Confidence Intervals: 42 Solved Exercises

Confidence intervals turn a sample estimate into a range that expresses statistical uncertainty. However, a correct answer requires more than inserting numbers into a formula. You must identify the target parameter, choose the appropriate standard error, check the sampling assumptions, and interpret the result without overstating what the data prove.

This guide works through exercises 4.13–4.54 from the supplied pages. Along the way, it compares normal and Student’s t methods, explains finite population corrections, and derives sample sizes for means and proportions. Moreover, it audits rounding choices and highlights limitations that a short answer key can hide.

How confidence intervals work

A point estimate gives one number, such as a mean wage of $500. By contrast, an interval estimate adds a range around that number. The range reflects the sampling variability that the chosen method can quantify; it does not automatically account for nonresponse, poor measurement, or an incomplete sampling frame.

What 95% confidence actually means

Under the model’s assumptions, a 95% confidence procedure captures the fixed population parameter in about 95% of repeated samples. After you calculate a particular interval, that interval either contains the parameter or does not. Therefore, the confidence level describes the procedure’s long-run performance, rather than a new probability assigned to a fixed parameter. See NIST’s explanation of confidence intervals.

For example, imagine repeating the same valid sampling process 1,000 times. A procedure with exact 95% coverage would produce 950 covering intervals on average. Nevertheless, one particular set of 1,000 repetitions need not contain exactly 950 successes. Randomness affects the coverage count too.

Standard deviation, standard error, and margin of error

Consequently, a standard deviation of 30 does not imply a margin of error of 30. The sample size and confidence level both matter. Similarly, a confidence interval for a mean does not describe the range containing 95% of individual people, wages, or product lifetimes.

Symbols you will use throughout the exercises

In this article, s uses the usual n − 1 denominator. However, an older textbook may define a sample dispersion statistic differently. That distinction becomes important when checking claims about unbiased estimators.

Choose the right confidence interval formula

Normal critical values and two-sided confidence levels

For a two-sided 95% interval, each tail contains 2.5%. Therefore, use the 97.5th percentile of the normal distribution. A one-sided 95% bound instead uses its 95th percentile, about 1.645; confusing these two settings changes the answer.

Finite population correction: use the right variance convention

When you sample without replacement, observing a substantial share of the population reduces the remaining sampling uncertainty. The supplied exercises use the population variance with denominator N. Consequently, their known-σ correction takes this form:

Survey sampling texts often define the finite population variance S² with denominator N − 1 instead. With that convention, the same variance becomes (1 − n/N)S²/n. If you estimate S² using the ordinary sample variance s², use (1 − n/N)s²/n. These formulas agree when you track the variance definition consistently. See Penn State’s lesson on simple random sampling.

What happens when the sample size increases?

Holding everything else fixed, the margin of error follows an inverse square-root relationship. Thus, doubling n reduces the margin by about 29.3%, rather than 50%. To cut the margin in half, you usually need four times as many observations.

These percentages follow directly from E ∝ 1/√n. However, changing the confidence level, variance, sampling design, or finite population correction can alter that comparison.

Exercises 4.13–4.22: normal intervals and sample size

Exercise 4.13: point estimates, unbiasedness, and interval estimates

Task: Explain a point estimate, an unbiased estimator, and an interval estimate. A point estimate is one observed value used to estimate a population parameter. For example, x̄ = 50 estimates μ, while p̂ = 0.70 estimates p.

An unbiased estimator has a sampling expectation equal to its target. Nevertheless, any one estimate can lie above or below that target. An interval estimate adds lower and upper limits together with a confidence level and the assumptions supporting the method.

Exercise 4.14: a mean of 50 from 64 observations

Given: n = 64, x̄ = 50, sample standard deviation s = 20, and N = 800. The question asks for a 95% interval and its interpretation. Because 64/800 = 8%, the book applies a finite population correction.

Thus, the textbook’s rounded answer of 45.3 to 54.7 checks out. Its method estimates σ with s and uses a large-sample normal approximation. For repeated samples under an adequate approximation, roughly 95% of the resulting intervals would cover μ.

