Unit 05 β Sampling Theory I
Management examples of sampling, principles of sampling, simple random sample, Excel functions for selecting a random sample. Sampling distributions for the statistic arithmetic mean β the concept of standard error.
Every question shape in this unit is drilled on Unit 05 β How to Solve Anything: four topics, each with a "how to recognise this topic" card, a master method, and every exercise Levin sets solved with its page number. Topic 4 is Q1 (a) and Q5 (b) β 7 marks.
Two slots, both papers, and they are the same calculation asked twice:
- Q1 (a) β 2 marks. "Β΅ = 520, Ο = 25, n = 36. Find P(515 < xΜ < 525)." Compute the standard error, standardise both ends, read the table.
- Q5 (b) β 5 marks. Same thing with a business wrapper, and it explicitly asks you to state the standard error first as its own sub-part.
If you can compute Ο/βn and standardise xΜ with it, you have 7 marks in the bag before you read the story. Final QP pp. 1, 3 Β· Re-Exam QP pp. 1, 3. Both solved β
1. Why We Sample, and How Core syllabus concept
1Statistic vs parameter β the distinction everything rests on
The deck opens with it slide 2:
Parameter β a numerical value associated with the population.
"Essentially, we would like to know the parameter. But in most cases it is hard to know the parameter since the population is too large. So we have to estimate the parameter by some proper statistic computed from the sample."
That paragraph is the answer to "why do we sample?" and it is also the reason for the symbol convention used everywhere below:
| Quantity | Population (parameter) | Sample (statistic) | Sampling distribution |
|---|---|---|---|
| Mean | ΞΌ | xΜ | ΞΌxΜ and ΟxΜ |
| Standard deviation | Ο | s | ΟxΜ = the standard error |
| Proportion | p | pΜ | ΞΌpΜ and ΟpΜ |
| Size | N | n | β |
Your class notes fix the convention in one line: "ΞΌ ke jagah ΞΌxΜ, Ο ke jagah ΟxΜ" β in sampling, ΞΌ is replaced by ΞΌxΜ and Ο by ΟxΜ Class notes p. 2.
2Types of sampling
The syllabus names simple random sampling specifically; Levin Β§6.2β6.3 gives the full set pp. 271β282:
| Method | How it works | Use it when |
|---|---|---|
| Simple random | Every element has an equal chance of selection; every sample of size n is equally likely. | The population list is available and reasonably homogeneous. |
| Systematic | Take every kth element after a random start. | You have an ordered list and want speed. Risky if the list has a hidden cycle. |
| Stratified | Split the population into homogeneous strata, then sample randomly within each. | The population has distinct subgroups you must represent (e.g. all four year-groups). |
| Cluster | Split into clusters, randomly choose whole clusters, survey everyone in them. | The population is geographically spread and travel is the main cost. |
| Non-random (judgement / convenience / quota) β cheap and fast, but the sampling error cannot be quantified, so none of the formulas below apply. Levin Β§6.3 p. 279. | ||
=RAND() gives a uniform random number in [0,1) β put it beside each record, sort by it, take the top n. =RANDBETWEEN(1, N) picks a random record number directly. The Data Analysis ToolPak's Sampling tool does both periodic and random sampling in one dialog.2. The Sampling Distribution of the Mean Core syllabus concept
1The definition β write it exactly like this
The deck slide 3 and your class notes Class notes p. 1 agree word for word:
The crucial thing to grasp: xΜ is itself a random variable. Draw a different sample and you get a different xΜ. Since it is a random variable, it has its own distribution, its own mean, and its own standard deviation β and that standard deviation is what we call the standard error.
2Worked example β building one by hand
Your class built a sampling distribution from scratch, which is the fastest way to see why it exists Class notes p. 1. Five employees β Carol 3, Debbie 3, Elizabeth 7, Frank 9, George 14 years of service. Take all possible samples of size 3.
- Step 1 β How many samples are there? β΅Cβ = 5! / (3! Β· 2!) = 10 samples
- Step 2 β List every sample mean and how often it occurs
Sample mean xΜ 4β 5 6β 6β 8 8β 10 Total Frequency f 1 1 2 1 2 2 1 10 That frequency table is the sampling distribution of the mean.
- Step 3 β Check the population mean against the mean of the sampling distribution
ΞΌ = (3 + 3 + 7 + 9 + 14) / 5 = 36/5 = 7.2
ΞΌxΜ = Ξ£ fΒ·xΜ / Ξ£f = 72 / 10 = 7.2 β
They are equal β which is the first clause of the Central Limit Theorem, and no coincidence.
3. The Central Limit Theorem Core syllabus concept
1The three statements
The deck states it in three bullets slide 8:
- The mean of the population is always equal to the mean of the sampling distribution of the mean.
ΞΌxΜ = ΞΌ - As the sample size increases, the sampling distribution of the mean approaches normality, regardless of the shape of the population distribution.
- That relationship between the shape of the population and the shape of the sampling distribution is what we call the Central Limit Theorem.
