STAT101 - Introduction to Probability Theory and Statistics
A slide-deck replacement: sampling error, the Central Limit Theorem, and the sampling distributions of the mean and of a proportion — with full worked examples and exam drills.
Understand the concept of a sampling distribution and why sampling error occurs.
Determine the mean and standard deviation for the sampling distribution of the sample mean x̄, and apply the Central Limit Theorem.
Sampling error is the difference between a measure computed from a sample (a statistic) and the corresponding measure computed from the population (a parameter). It happens purely because of which particular sample was randomly selected — it is not a mistake, and it is not the same thing as a non-sampling error (like a data-entry mistake, a biased survey question, or a broken measuring instrument, which ARE avoidable mistakes).
Sampling error is unavoidable and random — it exists purely because a sample is not the whole population. Non-sampling error comes from mistakes in the data collection process itself (bad wording, non-response bias, data entry errors, a broken scale) and CAN be avoided with better survey/experiment design. If an exam question asks "why did the sample mean differ from the population mean, assuming the survey was conducted correctly," the answer is sampling error.
62 students took a midterm exam (this is the entire population of interest, N = 62). The population parameters are:
A random sample of n = 5 scores is drawn: 100, 44, 55, 90, 75. The sample mean is x̄ = 72.80.
A second random sample of n = 5 gives x̄ = 54.80.
Answer / key insight: Just by chance, the two samples produced very different sampling errors (+1.37 vs. −16.63) even though both were drawn from the exact same population using the exact same procedure. The sampling error depends entirely on which specific sample happens to get selected — it can be small or large, positive or negative.
If you take 200 different random samples of n = 5 from this population and plot a histogram of the 200 sample means, you get a sampling distribution that is:
Increasing the sample size to n = 10 tightens the sampling distribution further: μ_x̄ ≈ 71.32 and σ_x̄ ≈ 5.12 — smaller spread than with n = 5. Larger samples produce sampling distributions with less variability.
For any population, the average value of all possible sample means computed from all possible random samples of a given size equals the population mean:
For any population, the standard deviation of all possible sample means computed from all possible random samples of size n equals the population standard deviation divided by the square root of the sample size:
This quantity, σ/√n, is called the standard error of the mean — it is the standard deviation of the sampling distribution, NOT the standard deviation of the raw population.
If a population is normally distributed with mean μ and standard deviation σ, the sampling distribution of x̄ is also exactly normally distributed, with mean μ_x̄ = μ and standard deviation σ_x̄ = σ/√n — for ANY sample size n, even small ones.
For simple random samples of n observations taken from a population with mean μ and standard deviation σ, regardless of the shape of the population's distribution, provided the sample size is sufficiently large, the distribution of the sample means x̄ is approximately normal with:
The larger the sample size, the better the normal approximation. Even if the population is uniform, skewed, or otherwise strangely shaped, the sample means will pile up into a bell curve once n is big enough. This is why the CLT is so powerful: it lets us use the normal distribution's math (z-scores, standard normal table) for the sample mean even when we know nothing about the shape of the population it came from.
French fry length is normally distributed with mean μ = 4.0 inches and standard deviation σ = 1.0 inch. What is the probability that a random sample of n = 100 fries has a sample mean exceeding 4.10 inches?
Step 1 — compute the standard error: σ_x̄ = σ/√n = 1.0/√100 = 1.0/10 = 0.10
From the standard normal table: P(0 ≤ z ≤ 1.00) = 0.3413, so P(z > 1.00) = 0.5000 − 0.3413 = 0.1587.
Answer: P(x̄ > 4.10) = 0.1587, about 15.9%.
Same French fry population (μ = 4.0, σ = 1.0). What is the probability that the mean of a sample of n = 100 fries is less than 3.85 inches?
Standard error: σ_x̄ = σ/√n = 1.0/√100 = 1.0/10 = 0.10
From the standard normal table: P(0 ≥ z ≥ −1.50) = 0.4332, so P(z < −1.50) = 0.5000 − 0.4332 = 0.0668.
Answer: P(x̄ < 3.85) = 0.0668, about 6.68%.
When standardizing a sample mean, you must use the standard error σ/√n in the denominator of the z formula — NOT the raw population standard deviation σ by itself. Using z = (x̄ − μ)/σ instead of z = (x̄ − μ)/(σ/√n) is the single most common error on Chapter 7 exams. Always ask yourself: "am I working with one single observation x, or an average of a sample x̄?" If it's a sample mean, you divide σ by √n first.
Look at the wording: "the mean of a sample of n observations," "a random sample of size n," "the average of n items" → use σ/√n. If the question is about a single individual measurement ("one fry," "one student's score") with no sample/averaging language, you use the raw σ with no division by √n (that's an ordinary normal-distribution problem from an earlier chapter, not a sampling-distribution problem).
Population proportion (p): the fraction of values in a population that have a specific attribute of interest, p = X/N, where X = number in the population with the attribute, N = population size.
Sample proportion (p̄): the fraction of items in a sample that have the attribute of interest, p̄ = x/n, where x = number in the sample with the attribute, n = sample size.
The best estimate of the true population proportion p is p̄, the sample proportion. Any inference about how close p̄ is likely to be to the true p requires knowledge of the sampling distribution of p̄.
For a sufficiently large sample size, the sampling distribution of the sample proportion p̄ is approximately normal, with:
The rule of thumb for "sufficiently large" here is different from the CLT's n ≥ 30: for proportions, we require np ≥ 5 AND nq ≥ 5 (where q = 1 − p) before treating the sampling distribution of p̄ as approximately normal.
