Statistical Inference: Estimation and Solved Exercises

A bank rarely needs to inspect every account to study average savings. Similarly, a factory can learn about battery life without destroying its entire inventory. The challenge, however, goes beyond calculating an average: researchers must explain how much uncertainty surrounds that average.

Throughout this article, textbook quantities serve as teaching examples rather than observations from a current bank, factory, or education survey. Tables and figures identify original calculations. Consequently, readers can separate mathematical results from published reference material.

Statistical inference: what estimation actually answers

Descriptive statistics summarize the data you have. Statistical inference asks what those data imply about a broader population. For example, a sample average of $3,000 describes selected accounts; an estimate of the average across all eligible accounts concerns a population parameter.

Population, sample, parameter, and statistic

Before sampling, X̄ varies because different selections produce different means. After observation, x̄ is the particular number you calculated. Therefore, distinguishing an estimator from an estimate helps explain why sampling distributions exist.

Point estimation versus interval estimation

A point estimate reports one value, such as a mean battery life of five hours. By contrast, an interval estimate reports a range together with a procedure’s confidence level. Both approaches target the same population feature, but the interval also communicates sampling uncertainty.

Hypothesis testing asks whether the data conflict with a specified parameter value under a model. A confidence interval instead displays a collection of parameter values compatible with the corresponding test. For matching two-sided procedures, a 95% interval excludes exactly the null values that a 5% test rejects.

Reference: NIST: relationship between tests and confidence intervals.

Why a random sample matters

A simple random sample without replacement gives every subset of the requested size the same selection probability. Consequently, each population member has the same inclusion probability n/N. Equal individual chances alone, however, do not fully define simple random sampling: some designs give equal individual chances while restricting which groups can appear together.

Random selection supports valid probability calculations, but it does not guarantee a perfectly representative realized sample. Chance can still produce imbalance. Moreover, an incomplete sampling frame or selective nonresponse can introduce bias even when the initial selection follows a random procedure.

For instance, sampling every Monday’s production may repeatedly capture a special maintenance schedule. A larger sample from that schedule would still miss other operating conditions. Therefore, defining the population and selection process comes before choosing an impressive sample size.

Sampling distributions and the standard error

A distribution of means differs from a distribution of observations

Imagine repeatedly drawing 25 accounts and recording the average balance each time. The collection of possible averages forms the sampling distribution of the mean. Individual balances may vary widely; nevertheless, averages tend to cluster more tightly because positive and negative deviations partly offset one another.

Under independent sampling from a common population with finite variance, the mean’s variance equals σ²/n. Taking a square root gives its standard error. Notice that the standard error describes the spread of repeated estimates, whereas the population standard deviation describes the spread of individual observations.

Reference: Penn State STAT 500: sampling distributions.

How sample size changes precision

Increasing n from 25 to 100 cuts the standard error from 12 to 6. In contrast, doubling n reduces the error only to about 70.71% of its previous value. Accordingly, achieving twice the precision generally requires four times as many independent observations.

A lower standard error does not prove a lower total error. For an estimator with bias b, mean squared error equals its variance plus b². As a result, collecting more observations can leave a persistent selection bias almost untouched.

Finite population correction: when sampling changes dependence

Sampling without replacement creates dependence between selected units. Once you draw a particular account, you cannot draw it again. Therefore, the sample contains an increasing share of the available information as n approaches N.

Here, σ² means the finite population variance calculated with denominator N. The correction becomes zero when n = N because a complete census leaves no sampling variation in the mean. However, a census can still contain recording errors or missing measurements.

Reference: OpenStax: sampling-distribution and finite-population formulas.

Many courses ignore the correction when the sample covers less than 5% of the population. This convention simplifies arithmetic; it does not switch the mathematics on at exactly 5%. In fact, the correction applies to every simple random sample without replacement, although a small sampling fraction often makes its practical effect minor.

Why some references use a different-looking correction

Survey sampling often defines the finite population variance with denominator N − 1 and calls it S². Under that convention, Var(X̄) = (1 − n/N)S²/n. Because S² = Nσ²/(N − 1), both conventions produce the same answer.

When you estimate uncertainty from an observed sample variance s², the design-based estimated standard error is s√[(1 − n/N)/n]. Consequently, inserting s directly into the σ formula introduces a small denominator mismatch. The difference often looks tiny, but identifying the convention prevents hidden inconsistencies.