With the ordinary sample variance convention, a design-based estimated standard error is (20/8)√(1 − 64/800) = 2.397916. Moreover, using t with 63 degrees of freedom gives approximately (45.21, 54.79). This is a useful approximate alternative, not an exact finite-population t guarantee.

Exercise 4.15: 90%, 95%, and 99% intervals with known σ

Given: x̄ = 80, n = 25, σ = 30, and N = 1,000, with a normal population model. Because the sample fraction is 2.5%, the exercise ignores the small finite correction. Hence, SE = 30/5 = 6.

As confidence rises, the interval widens. Therefore, stronger coverage comes at the cost of lower precision when the sample stays fixed. The normal population assumption also matters: it justifies the normal mean calculation despite n being below 30.

Exercise 4.16: test scores from a finite school population

Given: N = 500, n = 36, x̄ = 380, and known σ = 40. Since 36/500 = 7.2%, include the finite population correction. First, calculate the standard error without rounding the correction prematurely.

The book reports (367.46, 392.54) because it rounds the correction to 0.96 and the standard error to 6.4. Consequently, the difference reflects arithmetic precision rather than a different statistical method. The normal approximation remains an assumption about the sampling distribution.

Exercise 4.17: minimum sample size for weekly wages

Given: Known σ = $40, target margin E = $20, a normal wage population model, and 99% confidence. The factory has several thousand workers, so the planned sample is a small population share. Start with the margin-of-error equation.

The textbook uses 2.58 and obtains 26.6256, which also rounds upward to 27. However, rounding to the nearest integer would be the wrong general rule. Always round a minimum required sample size upward.

Exercise 4.18: derive the formula and examine its assumptions

(a) Formula: Rearrange E = z*σ/√n to obtain n = (z*σ/E)². Therefore, the wage example again requires 27 workers. (b) Why planning matters: Too few observations fail the precision target; too many use resources without a corresponding requirement.

(c) Total width: A margin of ±$20 produces a total width of $40. (d) Unknown shape: The textbook suggests increasing n to at least 30. However, 30 is not a universal guarantee of a good normal approximation, especially with strong skewness or extreme observations.

(e) Unknown σ: A pilot sample, relevant historical data, or a conservative planning value can support sample-size planning. Subsequently, a t-based planning calculation may require iteration. Dividing an observed range by six is only a rough heuristic, not a reliable substitute for a defensible variance estimate.

Exercise 4.19: counts and proportions in a binomial model

Let X count successes in n independent Bernoulli trials with common success probability p. Then p̂ = X/n. As a result, a count and a proportion have related means but different units and standard deviations.

For sampling without replacement from a finite binary population, multiply the population-based standard error by √[(N − n)/(N − 1)]. In that setting, the exact count model is hypergeometric rather than binomial. Nevertheless, the binomial approximation can be useful when the sampling fraction is small.

Exercise 4.20: workers’ retirement preferences

Given: Of n = 100 sampled workers, 70 prefer their own retirement arrangements; the plant has N = 1,200 workers. Therefore, p̂ = 70/100 = 0.70. The 8.33% sampling fraction motivates a finite correction.

Thus, the estimated share lies between about 61.40% and 78.60% under the exercise’s approximate method. The book rounds this to 61%–79%. This interval concerns the population preference proportion, not the probability that one particular worker prefers an individual plan.

Exercise 4.21: polling sample size with a prior proportion

Given: 90% confidence, a target margin of 0.06, and a planning proportion p₀ = 0.30. Treat 0.06 as six percentage points. Next, solve for n using the planned variance p₀(1 − p₀) = 0.21.

By contrast, the book uses z = 1.64 and obtains 156.8933, leading to 157. Early rounding changes the integer answer here. Moreover, this planning guarantee depends on the assumed proportion; if the true proportion is closer to 0.50, the required sample can increase.

Exercise 4.22: polling without a prior proportion

(a) The general formula is n = ceiling[z*²p₀(1 − p₀)/E²]. Substituting p₀ = 0.30 reproduces the precise answer of 158. (b) When no useful prior estimate exists, maximize p₀(1 − p₀) by choosing p₀ = 0.50.

