standard deviation 216.4 miles. (a) Sampling 800 tires, what is the probability the sample mean lies between the true mean and 300 miles above it? (b) How large a sample makes you 95% sure the sample mean is within 100 miles of the true mean?

(a) Standard error, then standardise 300 miles.
\[ \sigma_{\bar{x}}=\frac{216.4}{\sqrt{800}}=7.6509 \] \[ z=\frac{300}{7.6509}=39.2 \]
That \(z\) is far off the end of any table, so essentially the whole upper half is captured.
\[ P(\mu<\bar{x}<\mu+300)=A(39.2)\approx\mathbf{0.5000} \] With \(n=800\) the sample mean is within a few miles of \(\mu\); 300 miles is unreachable.
(b) Sample size for a mean.
\[ n=\left(\frac{1.96\times216.4}{100}\right)^2=(4.2414)^2=17.99 \] \[ n=\mathbf{18 \text{ tires}} \]
6-41Size from a target SELevin Ex. 6-41 · p. 307

Q. Radiation readings have standard deviation 150 millirems. How many readings are needed so the standard error is no higher than 25 millirems?

Why: a variant with no confidence level — the target is stated directly as a standard error, so there is no \(z\) at all. Set \(\sigma/\sqrt{n} \le 25\) and solve.
Write the requirement as an inequality.
\[ \frac{150}{\sqrt{n}} \le 25 \]
Rearrange and square.
\[ \sqrt{n} \ge \frac{150}{25}=6 \;\Rightarrow\; n \ge 36 \] \[ n=\mathbf{36 \text{ readings}} \]
Note there is no \(z\) here — the question fixed the standard error itself rather than a confidence level. If it had said "95% confident within 25 millirems", \(z=1.96\) would enter and \(n\) would become \((1.96\times150/25)^2 = 139\).
6-44Mean, unitsLevin Ex. 6-44 · p. 307

Q. Reading time has population standard deviation 90 minutes. How large a sample lets the researcher assert with 95% confidence that the sample mean differs from the true mean by at most half an hour?

Why: the units trap — "half an hour" must become 30 minutes to match the σ, which is in minutes.
Convert the tolerance into the same units as σ.
\(E = \tfrac12\) hour \(= \mathbf{30}\) minutes; \(\sigma = 90\) minutes; \(z = 1.96\)
Substitute and round up.
\[ n=\left(\frac{1.96\times90}{30}\right)^2=(5.88)^2=34.57 \] \[ n=\mathbf{35 \text{ students}} \]
6-45ProportionLevin Ex. 6-45 · p. 307

Q. About 40% of subscribers are believed to be aware of a campaign. How large a sample is needed to be 95% confident the true proportion is within 3% of the sample proportion?

A prior estimate is given, so use it rather than 0.5. Convert 3% to a decimal.
\(p=0.40,\ q=0.60,\ E=0.03,\ z=1.96\)
Substitute and round up.
\[ n=\frac{(1.96)^2(0.40)(0.60)}{(0.03)^2}=\frac{3.8416\times0.24}{0.0009}=1024.43 \] \[ n=\mathbf{1025 \text{ subscribers}} \]
6-46Proportion + finite populationLevin Ex. 6-46 · p. 308

Q. An insurer has 6,000 policy holders. How large a sample is needed to be 95% confident the sample proportion differs from the true proportion by at most 4%?

Why: two things at once — no prior estimate (so \(p=0.5\)) and a stated finite population, which lets you shrink the answer.
Why we do itWhat to write
No estimate of \(p\) ⇒ take 0.5 and say why.
No prior information, so \(p=q=0.5\) — the value maximising \(pq\), giving the safest size.
Uncorrected size first.
\[ n_0=\frac{(1.96)^2(0.25)}{(0.04)^2}=\frac{0.9604}{0.0016}=600.25 \;\Rightarrow\; 601 \]
\(n_0/N = 601/6000 = 0.10 > 0.05\), so the finite population correction is worth applying.
\[ n=\frac{n_0}{1+\dfrac{n_0-1}{N}}=\frac{600.25}{1+\dfrac{599.25}{6000}}=\frac{600.25}{1.0999}=545.74 \]
Round up, and give both answers.
\[ n=\mathbf{546 \text{ policy holders}} \] Without the correction: 601. Knowing the population is only 6,000 saves 55 interviews. If the question does not mention \(N\), quote 601.
SC 6-8Both directionsLevin SC 6-8 · p. 308

Q. \(\sigma = 1.25\). (a) With \(n = 36\), find \(P(\mu-0.5 \le \bar{x} \le \mu+0.5)\). (b) How large a sample makes that probability 0.98?

(a) Standard error, then a symmetric interval.
\[ \sigma_{\bar{x}}=\frac{1.25}{6}=0.2083 \qquad z=\pm\frac{0.5}{0.2083}=\pm2.40 \] \[ P=2A(2.40)=2(0.4918)=\mathbf{0.9836} \]
(b) Now run it backwards: 0.98 two-sided ⇒ \(A(z)=0.49\) ⇒ \(z=2.33\).
\[ 2.33=\frac{0.5}{1.25/\sqrt{n}} \;\Rightarrow\; \sqrt{n}=\frac{2.33\times1.25}{0.5}=5.825 \]
Square and round up.
\[ n=33.93 \;\Rightarrow\; \mathbf{34 \text{ households}} \]
Both parts match Levin's printed answers (0.9836 and "at least 34") exactly.
SC 7-12Proportion, p = 0.5Levin SC 7-12 · p. 357

Q. Find the sample size for a proportion within 0.04 at 90% confidence, with no prior estimate.

\(p=q=0.5\), \(z=1.64\), \(E=0.04\).
\[ n=\left(\frac{1.64\times0.5}{0.04}\right)^2=(20.5)^2=420.25 \] \[ n\ge\mathbf{421} \]
Matches Levin's printed answer (420.25, so \(n \ge 421\)) exactly. Note Levin writes the proportion formula in the equivalent form \(E = z\sqrt{pq/n}\) and solves for \(n\) — identical algebra.
SC 7-13Proportion, 98%Levin SC 7-13 · p. 357

Q. Find the sample size for a proportion within 0.05 at 98% confidence, with no prior estimate.

\(p=q=0.5\), 98% ⇒ \(z=2.33\), \(E=0.05\).
\[ n=\left(\frac{2.33\times0.5}{0.05}\right)^2=(23.3)^2=542.89 \] \[ n\ge\mathbf{543 \text{ records}} \]
Matches Levin's printed answer (542.89, so at least 543) exactly.
THE 90-SECOND VERSION — WHAT TO CARRY INTO THE HALL
Unit 6 Summary Rules
\[ \text{FPM: } \sqrt{\frac{N - n}{N - 1}} \qquad (n/N > 0.05) \] \[ \text{Proportion SE: } \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}, \qquad z = \frac{\hat{p} - p}{\sigma_{\hat{p}}} \] \[ \text{Sample Size Mean: } n = \left(\frac{z \cdot \sigma}{E}\right)^2, \qquad \text{Sample Size Proportion: } n = \frac{z^2 p q}{E^2} \]
◆ Where this sits in the rest of the site

Concepts: Unit 06 notes. Q3 (b) is worked in both its versions — for a mean and for a proportion — at Solved Papers, and the finite population correction reappears inside Q4 (a).

With Units 4, 5 and 6 done you have covered 21 of the 31 in-scope marks. The rest sit in Unit 3 (Bayes, Q5 a — 5 marks) and Units 1–2 (the Q3 a interpretation — 5 marks).