Reference: Penn State STAT 506: estimating population means under simple random sampling.

The central limit theorem and the limits of n = 30

For independent, identically distributed observations with a finite mean and finite variance, the standardized sample mean approaches a standard normal distribution as n grows. This result explains the broad usefulness of normal approximations. However, it does not guarantee an accurate approximation for every population at n = 30.

Strong skewness, rare extreme values, and dependence can require additional care. Conversely, independent observations from a normal population produce a normal sample mean at every sample size. Thus, the population shape and sampling design matter alongside n.

Reference: Penn State STAT 414: central limit theorem and finite-variance assumptions.

A useful counterexample uses independent Bernoulli observations with p = 0.001. Even at n = 30, the chance of observing zero successes equals 0.999³⁰ ≈ 97.04%. Consequently, the sample mean piles up at zero rather than resembling a smooth bell curve. This calculation shows why a universal 30-observation promise fails.

Confidence intervals, z scores, and sample size

Standardize the mean with its standard error

To calculate a probability between two possible sample means, standardize both endpoints. Then subtract their standard normal cumulative probabilities. Throughout, Φ(z) denotes the probability that a standard normal variable falls at or below z.

An individual-observation probability uses σ in the denominator. A sample-mean probability instead uses SE. Therefore, confusing these quantities can change an answer dramatically, even when the numerator remains identical.

Known population standard deviation: the z interval

A two-sided interval places α/2 in each tail. For example, 95% confidence uses α = 0.05 and z₀.₉₇₅ ≈ 1.959964. If sampling without replacement accounts for a material share of a finite population, apply the appropriate finite population correction to the standard error.

Reference: NIST: confidence intervals with known population standard deviation.

Unknown population standard deviation: the t interval

When σ is unknown, the conventional one-sample mean interval uses Student’s t distribution. Under independent normal observations, this procedure has exact coverage. With suitable large-sample conditions, it can also provide an approximation for nonnormal populations.

Degrees of freedom equal n − 1 because estimating the sample mean imposes one constraint on deviations from that mean. Furthermore, t critical values exceed the corresponding normal values at finite degrees of freedom. As n increases, the difference shrinks.

Reference: NIST: confidence limits for the mean.

What 95% confidence means

A 95% confidence procedure captures the fixed population parameter in about 95% of repeated samples when its assumptions hold. Once you calculate a particular interval, that interval either contains the parameter or it does not. Consequently, the frequentist confidence level describes the procedure’s long-run behavior.

The interval does not imply that 95% of observations lie between its endpoints. Similarly, it does not automatically describe the next observation. A prediction interval answers that separate question and generally includes substantially more uncertainty.

Planning sample size for a population mean

Here, E represents the desired margin of error, or half-width. Always round the required sample size upward. Otherwise, the planned interval can exceed the requested margin even before practical data-collection problems arise.

For a finite population sampled without replacement, solve E² = z²σ²(N − n)/[n(N − 1)]. Rearranging gives n = Nz²σ²/[E²(N − 1) + z²σ²], followed by upward rounding. This expression assumes σ² uses denominator N; sample-based planning needs a justified variance estimate.

Estimating a population proportion

The familiar Wald interval adds and subtracts z times this estimated standard error. Nevertheless, it can perform poorly near zero or one, especially with sparse successes or failures. Wilson’s score interval offers a useful alternative and keeps endpoints within the parameter’s valid range.

Reference: NIST: Wilson and exact binomial confidence intervals.

For proportion planning without a prior estimate, p = 0.5 maximizes p(1 − p). As a result, a 95% margin of ±5 percentage points requires ceil(1.959964² × 0.25/0.05²) = 385 observations under simple independent sampling. This figure excludes design effects, finite population gains, and nonresponse adjustments.

Recalculated worked examples from the supplied pages

The following solutions retain the numerical inputs in Examples 1–8 while explaining the reasoning in original language. They also separate printed rounding conventions from calculations that retain more precision. Therefore, small differences have an explicit mathematical explanation.

Example 1: choosing 5 employees from 80

Label employees 1 through 80 and select five distinct labels uniformly without replacement. Each employee has inclusion probability 5/80 = 6.25%. One valid selection is {13, 19, 54, 59, 71}; however, many other selections have exactly the same probability.