The textbook’s rounded z = 1.64 instead gives 186.7778 and an answer of 187. Consequently, full precision adds one observation in both planning cases. Relative to the p₀ = 0.30 plan, the conservative precise plan adds 30 respondents.

Exercises 4.23–4.28: Student’s t and Chebyshev

Exercise 4.23: when and why to use Student’s t

(a) For independent observations from a normal population with unknown σ, use the t statistic. (b) Its distribution is symmetric with heavier tails than the standard normal; the shape depends on degrees of freedom. (c) Compare the two standardizations below.

(d) The ordinary one-sample t statistic has n − 1 degrees of freedom. Once you estimate the sample mean, the deviations must sum to zero, so only n − 1 deviations vary freely. As degrees of freedom increase, the t distribution approaches the standard normal.

Importantly, unknown σ does not stop being unknown when n reaches 30. A t interval remains the usual mean interval when you estimate σ with s. See NIST’s confidence limits for a mean for the standard t formula.

Exercise 4.24: tail probabilities with nine degrees of freedom

Given: ν = 9. A right-tail area a requires the quantile t_1−a,9; symmetry gives the corresponding negative left-tail value. Therefore, distinguish a tail area from the area between zero and a positive cutoff.

For example, 10% in each tail leaves 80% in the center and gives ±1.383. Meanwhile, 2.5% in each tail leaves 95% in the center and gives ±2.262. At very large degrees of freedom, the positive values approach 1.282, 1.645, 1.960, and 2.576, respectively.

Exercise 4.25: the same mean with an unknown population standard deviation

Given: x̄ = 80, n = 25, s = 30, and a normal population model. Here σ is unknown, unlike exercise 4.15. Thus, use 24 degrees of freedom and estimated SE = 30/√25 = 6, ignoring the small finite correction as the book does.

These intervals are wider than the known-σ intervals in exercise 4.15. Specifically, the 95% margin increases from about 11.760 to 12.383, a rise of approximately 5.30%. The additional width accounts for uncertainty in estimating the population standard deviation.

Exercise 4.26: mean lightbulb life from nine observations

Given: n = 9, x̄ = 300 hours, s = 45 hours, and a normal lifetime population model. Because σ is unknown, use a t interval with eight degrees of freedom. For 90% confidence, t* = 1.859548.

Rounding the limits to whole hours yields the book’s 272–328 hours. However, the interval estimates the mean lifetime, not the range containing 90% of individual bulbs. The central area in the following graph belongs to the t pivot.

Exercise 4.27: a distribution-free interval with known σ

Given: n = 25, x̄ = 80, known σ = 30, N = 1,000, and a nonnormal population. The exercise requests 95% coverage without relying on a normal approximation. Chebyshev’s inequality supplies a conservative bound when the variance is known and finite.

This interval has coverage of at least 95% under the stated variance assumptions, rather than exactly 95%. Its width exceeds both the z and t intervals because it makes far fewer distributional assumptions. Including the known-σ finite correction would narrow it slightly to approximately (53.49, 106.51).

For the underlying inequality and its sample-mean derivation, see Cornell’s probability lecture on Chebyshev’s inequality. The calculation requires the variance of the estimator, not merely an observed spread with no uncertainty allowance.

Exercise 4.28: choose among normal, t, and Chebyshev methods

Accordingly, the sampling design and the source of the variance matter as much as the sample size. A small nonnormal sample does not support an exact t interval merely because software can calculate one. Conversely, a known normal population permits a z calculation with a small sample when σ is known.

Exercises 4.29–4.38: sampling distributions

Exercise 4.29: statistical inference, parameters, and statistics

(a) Statistical inference uses sample information to estimate population features or evaluate hypotheses. (b) A parameter describes a population, while a statistic describes a sample. For example, μ is a parameter and x̄ is a statistic.

(c) Random sampling supplies a probability-based foundation for representative inference. However, randomness does not guarantee that every realized sample mirrors the population perfectly. Coverage errors, nonresponse, and chance imbalances can still affect an analysis.

