STAT101 - Introduction to Probability Theory and Statistics
Chapter 8 Slide Breakdown — Groebner, Shannon & Fry, Business Statistics, 10th Global Edition
This chapter is the first real "inferential statistics" chapter: instead of just describing a sample, we use it to make a statement about the unknown population parameter. It has three parts: 8.1 building a confidence interval for a population mean μ (when σ is known); 8.2 figuring out how large a sample you need before you collect data, to hit a target margin of error; and 8.3 doing the same confidence-interval logic for a population proportion p instead of a mean. Every formula in this chapter is a variation on the same idea: point estimate ± (critical value) × (standard error).
A point estimate is a single statistic, determined from a sample, that is used to estimate the corresponding population parameter.
| Parameter | Population | Sample (Point Estimate) |
|---|---|---|
| Mean | μ | x̄ |
| Proportion | p | p̄ |
Point estimates are subject to sampling error — they are almost always at least a little bit wrong, because a different sample would almost certainly give a different x̄. A single number gives you no sense of how far off it might be. This is exactly why we build confidence intervals instead of reporting a bare point estimate.
A confidence interval estimate is an interval developed from a statistical sample such that, if all possible intervals of a given width were constructed, a percentage of these intervals — known as the confidence level — would include the true population parameter.
Every confidence interval has the same skeleton: Lower Confidence Limit ← Point Estimate → Upper Confidence Limit.
It does NOT mean "there's a 95% chance μ is in this particular interval." μ is a fixed (if unknown) number — it either is or isn't in the interval, with no probability about it. It means: "if we repeated this sampling process many times and built an interval each time, 95% of those intervals would capture the true μ." This is a subtle but very common exam trap — always describe the confidence level as a property of the procedure, not of one specific interval.
where x̄ = sample mean, z = critical value from the standard normal table, σ = population standard deviation, n = sample size. The quantity σ/√n is the standard error of the mean.
| Confidence Level | Critical Value (z) |
|---|---|
| 80% | z = 1.28 |
| 90% | z = 1.645 |
| 95% | z = 1.96 |
| 99% | z = 2.575 |
Critical values can be found using the standard normal table, or Excel's NORM.S.INV
function. For a 95% confidence interval, the middle 95% of the standard normal curve is split into two
tails of 2.5% (0.025) each, which corresponds to z = ±1.96 — i.e., 0.475 of the area lies between z = 0
and z = 1.96 on each side.
Higher confidence (99% vs. 90%) means you want to "catch" the true parameter more often, so you need a wider net — a bigger z-value, and therefore a wider interval. This is the direct trade-off explored later in this chapter.
Problem: Assuming that the population distribution for credit card balances is normal with a population standard deviation σ = $400, construct and interpret a 95% confidence interval estimate for the population mean balance if a random sample of n = 100 accounts has a sample mean x̄ = $2,000.
Step 1 — Identify pieces: x̄ = 2000, σ = 400, n = 100, confidence level = 95% → z = 1.96.
Step 2 — Standard error: σ/√n = 400/√100 = 400/10 = 40.
Step 3 — Margin of error: e = z · (σ/√n) = 1.96 × 40 = 78.40.
Step 4 — Confidence interval: x̄ ± e = 2000 ± 78.40 → [$1,921.60, $2,078.40].
Interpretation: Based on the sample data, with 95% confidence, we conclude that the population mean balance is somewhere between $1,921.60 and $2,078.40. The margin of error tells us the sample mean of $2,000 is within $78.40 of the true population mean.
Problem: The distribution of hours worked by students at a university is normally distributed. If the population standard deviation is known to be 5 hours, construct and interpret a 90% confidence interval estimate for the mean hours worked by all students if a random sample of 64 students has a mean equal to 23 hours.
Step 1: x̄ = 23, σ = 5, n = 64, confidence level = 90% → z = 1.645.
Step 2 — Standard error: σ/√n = 5/√64 = 5/8 = 0.625.
Step 3 — Margin of error: e = 1.645 × 0.625 = 1.028 ≈ 1.03.
Step 4 — Confidence interval: 23 ± 1.03 → [21.97, 24.03] hours.
Interpretation: With 90% confidence, the mean hours worked by all students at the university is between 21.97 and 24.03 hours. The margin of error indicates the sample mean of 23 hours is within 1.03 hours of the true population mean.
Using the university hours example (x̄=23, σ=5, n=64): a 90% CI is narrowest, 95% is wider, and 99% is widest. Increasing the confidence level increases the margin of error (and thus widens the interval), because a bigger z-value is needed to "guarantee" catching μ more often.