A random-digit procedure must reject 00, values above 80, and repeated labels. Alternatively, a computer can shuffle the complete list and take the first five labels. Neither method should replace inconvenient selections after seeing employee characteristics.

Example 2: comparing sampling distributions

Both n = 5 and n = 20 target the same mean μ. Under independent sampling, their standard errors equal σ/√5 and σ/√20. Consequently, the second standard error is exactly half the first, even though its sample contains four times as many observations.

Example 3: N = 900, μ = 20, σ = 12

For n = 36, the sampling fraction equals 4%, so the chapter uses SE = 2 under its 5% convention. At n = 64, the fraction exceeds 5%, and the corrected error equals 1.446487. Accordingly, rounding the correction only at the end gives 1.45 rather than the printed 1.44.

Example 4: probability that the sample mean falls between 18 and 24

Using the chapter’s negligible-fraction convention gives approximately 81.86%. Its printed 81.85% reflects rounded table entries. If the population is genuinely finite and sampling occurs without replacement, the exact finite-N standard error gives a slightly different normal approximation; the population’s shape still determines approximation quality.

Example 5: estimating a mean with n = 144, x̄ = 100, s = 60, N = 1,000

The printed interval [90.89, 109.11] comes from rounding the correction to 0.93 early. Keeping precision produces approximately [90.93, 109.07] under the same plug-in convention. However, the given spread is a sample standard deviation, so this z interval remains a large-sample approximation.

Under the design-based sample-variance convention, the estimated SE is 5√0.856 ≈ 4.626013. Combining that estimate with t₀.₉₇₅,₁₄₃ gives a survey-style approximate interval of [90.8558, 109.1442]. This second calculation is not an exact finite-population t theorem; it illustrates how a stated convention changes the result.

Example 6: sample size for a ±3-minute margin at 90% confidence

The rounded critical value 1.64 gives 67.24, but both calculations round upward to 68. Furthermore, the requested ±3 minutes represents the half-width, so the complete planned interval spans six minutes. If substantial nonresponse occurs, invite more than 68 units to obtain 68 completed observations.

Example 7: 40 graduates among 100 college attendees

The target population consists of people who attended college, rather than all adults. Therefore, the estimate describes the graduate share conditional on college attendance. Changing that denominator would change the research question, even if the arithmetic remained identical.

Example 8: battery life with n = 10, x̄ = 5 hours, s = 1 hour

Rounding the endpoints at the end yields [4.28, 5.72] hours. The supplied page reports [4.29, 5.71] after rounding the margin early. Under the stated normality and independence assumptions, the t method addresses uncertainty about the population mean, rather than the lifetime of each battery.

Solved exercises: statistical inference problems 4.1–4.12

Exercise 4.1: inference, parameters, estimation, and testing

Part (a). Statistical inference uses a sample to learn about a defined population. Its value comes from making uncertainty explicit when a census would cost too much, take too long, or destroy the products under inspection. For example, destructive battery testing can estimate average life while preserving most inventory.

Part (b). A parameter describes the population, while a statistic describes a sample. Thus, μ and σ summarize the population; x̄ and s summarize observed units. A statistic becomes useful for inference when its sampling behavior connects it to the parameter of interest.

Part (c). Estimation supplies a point or interval for an unknown parameter. In contrast, testing evaluates a stated hypothesis using sample evidence and a chosen error criterion. Failure to reject a null hypothesis does not prove that the hypothesis is true.

Exercise 4.2: random sampling and its importance

A random sampling design uses a probability mechanism to select observations. For simple random sampling without replacement, every subset of size n has probability 1/C(N,n). Consequently, each individual has probability n/N of inclusion.

This design allows researchers to derive sampling variances and confidence procedures. Nevertheless, randomization does not guarantee balance in every sample. Stratification can ensure that specified groups enter the sample while preserving probability-based selection within those groups.

Exercise 4.3: obtaining samples from employees and manufactured parts

Part (a). Construct a complete frame, assign unique labels, and use a random-number generator or a properly read random-digit table. When sampling without replacement, reject duplicates. In addition, document the starting rule or random seed when reproducibility matters.