Exercise 4.30: read a random-number table

Task: Read specified starting positions and directions in Appendix 4 to select samples. The supplied pages show the printed answers but do not include the random-number table itself. Therefore, the sequences below reproduce the visible answer key, rather than an independently verified table traversal.

To reproduce the exercise, first label the eligible units consistently. Next, read two-digit groups for a 99-unit frame or three-digit groups for a 400-unit frame, using the appendix’s convention. Reject values outside the allowed range.

Exercise 4.31: construct a theoretical sampling distribution

(a) Finite population: For simple random sampling without replacement, enumerate every possible sample of the required size, calculate its mean, and attach its sampling probability. The resulting distribution describes x̄ across the sampling process. If the sample size is n and the population size is N, there are C(N,n) unordered samples.

(b) Infinite population model: Conceptually repeat the sampling process indefinitely and record the mean each time. In practice, probability theory derives the distribution, while simulation approximates it. Therefore, the sampling distribution is not the same object as the histogram of individual observations in one sample.

Exercise 4.32: the mean and standard error of x̄

The first equation expresses the unbiasedness of the sample mean under the stated sampling scheme. Meanwhile, the second quantifies how much sample means vary. The finite correction applies at every sampling fraction mathematically, although the textbook ignores it when the fraction is small.

Exercise 4.33: sample means from a population of 1,000

Given: N = 1,000, μ = 50, and σ = 10. In both parts, the expected sample mean remains 50. Next, compute its standard error using the exercise’s finite-population convention.

Thus, the book reports standard errors of 2 and approximately 1.07. The exact finite-population calculation for n = 25 is about 1.976, so ignoring the correction slightly overstates uncertainty. Increasing n from 25 to 81 reduces the standard error substantially without changing its center.

Exercise 4.34: the shape of a sampling distribution

(a) Under independent sampling from a normal parent distribution, the sample mean is exactly normal even when n = 10. (b) With n = 50 and a nonnormal parent, a normal approximation may be reasonable. However, its quality depends on skewness, tails, and dependence.

(c) The central limit theorem motivates the second answer under its conditions. A fixed finite population sampled without replacement requires the corresponding finite-population approximation conditions. Therefore, the familiar textbook answers summarize a model, rather than a universal sample-size rule.

Exercise 4.35: standardize an observation and a sample mean

For independent sampling with known σ, substitute SE(x̄) = σ/√n. Consequently, an average of many observations can be several standard errors from μ even when it differs by less than one individual standard deviation. Confusing these two denominators produces incorrect probability calculations.

Exercise 4.36: probability that a sample mean lies between 49 and 50

Given: μ = 48, σ = 12, and n = 36. First, SE = 12/6 = 2. Next, the standardized endpoints are (49 − 48)/2 = 0.5 and (50 − 48)/2 = 1.

The printed answer of 14.98% follows a rounded normal table. However, the population shape is unspecified in the visible question. Hence, the numerical result uses a normal approximation; the mean and standard deviation alone do not establish an exact probability.

Exercise 4.37: accounts receivable and a finite correction

Given: N = 2,000 accounts, μ = $10,000, σ = $4,000, and n = 144. The requested event is $9,500 < x̄ < $10,500. Because the sampling fraction is 7.2%, include the finite correction.

The answer key gives 88.12%, which matches rounding the cutoff to z = 1.56 before reading a normal table. Therefore, the small difference is reproducible. As in the previous exercise, the calculation relies on an adequate normal approximation to the sampling distribution.

Exercise 4.38: which estimators are unbiased?

For ordinary independent identical sampling, x̄ is unbiased for μ, p̂ is unbiased for p, and the sample variance with denominator n − 1 is unbiased for σ². However, the ordinary sample standard deviation s is generally biased downward for σ. The square-root transformation breaks the inference from unbiased variance to unbiased standard deviation.

Under normal sampling, E(s) = c₄(n)σ, where c₄(n) = √[2/(n − 1)]Γ(n/2)/Γ((n − 1)/2). Thus, s/c₄(n) supplies an unbiased normal-model estimator of σ. This correction depends on the model; it is not a universal adjustment for every distribution.