Using the credit card example (x̄=2000, σ=400, 95% confidence): as n grows from 100 → 200 → 400, the interval gets narrower. Increasing the sample size decreases the margin of error, because σ/√n shrinks as n grows.
When designing a study you generally want all three of: (1) a small margin of error, (2) high confidence, and (3) a small sample size. Unfortunately these are conflicting objectives — you cannot maximize all three at once. Shrinking the margin of error while keeping confidence high requires a larger sample; keeping the sample small while wanting high confidence forces a bigger margin of error.
Students often forget to divide σ by √n before multiplying by z — i.e., they compute z·σ instead of z·(σ/√n). Always compute the standard error (σ/√n) as its own step first, then multiply by the critical value to get the margin of error, and only then apply it to x̄.
Section 8.1 assumed the sample size n was already fixed and asked "how wide is my interval?" Section 8.2 flips the question around: before collecting any data, a researcher decides how small they want the margin of error to be (and how confident they want to be), and then solves for the minimum sample size n needed to achieve that.
where z = critical value for the desired confidence level, σ = population standard deviation (or a reasonable estimate of it), e = the desired margin of error. Always round n UP to the next whole number — even n = 100.02 requires n = 101, because a fractional sample size doesn't exist and rounding down would under-shoot your target precision.
It's just the margin-of-error formula e = z·(σ/√n) solved for n. Multiply both sides by √n, divide by e, then square: √n = z·σ/e → n = (z·σ/e)².
Problem: A bank wants to estimate the average checking account balance for its customers. Based on past data, the population standard deviation is believed to be σ = $300. The bank wants to be 95% confident that the sample mean is within e = $50 of the true population mean. How large a sample is required?
Step 1 — Identify pieces: z = 1.96 (95% confidence), σ = 300, e = 50.
Step 2 — Apply the formula: n = (z·σ/e)² = (1.96 × 300 / 50)² = (588/50)² = (11.76)² = 138.2976.
Step 3 — Round UP: n = 139.
Interpretation: The bank needs a random sample of at least 139 customers to be 95% confident that the sample mean is within $50 of the true population mean balance.
Never round the required sample size DOWN, even if the decimal is tiny (e.g., n = 138.02 still becomes 139, not 138). Rounding down would give you a sample slightly too small to guarantee the requested margin of error at the requested confidence level.
Sections 8.1–8.2 dealt with a population mean (a quantitative measure, like dollars or hours). Section 8.3 applies the exact same logic to a population proportion p — the fraction of a population with some categorical characteristic (e.g., the proportion of customers who are "satisfied," or the proportion of parts that are defective). The sample proportion p̄ is the point estimate of p.
where p̄ = sample proportion, z = critical value for the desired confidence level, n = sample size. The quantity √(p̄(1−p̄)/n) is the standard error of the proportion.
This is the proportion analog of z = (x̄ − μ)/(σ/√n) — it standardizes how far a sample proportion is from a hypothesized population proportion p, in standard-error units.
Problem: A quality manager samples n = 250 parts from a production line and finds that 20 of them are defective. Construct and interpret a 95% confidence interval estimate for the true proportion of defective parts in the whole production run.
Step 1 — Point estimate: p̄ = 20/250 = 0.08.
Step 2 — Standard error: √(p̄(1−p̄)/n) = √(0.08 × 0.92 / 250) = √(0.0736/250) = √0.0002944 ≈ 0.01716.
Step 3 — Margin of error: e = z × standard error = 1.96 × 0.01716 ≈ 0.0336.
Step 4 — Confidence interval: 0.08 ± 0.0336 → [0.0464, 0.1136], i.e., about 4.64% to 11.36%.
Interpretation: With 95% confidence, the true proportion of defective parts produced by this line is between about 4.6% and 11.4%.
Derived the same way as the mean version, using p̄(1−p̄) in place of σ². If no estimate of p̄ is available, using p̄ = 0.5 gives the most conservative (largest) required sample size, since p̄(1−p̄) is maximized at p̄ = 0.5.
Every formula for a proportion mirrors the mean version with p̄(1−p̄) standing in for σ²: standard error σ/√n ↔ √(p̄(1−p̄)/n); margin of error z·(σ/√n) ↔ z·√(p̄(1−p̄)/n); z-statistic (x̄−μ)/(σ/√n) ↔ (p̄−p)/√(p(1−p)/n). If you memorize the mean formulas well, the proportion formulas are "free" — just swap the variance term.