Part (b). For 10 employees from 95, read two-digit labels and retain distinct values from 01 through 95. The visible solution gives {60, 39, 4, 34, 76, 43, 52, 14, 8, 95}. Each employee’s inclusion probability equals 10/95 ≈ 10.53%.

Part (c). For 12 parts from 240, read three-digit labels and retain distinct values from 001 through 240. The page’s solution lists only eight labels: {215, 182, 51, 9, 127, 177, 53, 186}. Therefore, that displayed answer does not satisfy the requested sample size.

In a complete simple random sample of 12 from 240, every part has inclusion probability 12/240 = 5%. However, merely writing twelve distinct labels does not establish that a random mechanism selected them. The selection procedure supplies that justification.

Exercise 4.4: constructing the sampling distribution of the mean

Part (a). List all possible samples under the specified design, calculate each sample mean, and assign probabilities using the design. For a small finite population, exact enumeration works. Alternatively, repeated simulation approximates the same distribution when exhaustive enumeration becomes impractical.

Part (b). The distribution’s mean equals E(X̄), and its standard deviation equals SE(X̄). Consequently, a mean describes the center of repeated estimates, while a standard error describes their spread. Neither quantity requires the realized sample itself to resemble a bell curve.

Exercise 4.5: deriving the mean and standard error

For independent observations, variances add without covariance terms. Therefore, Var(X̄) = (1/n²)(nσ²) = σ²/n. Taking a square root yields σ/√n.

Without replacement, different draws have covariance −σ²/(N − 1). Incorporating the n(n − 1) ordered covariance terms gives the finite-population variance below. This derivation also explains why the correction reduces uncertainty.

Exercise 4.6: enumerate samples from {1, 3, 5, 7, 9}

Part (a). The population mean equals 25/5 = 5. Squared deviations from five are 16, 4, 0, 4, and 16, so the population variance equals 40/5 = 8. Accordingly, σ = √8 ≈ 2.828427.

Part (b). Select two different values without replacement and ignore order. There are C(5,2) = 10 equally likely samples. Their means form the following exact sampling distribution.

Part (c). The mean of the sampling distribution equals 5. Its standard error uses N = 5 and n = 2, so the correction cannot reasonably be ignored. Consequently, SE = (√8/√2)√(3/4) = √3 ≈ 1.732051.

Exercise 4.7: verify the distribution directly

Therefore, the standard error again equals √3 ≈ 1.732051. This direct calculation confirms the finite-population formula. Notice that the printed working contains a deviation term (7 − 8)²; the correct term is (7 − 5)², which matches its final sum.

An additional comparison: what changes with replacement?

With replacement, all 25 ordered pairs become possible, including repeated values such as (1,1). Consequently, the mean can now range from 1 to 9. The five original values still have mean five, but the sampling variance changes.

Consequently, the without-replacement standard error is about 13.40% smaller than the with-replacement value. This side-by-side calculation isolates the sampling design’s effect while holding the population and sample size fixed. It also shows why “finite population” alone does not justify a correction: replacement matters.

Exercise 4.8: N = 12,000, μ = 100, σ = 60

Part (a), n = 100. The sampling distribution has mean 100. Ignoring the small 0.8333% sampling fraction gives SE = 60/√100 = 6, as in the chapter. Keeping the finite correction instead gives approximately 5.975197.

Part (b), n = 900. The sampling fraction equals 7.5%. Therefore, the chapter applies the correction, producing the following result.

Thus, the expected sample mean remains 100, while the corrected standard error rounds to 1.92. Relative to the uncorrected SE of 2, the correction lowers uncertainty by approximately 3.82%. A larger sample narrows the distribution without shifting its expected center.

Exercise 4.9: when is the sampling distribution normal?

Part (a). Independent observations from a normal population yield a normal sample mean at every n. For a nonnormal population, an appropriate central limit theorem may justify an approximate normal shape as n grows. Finite variance and suitable independence or design conditions remain essential.

Part (b). Normal approximations let researchers calculate probabilities and construct intervals using a standardized scale. However, n ≥ 30 is a classroom guideline, rather than a universal guarantee. For a finite population sampled without replacement, exact normality does not follow merely from the usual independent normal argument; the finite design requires separate consideration.