Exercises 4.39–4.48: more intervals and sample sizes

Exercise 4.39: state the central normal probability statements

These expressions describe random intervals before sampling under an exact normal pivot. If the pivot is only approximately normal, replace equality by an approximation. After observing the data, report the interval with its confidence level rather than assigning a sampling probability to the fixed parameter.

Exercise 4.40: a 90% interval from 144 observations

Given: x̄ = 300, n = 144, sample standard deviation s = 100, and N = 5,000. The sample fraction is 2.88%, so the book ignores the finite correction. Following its large-sample z method, estimated SE = 100/12 = 8.333333.

The printed limits are 286.34 and 313.66, a small rounding difference. Using the precise 90% normal quantile gives (286.29, 313.71). Since the problem supplies s rather than known σ, a t-based interval with 143 degrees of freedom is also appropriate under suitable assumptions; without the FPC it is approximately (286.20, 313.80).

Exercise 4.41: 95% and 99% intervals for the same sample

Keep the data from exercise 4.40 and change only the confidence level. Consequently, the point estimate stays at 300 and the estimated standard error remains 8.333333. The higher critical value expands each interval.

Using the book’s rounded constants gives (283.67, 316.33) and (278.50, 321.50). Its printed 99% limits of 278.51 and 321.49 reflect further rounding. Nevertheless, the substantive conclusion stays the same: higher confidence requires greater width.

Exercise 4.42: mean weight of army recruits

Given: n = 400 recruits, x̄ = 170 lb, known σ = 40 lb, and a population exceeding 100,000. The sample fraction is below 0.4%, so the correction is negligible for this exercise. Therefore, SE = 40/20 = 2 lb.

Rounding to one decimal place gives the printed 166.7–173.3 lb answer. This result estimates the mean weight of the target recruit population. It does not imply that most individual recruits weigh within this narrow interval.

Exercise 4.43: plan a lightbulb study within ten hours

Given: Known σ = 30 hours, target margin E = 10 hours, and 95% confidence. The exercise assumes that a normal approximation is adequate for the eventual sample. Apply the mean sample-size equation.

For comparison, n = 34 gives a margin above ten hours, whereas n = 35 lowers it below ten. Thus, the ceiling operation is essential to the claimed planning target. A prior estimate of σ would make this a planning approximation rather than a guarantee about the realized interval width.

Exercise 4.44: sample-size formula, total width, and missing assumptions

(a) The formula is n = ceiling[(z*σ/E)²]. (b) A ±10-hour margin produces a 20-hour total width. These quantities should never share the same symbol without explanation.

(c) If the calculated n were below 30, the book would raise it to 30 to support its normal approximation. However, there is no universal cutoff that guarantees adequate coverage. For known normal lifetimes, a smaller n can be valid; for highly skewed lifetimes, even 30 may need scrutiny.

Exercise 4.45: binomial count and proportion formulas

In the finite-population version requested by the book, multiply each proportion standard error by √[(N − n)/(N − 1)]. However, sampling without replacement introduces dependence and leads to a hypergeometric count model. This distinction explains why the same proportion formulas can need different variance adjustments.

Exercise 4.46: mathematics backgrounds among graduate students

Given: Eight of n = 36 sampled economics graduate students hold mathematics degrees; N = 880. Thus, p̂ = 8/36 = 0.222222. The 4.09% sampling fraction falls below the book’s correction threshold.

Therefore, the direct calculation gives about 10.83%–33.62%. The printed answer of 11%–33% appears to use coarser intermediate rounding. With eight successes, a basic normal approximation deserves extra care even though the book’s older five-success rule passes.

For a binomial-model comparison, the Wilson interval is approximately (0.130, 0.353) at 90% confidence, ignoring the small finite correction. It does not simply center a symmetric interval on p̂. See NIST’s Wilson and exact proportion interval reference for the alternative construction.