Past records for a national retail chain show that the true proportion of customers who pay with a debit card is p = 0.60. A particular store samples n = 100 purchases and observes 48 debit-card users (p̄ = 0.48). What is the probability of a sample proportion of 0.48 or less, if the true population proportion really is 0.60?
Standard error: σ_p̄ = √( (0.60)(0.40)/100 ) = √0.0024 ≈ 0.049
From the standard normal table: P(0 ≥ z ≥ −0.77) = 0.2794, so P(z < −0.77) = 0.5000 − 0.2794 = 0.2206.
Answer: P(p̄ ≤ 0.48) = 0.2206, about 22.1%. So it's not extremely unusual (about a 1-in-4.5 chance) to see a sample proportion this low even when the true proportion is 0.60 — sampling error again at work.
The IRS believes p = 0.15 (15%) of all taxpayers will have a fraudulent return filed in their name. In a random sample of n = 200 taxpayers, what is the probability that more than 40 of them will have had this experience?
First convert the count to a sample proportion: p̄ = x/n = 40/200 = 0.20. We want P(p̄ > 0.20).
From the standard normal table: P(0 ≤ z ≤ 1.98) = 0.4761, so P(z > 1.98) = 0.5000 − 0.4761 = 0.0239.
Answer: P(p̄ > 0.20) = 0.0239, about 2.4%. It would be fairly unusual — only about a 2.4% chance — to see more than 40 out of 200 taxpayers affected if the true rate really is 15%.
| Clue in the wording | Use... |
|---|---|
| "average length/weight/score/time," a continuous measurement, given μ and σ | Sampling distribution of the mean (x̄, σ/√n) |
| "proportion/percentage/fraction who...", "x out of n have an attribute," given p (or p̄ = x/n) | Sampling distribution of a proportion (p̄, √(p(1−p)/n)) |
| Yes/no, success/failure, categorical attribute being counted | Proportion |
| A numeric quantity that's averaged across the sample | Mean |
1. Sampling error is best described as:
2. According to the Central Limit Theorem, as the sample size n increases, the sampling distribution of x̄:
3. What is the standard error of the sample mean, σ_x̄?
4. A population has σ = 20 and a sample of n = 25 is taken. What is the standard error of the mean?
5. The conservative rule-of-thumb sample size needed for the Central Limit Theorem to guarantee an approximately normal sampling distribution, regardless of population shape, is:
6. A student computes z = (x̄ − μ)/σ (without dividing by √n) for a sample-mean problem. What is wrong with this calculation?
7. Which conditions must hold for the sampling distribution of a sample proportion p̄ to be approximately normal?
8. What is the standard error of a sample proportion, σ_p̄?
9. A problem states "60 out of 150 sampled customers preferred Brand A." Which sampling distribution applies, and what is p̄?
10. A population of delivery times is normally distributed with μ = 30 minutes and σ = 8 minutes. Find the probability that the mean delivery time of a random sample of n = 16 deliveries is greater than 33 minutes.
Answer:
σ_x̄ = σ/√n = 8/√16 = 8/4 = 2
z = (33 − 30)/2 = 1.50
P(0 ≤ z ≤ 1.50) = 0.4332 (from standard normal table)
P(z > 1.50) = 0.5000 − 0.4332 = 0.0668
There is about a 6.68% chance the sample mean delivery time exceeds 33 minutes.
11. A machine fills bottles with a mean of μ = 500 mL and σ = 12 mL (population approximately normal). Find P(x̄ < 495) for a sample of n = 36 bottles.
Answer:
σ_x̄ = 12/√36 = 12/6 = 2
z = (495 − 500)/2 = −2.50
P(0 ≥ z ≥ −2.50) = 0.4938
P(z < −2.50) = 0.5000 − 0.4938 = 0.0062
Only about a 0.62% chance the sample mean fill is below 495 mL — this would be a red flag for the filling process.
12. A polling firm believes 45% of voters (p = 0.45) support a proposition. In a random sample of n = 120 voters, what is the probability that the sample proportion p̄ is between 0.40 and 0.50?
Answer:
Check conditions: np = 120(0.45) = 54 ≥ 5 and nq = 120(0.55) = 66 ≥ 5. OK to use normal approximation.
σ_p̄ = √( (0.45)(0.55)/120 ) = √(0.2475/120) = √0.0020625 ≈ 0.0454
z for p̄ = 0.40: (0.40 − 0.45)/0.0454 = −1.10
z for p̄ = 0.50: (0.50 − 0.45)/0.0454 = 1.10
P(−1.10 ≤ z ≤ 1.10) = 2 × P(0 ≤ z ≤ 1.10) = 2 × 0.3643 = 0.7286
There is about a 72.9% chance the sample proportion falls between 0.40 and 0.50.
| Term | Definition |
|---|---|
| Sampling error | The difference between a sample statistic (x̄ or p̄) and the true population parameter (μ or p), occurring purely because of which sample was randomly selected. |
| Non-sampling error | Error introduced by mistakes in data collection (bad wording, non-response, data entry errors), which is avoidable — unlike sampling error. |
| Sampling distribution | The probability distribution of a sample statistic (like x̄ or p̄) across all possible random samples of a given size from the population. |
| Central Limit Theorem (CLT) | For sufficiently large n, the sampling distribution of x̄ is approximately normal with mean μ and standard deviation σ/√n, regardless of the population's shape. |
| Standard error | The standard deviation of a sampling distribution — σ/√n for the mean, or √(p(1−p)/n) for a proportion. |