Exercise 4.10: interval probabilities and the empirical rule

Part (a). Calculate the sampling standard error, convert each endpoint to a z score, and subtract cumulative probabilities. For a single observation, replace SE with σ. Therefore, the same normal table can answer both questions, provided you use the correct scale.

Approximately 68.27%, 95.45%, and 99.73% of normal sample means lie within one, two, and three standard errors of μ. The chapter’s nearby values reflect its normal-table rounding. Furthermore, the familiar 95% confidence multiplier is 1.96, whereas two standard errors correspond to about 95.45%.

Exercise 4.11: P(X̄ > 100) with μ = 90, σ = 60, n = 25

The exercise specifies a normal population, so the independent-sampling mean is normal despite n being less than 30. Assume a negligible sampling fraction or sampling with replacement, since no finite N appears. First, calculate SE = 60/5 = 12.

The printed result of 20.33% uses z = 0.83. Retaining the unrounded z score gives 20.23%. In addition, the source figure labels the center 95 even though the exercise specifies μ = 90; the graph’s correct center is 90.

Exercise 4.12: bank accounts and a finite population

Use N = 1,450 accounts, μ = $3,000, σ = $1,200, and n = 100. The sample covers about 6.90% of accounts, so retain the finite correction. This assumes a simple random sample without replacement from the specified account population.

The supplied answer of 4.18% rounds the z score to −1.73 before consulting a table. Without the finite correction, the probability would be about 4.78%. However, knowledge of only N, μ, and σ cannot determine the exact finite-population probability; an extremely skewed balance distribution can weaken the approximation.

Numerical audit: corrections that improve the lesson

Not every difference changes the practical conclusion. Nevertheless, explaining its source helps students distinguish a conceptual mistake from a rounding artifact. A reliable workflow keeps intermediate precision and rounds only the reported answer.

Unbiased variance does not imply unbiased standard deviation

The sample variance with denominator n − 1 is unbiased for σ² under independent identically distributed sampling with finite variance. However, taking a square root does not preserve unbiasedness. As a result, the familiar sample standard deviation s generally underestimates σ on average.

Under normal sampling, E(s) = c₄(n)σ, where c₄(n) = √[2/(n − 1)] Γ(n/2)/Γ[(n − 1)/2]. For n = 10, c₄ ≈ 0.972659, so the expected sample standard deviation is about 2.73% below σ. This calculation clarifies a distinction that many introductory treatments compress into one sentence.

Reference: NIST: control-chart constants and estimation of process standard deviation.

Why a confidence interval can miss the real problem

Suppose a satisfaction survey estimates 90% approval with a narrow interval, but only delighted customers answer. The interval may describe random error under an assumed sampling model while leaving selection bias unresolved. Consequently, numerical precision cannot compensate for a poor connection between respondents and the target population.

Similarly, averaging correlated daily sales as if every day supplied independent information can understate uncertainty. A seasonal pattern or common advertising campaign may connect observations. Therefore, check the unit of analysis before treating the row count as the effective sample size.

Practical applications in economics and business

Estimating an average is different from forecasting a person

A bank might use a confidence interval to estimate average account balances. However, a marketing team might need a prediction interval for the balance of a newly acquired customer. The first target concerns a population mean, while the second includes individual variation.

For example, the battery calculation makes the difference visible. With n = 10, x̄ = 5, and s = 1, the mean interval is about [4.28, 5.72] hours. Under independent normal sampling, a 95% prediction interval for one additional battery uses s√(1 + 1/n), producing a much wider interval of approximately [2.63, 7.37] hours.

Reference: NIST: prediction intervals.

Report units and percentage points clearly

An interval from 27.38% to 52.62% has a half-width of about 12.62 percentage points. That is not the same as a 12.62% relative change. For example, increasing a share from 40% to 45% means a five-percentage-point rise, or a 12.5% relative increase.

Means and standard errors share the original measurement units. In contrast, variance uses squared units and z scores have no units. Accordingly, adding a dollar-valued standard deviation to a proportion should immediately signal a formula or data-entry error.

A repeatable estimation workflow

1. Define the target. State the population, time period, unit, and parameter.
2. Check selection. Identify the sampling frame, replacement rule, and nonresponse.
3. Inspect observations. Look for skewness, outliers, dependence, and missing values.
4. Choose the standard error. Match it to the estimator and actual sampling design.
5. Select the procedure. Explain why z, t, Wilson, or another method fits the problem.
6. Calculate without early rounding. Preserve precision through the final endpoint or probability.
7. Interpret the target correctly. Distinguish mean uncertainty from individual prediction.
8. Report limitations. Explain approximation quality and sources of bias beyond sampling error.