Exercise 4.47: sample size for a defective-bulb proportion

Given: Planned defective proportion p₀ = 0.20, target margin E = 0.10, and 95% confidence. Here the margin is ten percentage points. First, compute the planned Bernoulli variance: 0.20 × 0.80 = 0.16.

This result matches the printed answer. However, the planned precision depends on p₀ = 0.20. A different true defect rate can change the achieved standard error, and the Wald planning formula does not guarantee exact finite-sample coverage.

Exercise 4.48: plan the same study without a prior defect rate

(a) Use n = ceiling[z*²p₀(1 − p₀)/E²]. (b) Without a defensible prior estimate, choose p₀ = 0.50 to maximize the planned variance. Consequently, the required sample grows.

The conservative plan adds 35 bulbs relative to the 62-bulb plan. That is approximately a 56.45% increase in sample size. This difference illustrates the practical value of reliable preliminary information about the expected proportion.

Exercises 4.49–4.54: t intervals and method selection

Exercise 4.49: t quantiles and an apparent percentage typo

Given: 29 degrees of freedom. The visible question asks for right-tail areas of 10%, 5%, 2.5%, and 0.05%. To avoid ambiguity, convert every percentage to a decimal probability before locating the quantile.

Exercise 4.50: convert areas from zero into normal quantiles

A standard normal distribution places half its probability below zero. Therefore, an area A between zero and a positive cutoff corresponds to cumulative probability 0.5 + A. Apply z = Φ⁻¹(0.5 + A).

The t cutoffs exceed the corresponding normal cutoffs because of heavier tails. In particular, the final comparison is 2.575829 versus 2.756386. That comparison uses a 0.5% right tail, reinforcing the interpretation discussed in exercise 4.49.

Exercise 4.51: a 95% mean interval with sixteen observations

Given: n = 16, x̄ = 50, s = 10, and a very large normal population. (a) With unknown σ, use 15 degrees of freedom. The estimated standard error is 10/4 = 2.5.

(b) If the problem instead gives known σ = 10, use z rather than t. The standard error stays numerically equal to 2.5, but the critical value falls. Hence, the interval becomes narrower.

Exercise 4.52: mean exam grade from only four students

Given: n = 4, x̄ = 75, s = 8, and normally distributed population grades. Thus, ν = 3 and estimated SE = 8/2 = 4. A very small sample makes the t critical values large.

The results round to about 62–88 at 95% confidence and 52–98 at 99% confidence, matching the book. However, these calculations depend heavily on the stated normal model. Four observations provide little ability to assess that assumption from the sample alone.

Exercise 4.53: why a plug-in Chebyshev interval is not guaranteed

Given: n = 16, x̄ = 50, s = 10, and a very large nonnormal population. The book explicitly treats s as a rough estimate of σ. Following that approximation gives the arithmetic below.

This rounds to the printed 39–61 range. Nevertheless, Chebyshev’s inequality needs a known variance or a valid upper variance bound to guarantee coverage. Substituting the random quantity s into the bound does not automatically produce a valid distribution-free 95% interval.

Therefore, label (38.82, 61.18) as the textbook’s heuristic range, not a guaranteed confidence interval. Additional assumptions, a valid variance bound, or another justified inferential method would be necessary for a rigorous coverage claim. Compared with exercise 4.51, the range is much wider, but width alone cannot establish validity.

Exercise 4.54: choose a method for three scenarios

This final exercise brings the chapter’s main decision together. First, determine whether the dispersion comes from a known population quantity or an estimate. Next, examine the shape and sampling design. Only then choose a distribution or bound that supports the requested coverage claim.

Answer-key audit: details that change the conclusion

How much does the confidence level cost?

For fixed variance and margin, required sample size is proportional to z*². Therefore, changing from 90% to 95% confidence requires about 1.420 times the unrounded sample size. Moving from 95% to 99% requires about 1.727 times as many observations.

These comparisons are mathematical planning results, not survey observations. Moreover, integer rounding and a finite population can change the final head counts. Still, the ratios help explain why very high confidence can become expensive.