Don't confuse σ (population standard deviation of a quantitative variable) with the proportion's standard error term √(p̄(1−p̄)/n) — a proportion problem never has a given "σ" to plug in; the variability is built into p̄(1−p̄) itself. Also, a proportion must always be between 0 and 1 (or state it as a percentage 0–100%) — if your confidence interval math produces a bound below 0 or above 1, double check your arithmetic.
1. A single statistic, such as x̄, used to estimate an unknown population parameter is called a:
2. What is the correct critical value (z) for a 99% confidence interval?
3. If the confidence level is increased from 90% to 99% while the sample size stays the same, the margin of error will:
4. A sample of n = 100 accounts has x̄ = $2,000 and σ = $400. What is the margin of error for a 95% confidence interval?
5. A researcher wants a margin of error of e = 5 with 95% confidence, and estimates σ = 25. What sample size is required? (n = (zσ/e)², round up)
6. In the formula for the confidence interval of a proportion, p̄(1 − p̄)/n plays the same role that which quantity plays in the mean formula?
7. Which statement about a "95% confidence interval" is correct?
8. Explain why increasing the confidence level widens a confidence interval, while increasing the sample size narrows it. Use the formula x̄ ± z(σ/√n) in your answer.
Answer: Increasing the confidence level requires a larger critical value z (e.g., 1.645 → 1.96 → 2.575 as confidence rises from 90% to 95% to 99%), which directly increases the margin of error z(σ/√n) and widens the interval — you need a "wider net" to catch μ more often. Increasing the sample size n makes the denominator √n larger, which shrinks the standard error σ/√n and therefore shrinks the margin of error, narrowing the interval — more data means less sampling variability, so the point estimate becomes more precise.
9. A survey of 400 customers finds that 60 are dissatisfied with a product. Construct and interpret a 90% confidence interval for the true proportion of dissatisfied customers.
Answer: p̄ = 60/400 = 0.15. Standard error = √(0.15 × 0.85/400) = √(0.1275/400) = √0.00031875 ≈ 0.01785. Margin of error = z × SE = 1.645 × 0.01785 ≈ 0.02937. Confidence interval = 0.15 ± 0.0294 → [0.1206, 0.1794], i.e., about 12.1% to 17.9%. Interpretation: with 90% confidence, the true proportion of dissatisfied customers is between about 12.1% and 17.9%.
10. A researcher wants to estimate a population proportion with a margin of error of e = 0.03 at 95% confidence, and has no prior estimate of p̄. What sample size should be used, and why is p̄ = 0.5 chosen in this situation?
Answer: Using n = p̄(1−p̄)(z/e)² with p̄ = 0.5: n = 0.5 × 0.5 × (1.96/0.03)² = 0.25 × (65.33)² = 0.25 × 4268.4 ≈ 1067.1 → round up to n = 1068. p̄ = 0.5 is chosen because p̄(1−p̄) is maximized when p̄ = 0.5 (0.5 × 0.5 = 0.25, the largest possible value of that product), which gives the most conservative — i.e., the largest — required sample size. This guarantees the margin of error target is met no matter what the true p̄ later turns out to be.
Critical values: 80% → z=1.28, 90% → z=1.645, 95% → z=1.96, 99% → z=2.575.
| Term | Definition |
|---|---|
| Point Estimate | A single statistic, determined from a sample, used to estimate the corresponding population parameter. |
| Confidence Interval Estimate | An interval built from a sample such that, if all possible intervals of that width were constructed, a specified percentage (the confidence level) would contain the true parameter. |
| Confidence Level | The percentage of all possible confidence intervals (of a given design) that would contain the true population parameter. |
| Critical Value (z) | The number of standard errors from the point estimate needed to achieve a given confidence level, found from the standard normal distribution. |
| Standard Error of the Mean | σ/√n — the standard deviation of the sampling distribution of the sample mean. |
| Margin of Error | The half-width of a confidence interval; e = (critical value) × (standard error). |
| Sampling Error | The unavoidable difference between a sample statistic and the true population parameter due to sampling only a subset of the population. |
| Required Sample Size | The minimum sample size needed to achieve a target margin of error at a target confidence level; always rounded up. |
| Population Proportion (p) | The fraction of the population having a characteristic of interest. |
| Sample Proportion (p̄) | The point estimate of p, computed as the number of sample items with the characteristic divided by n. |
| Standard Error of the Proportion | √(p̄(1−p̄)/n) — the standard deviation of the sampling distribution of the sample proportion. |
| Parameter | A descriptive numerical measure (μ, p) computed from an entire population. |
| Statistic | A descriptive numerical measure (x̄, p̄) computed from a sample. |