Your class notes put statement 2 in the most usable form Class notes p. 1: "even if the data is NOT normal, in sample as per CLT we can consider data to be normal."
2Why it matters so much
The deck's own answer slide 9: "It is perhaps the most important theorem in inferential statistics. The significance of CLT is that without the knowledge of the frequency distribution of the population, it permits us to use the sample statistics to make inferences about the population parameters."
The conventional threshold is n β₯ 30: at that point the sampling distribution is close enough to normal for any population shape. If the population is already normal, xΜ is exactly normal for any n.
4. The Standard Error of the Mean Core syllabus concept Β· the highest-value formula in this unit
1Definition
The deck's definition slide 10:
It is not the standard deviation of the data. It measures how much the sample mean bounces around from sample to sample.
Your class notes circle ΟxΜ = Ο/βn and label it "standard error of mean" Class notes p. 2; the finite form is circled and labelled "finite population multiplier" on p. 5. Deck: slide 20.
2Standard error and sample size β the relationship the exam loves
The deck spells out the chain of consequences slide 19:
- Standard error is a measure of the dispersion of the sample means around the population mean.
- As it decreases, the values taken by sample means cluster more closely around the population mean.
- So if SE decreases, the value of any sample mean will probably be closer to the population mean.
- Putting it differently: if SE decreases, the precision with which the sample mean estimates the population mean increases.
And since ΟxΜ = Ο/βn, the only lever you control is n. Quadruple the sample and you halve the standard error β the diminishing returns that Unit 6 Concept 3 makes explicit.
3Method β the four lines that answer Q1 (a) and Q5 (b)
Dividing by Ο instead of ΟxΜ. In the Final's Q1 (a) that would give z = (515β520)/25 = β0.2 instead of β1.2, and an answer of 0.1586 instead of 0.7698. If the question is about a sample mean xΜ, the denominator is always Ο/βn. Your class notes write the z formula with ΞΌxΜ and ΟxΜ subscripted every single time Class notes p. 4 for exactly this reason. Copy that habit.
4Worked example β the deck's bank-employees problem
A population of annual earnings of class II employees in all nationalised banks has mean $19,000 and SD $2,000. A random sample of 30 employees is drawn. What is the probability that their average annual earnings will be more than $19,750? slide 11, worked in class Class notes pp. 3β5.
- Step 1 β Write down what you have ΞΌ = 19,000 Ο = 2,000 n = 30 βΉ ΞΌxΜ = 19,000
- Step 2 β Standard error ΟxΜ = Ο/βn = 2000/β30 = 2000/5.477 = 365.15
- Step 3 β Draw it, then standardise
Sketch the sampling-distribution curve centred at 19,000, mark 19,750, shade the right tail.
z = (19,750 β 19,000) / 365.15 = 750 / 365.15 = 2.05 - Step 4 β Read the table
P(xΜ > 19,750) = P(Z > 2.05) = 0.5 β A(2.05) = 0.5 β 0.4798 = 0.0202 = 2.02%
Only about a 2% chance that a sample of 30 averages above $19,750 β the sample mean rarely strays that far from $19,000 because the standard error is only $365.
Your class notes reach 0.0202 p. 5. An earlier attempt on p. 4 reads the table as 0.48991 and gets 0.02018 β the same number to three decimals.
5The finite population multiplier β when to bother
The deck's rule slide 21:
- The fraction n/N is called the sampling fraction.
- When N is very large relative to n, the FPM takes a value close to 1 and changes nothing.
- "As a rule of thumb, when the sampling fraction is less than 0.05, the finite population multiplier need not be used."
Your class notes write the threshold as n/N < 0.5 p. 5, p. 6. The deck says 0.05 slide 21, and so does Levin Β§6.7 p. 302. Almost certainly a dropped zero in the notes. Use 0.05. Practically it rarely matters: if a question hands you N at all, it usually wants the FPM used β the Final's Q4 (a) has n/N = 200/5000 = 0.04, below the threshold, and the official key applies the correction anyway Key p. 4. Safest rule: if N is given, use the FPM and say why.
6Worked example β the FPM in action
From the Re-Exam's Q4 (a) figures: N = 2,000 employees, n = 150, pΜ = 0.6 Key p. 3. Shown here for the mean-style structure; the proportion version is in Unit 6 Concept 2.
- Step 1 β Compute the sampling fraction n/N = 150 / 2000 = 0.075 βΉ 0.075 > 0.05, so apply the FPM
- Step 2 β Compute the multiplier β[(N β n)/(N β 1)] = β[(2000 β 150)/(2000 β 1)] = β(1850/1999) = β0.9255 = 0.9620
- Step 3 β Read what it did
It shrank the standard error by about 3.8%. That is the whole effect: sampling without replacement from a finite pool leaves less uncertainty than sampling from an infinite one, because each draw removes some of the population you were unsure about. As n β N the multiplier β 0, which is correct: measure everyone and there is no sampling error left at all.
Practice G1βG6 β standard error, sampling distribution of the mean, FPM β