When a finite population changes sample-size planning

Suppose you know σ under the population-variance convention used in the exercises and plan simple random sampling without replacement. Starting from E = z*σ√[(N − n)/(n(N − 1))], define n₀ = (z*σ/E)². Solving for n gives:

For example, if N = 500 and n₀ = 100, the threshold becomes 50,000/599 = 83.4725. Therefore, the finite-population plan requires 84 observations rather than 100. This result applies to estimating that finite population’s mean; a broader future population may require a different model.

A better comparison than simply asking which interval is shorter

A narrow interval is useful only when its coverage claim is credible. For example, an unjustified z interval can look impressively precise while understating uncertainty. Conversely, a conservative bound can remain valid while producing a range too wide for a practical decision.

In applied work, report the estimate, interval, confidence level, method, and sampling assumptions together. Also explain any important data limitations. A survey interval that excludes nonresponse bias should not appear to measure every source of uncertainty.

Frequently asked questions about confidence intervals

Does a 95% confidence interval contain 95% of observations?

No. A confidence interval for μ concerns the population mean. By contrast, an interval for individual observations addresses a different target and usually needs a wider construction. Therefore, identify the parameter before interpreting the endpoints.

Should I always use z when n is greater than 30?

No. When σ is unknown, t remains the standard choice for a mean under the usual normal model. As n grows, t and z become numerically similar. However, a large sample alone does not resolve selection bias, dependence, or extreme heavy tails.

Why does the sample size formula use 0.5 for an unknown proportion?

Because p(1 − p) reaches its maximum of 0.25 at p = 0.5. As a result, that choice gives the largest planned variance in the usual proportion formula. A reliable preliminary estimate can reduce the planned sample, but an unreliable estimate can understate the requirement.

Is the textbook Wald proportion interval always a good choice?

No. Its simple symmetric form can have poor coverage, especially for small samples or proportions near zero or one. Wilson and exact binomial methods address different limitations, so choose with the sampling model in mind. NIST provides both constructions in its proportion confidence interval guidance.

Why do my calculator and the answer key disagree slightly?

Common causes include rounded critical values, rounded standard errors, and different finite-variance conventions. Therefore, keep several decimal places until the final step. When planning a minimum sample size, use the full critical value and round the final n upward.

Can I claim a Chebyshev confidence interval using only s?

Not as a general distribution-free guarantee. The sample standard deviation is random and can underestimate the population spread. Consequently, plugging it into a known-variance bound needs additional justification; the arithmetic alone does not establish coverage.

Use the interval as a decision tool

Confidence intervals become useful when the method, arithmetic, and interpretation agree. Begin with the target parameter, distinguish known σ from estimated s, and account for the sampling design. Then calculate carefully and explain what repeated-sampling coverage means.

The exercises also show why checking an answer key can teach more than copying it. A single rounded critical value can alter a minimum sample size, while an unstated variance assumption can invalidate an entire coverage claim. Ultimately, a strong solution makes those assumptions visible and gives readers enough detail to reproduce the result.

Sources and calculation notes

The numerical scenarios above come from the user-supplied photographs of exercises 4.13–4.54, including the visible printed solutions. They are educational inputs, not current real-world survey or economic data. Exercise statements here are concise paraphrases; the explanations, recalculations, comparison tables, and figures are original to this guide.

1. NIST/SEMATECH: What are confidence intervals? Repeated-sampling interpretation and the known-σ interval.
2. NIST/SEMATECH: Confidence limits for the mean. Student’s t construction, degrees of freedom, and interval width.
3. Penn State STAT 506: Estimating population mean and total under SRS. Finite-population variance conventions and simple random sampling.
4. Penn State STAT 506: Confidence intervals and sample size. Sampling precision and sample-size planning.
5. NIST/SEMATECH: Confidence intervals for proportions. Wilson and exact binomial alternatives.
6. Cornell CS 2800: Chebyshev’s inequality and the weak law of large numbers. The variance-based probability bound.

Calculation convention: values labeled “precise” use numerical normal or t quantiles before final rounding. Figures depict the stated mathematical models rather than measured datasets. Unless explicitly noted, the worked textbook intervals retain the exercise’s choice to ignore finite corrections below a 5% sampling fraction.