A compact reporting template

For real studies, replace any known population inputs with justified estimates or a model appropriate to the research design. Furthermore, describe weighting or clustering when those features change the standard error. A short, accurate methods statement often adds more value than another decimal place.

Frequently asked questions about statistical inference

Is the standard error the same as the margin of error?

No. The standard error measures the sampling spread of an estimator. A margin of error multiplies that error by a critical value, such as 1.959964 for a two-sided 95% z interval. Therefore, changing the confidence level changes the margin without changing the underlying standard error.

Should I always use z when n exceeds 30?

No. When estimating an unknown population standard deviation from the sample, the conventional mean interval uses t. At larger sample sizes, t and z become close. However, sample size alone does not make σ known or remove distributional and design concerns.

Does a larger sample guarantee a better estimate?

A larger independent random sample generally lowers sampling variance. Nevertheless, it cannot guarantee that one realized estimate lies closer to the truth. Persistent nonresponse bias or systematic measurement error may also dominate any gain in sampling precision.

Can I reconstruct a probability from only a mean and standard deviation?

Usually, no. Many distributions share the same mean and variance but assign different probabilities to an event. A normal model or a justified approximation supplies the additional structure needed for the chapter’s probability calculations.

Can confidence intervals overlap when a difference is significant?

Yes. Separate intervals for two means do not directly equal an interval for their difference. The relevant standard error depends on both variances and any covariance between estimates. Consequently, compare the difference using its own interval or corresponding test.

What if a small sample comes from a strongly nonnormal population?

A t interval may lose its intended coverage, particularly with extreme skewness or outliers. Depending on the target and available information, researchers may consider a suitable model, a carefully assessed bootstrap, robust methods, or additional data. Chebyshev’s inequality alone does not automatically provide a convenient exact interval when the population variance is unknown.

Conclusion: connect every estimate to its sampling design

Statistical inference becomes clearer when you separate the population, the sample, the estimator, and its sampling distribution. From there, the standard error connects the estimate to an interval or probability. Finally, careful assumptions and transparent rounding make the result interpretable.

The worked exercises show that the same population can produce different uncertainty under different sampling designs. They also demonstrate why small arithmetic checks matter. Use these solutions as a model: define the question, choose the correct uncertainty formula, calculate carefully, and explain what the result actually describes.

References and calculation notes

The supplied photographs contain chapter pages 67–76 and exercises 4.1–4.12. This article paraphrases their prompts and supplies original explanations, calculations, tables, and figures. No book title or edition appears clearly enough in the provided pages to verify a full bibliographic citation.

• NIST: confidence limits for the mean
  https://www.itl.nist.gov/div898/handbook/eda/section3/eda352.htm
• NIST: confidence interval concepts
  https://www.itl.nist.gov/div898/handbook/prc/section1/prc14.htm
• NIST: tests and confidence intervals
  https://itl.nist.gov/div898/handbook/prc/section1/prc15.htm
• NIST: Wilson and exact binomial intervals
  https://www.itl.nist.gov/div898/handbook/prc/section2/prc241.htm
• Penn State: simple random sampling and finite population variance
  https://online.stat.psu.edu/stat506/Lesson01
• Penn State: the central limit theorem
  https://online.stat.psu.edu/stat414/Lesson27
• Penn State: sampling distributions
  https://online.stat.psu.edu/stat500/Lesson04
• OpenStax: sampling distribution formulas
  https://openstax.org/books/introductory-business-statistics-2e/pages/7-formula-review
• NIST: control charts and c₄
  https://www.itl.nist.gov/div898/handbook/pmc/section3/pmc32.htm
• NIST: prediction intervals
  https://www.itl.nist.gov/div898/software/dataplot/refman1/auxillar/predlimi.htm

Calculation notes: normal probabilities use the standard normal CDF; t critical values use the stated degrees of freedom. Numerical tables retain additional precision internally. The finite population formulas state their variance convention explicitly. All plotted values are theoretical or simulated teaching examples, rather than empirical claims about a current institution.