Statistics for Engineers

Chapter 1

Introduction to Statistics and Engineering Data

Engineers measure things: the diameter of a shaft, the breaking strength of a weld, the number of defective circuit boards in a shipment. Statistics is the science of collecting, organizing, analyzing, and interpreting those measurements so that good decisions can be made in the presence of variation. No two machined parts are ever exactly identical, and no two test results ever agree perfectly — statistics gives you the tools to describe that variation and to draw reliable conclusions from it.

1.1 Population, Sample, Parameter, Statistic

A population is the entire collection of individuals or items under study — every bearing produced this month, every concrete cylinder poured on a project. A sample is a subset of the population that is actually observed or measured. We use samples because measuring an entire population is usually too expensive, too slow, or physically impossible (destructive testing destroys the part!).

A number that describes a population is a parameter (for example, the true mean diameter μ of all bearings). A number computed from a sample is a statistic (for example, the sample mean x̄ of 50 inspected bearings). Greek letters (μ, σ, p) are traditionally used for parameters; Roman letters (x̄, s, p̂) for statistics.

1.2 Types of Data

Qualitative (categorical) data place items into categories: material type, pass/fail, vendor name. Quantitative data are numerical measurements or counts. Quantitative data are further split into:

• Discrete data — countable values, usually whole numbers: number of defects on a panel, number of trucks arriving per hour.
• Continuous data — any value in an interval, limited only by the precision of the instrument: length, mass, temperature, voltage, time.

1.3 Levels of Measurement

Data can also be classified by how much mathematical meaning the numbers carry:

• Nominal — names or labels only, no order (steel / aluminum / titanium).
• Ordinal — ordered categories, but differences are not meaningful (surface finish rated poor / fair / good / excellent).
• Interval — ordered with meaningful differences but no true zero; ratios are meaningless (temperature in °C or °F).
• Ratio — meaningful differences and a true zero, so ratios make sense (length, mass, force, time, temperature in kelvin).

1.4 Collecting Data: Sampling Methods

The value of any statistical conclusion depends on how the data were collected. Common sampling designs:

• Simple random sample — every group of n items in the population has the same chance of being chosen (e.g., select 50 serial numbers with a random number generator).
• Systematic sample — select every kth item after a random start (every 20th part off the line).
• Stratified sample — divide the population into subgroups (strata) such as machines or shifts, then randomly sample from each.
• Cluster sample — randomly select whole groups (entire pallets, entire crates) and inspect everything in the chosen groups.
• Convenience sample — take whatever is easiest. Fast, but prone to bias; avoid it for serious work.

In an observational study, the engineer records data without controlling conditions. In a designed experiment, the engineer deliberately changes one or more factors (the treatments) and observes the effect on a response variable. Only well-designed experiments can establish cause and effect.

Worked examples

Try each example with pencil and calculator first, then open the solution and check yourself line by line.

Practice problems

Each problem mirrors the matching worked example above. Reveal an answer only after you have attempted the problem.

1. 1.1 A civil engineer tests 12 concrete cylinders out of the 300 poured on a bridge project and finds a mean strength of 31.2 MPa for the 12 cylinders. Identify the population, the sample, the statistic, and the parameter of interest.

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2. 1.2 Classify each variable as qualitative or quantitative, and discrete or continuous where appropriate: (a) voltage output of a power supply, (b) number of warranty claims per month, (c) supplier name, (d) operating temperature of a motor, (e) number of rivets in a wing panel.

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3. 1.3 Give the level of measurement: (a) ZIP codes of customers, (b) Mohs hardness ranking of minerals, (c) boiling point in °F, (d) current in amperes, (e) customer satisfaction rated 1–5 (very poor to very good).

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4. 1.4 Identify the sampling method: (a) 15 parts randomly chosen from each of four CNC machines; (b) every 50th unit is pulled for inspection; (c) the technician grabs the 30 parts nearest the door; (d) three full pallets are randomly selected and every box on them is opened.

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5. 1.5 To test whether preheating reduces weld cracking, 40 identical joints are randomly split into two groups of 20; one group is preheated before welding and the other is not, and cracks are counted. Is this observational or experimental? Name the treatment and the response variable.

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Chapter 2

Organizing and Visualizing Data

Raw data are just a pile of numbers. The first job of the engineer is to organize them so that patterns become visible: Where is the center? How spread out are the values? Is the distribution symmetric or skewed? Are there gaps or outliers? Frequency tables, histograms, and stem-and-leaf displays answer these questions at a glance.

2.1 Frequency Distributions

A frequency distribution groups data into non-overlapping classes and counts how many observations fall in each. To build one with k classes (usually 5 to 10):

• Step 1. Compute the class width: (maximum − minimum) ÷ k, then round UP to a convenient number.
• Step 2. Choose the lower limit of the first class at or just below the minimum; add the width repeatedly to form the classes.
• Step 3. Tally the observations into the classes; the counts are the frequencies f.

The relative frequency of a class is f/n, the fraction of all data in that class. The cumulative frequency is the running total of frequencies up through that class. The class midpoint is the average of the class limits and represents the class in later calculations.

2.2 Histograms and Distribution Shape

A histogram displays classes on the horizontal axis and frequencies as touching bars. Watch for these common shapes:

• Symmetric (bell-shaped) — measurement error, many natural processes.
• Skewed right — long tail toward high values (repair times, lifetimes).
• Skewed left — long tail toward low values.
• Uniform — all classes roughly equal.
• Bimodal — two peaks, often a signal that two different processes or machines are mixed in one data set.

2.3 Stem-and-Leaf Displays

A stem-and-leaf display keeps every data value visible while still showing shape. Each value is split into a stem (leading digit or digits) and a leaf (final digit). Leaves are listed in increasing order beside their stem. It is a quick, pencil-friendly alternative to a histogram for small data sets.

2.4 Other Useful Graphs

• Pareto chart — a bar chart of defect categories sorted from most to least frequent; identifies the “vital few” causes.
• Time-series plot — data plotted in production order; reveals trends and shifts that a histogram hides.
• Scatter plot — pairs (x, y) plotted to reveal relationships between two variables (Chapter 13).

Worked examples

Try each example with pencil and calculator first, then open the solution and check yourself line by line.

Practice problems

Each problem mirrors the matching worked example above. Reveal an answer only after you have attempted the problem.

1. 2.1 Machine downtime (minutes) for 20 incidents: 5, 7, 8, 10, 12, 13, 15, 16, 18, 20, 21, 23, 25, 26, 28, 30, 32, 35, 37, 40. Construct a frequency distribution with 5 classes (round the class width up to 8) and list the frequencies.

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2. 2.2 For the distribution in Problem 2.1, compute the class midpoints and relative frequencies.

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3. 2.3 Construct a stem-and-leaf display for these 12 tensile strengths (ksi): 41, 43, 47, 52, 55, 55, 58, 60, 64, 69, 71, 73.

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4. 2.4 Surface roughness classes have frequencies 2, 5, 12, 5, 2 from lowest class to highest. Describe the shape of the distribution.

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5. 2.5 Using your answer to Problem 2.1, how many downtime incidents lasted 28 minutes or less, and what percentage of the sample is that?

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Chapter 3

Measures of Central Tendency

After organizing data, the next question is: where is the center? A single well-chosen number — an average — often summarizes hundreds of measurements. This chapter covers the three classic measures of center (mean, median, mode), plus the weighted mean and the mean of grouped data, and explains when each is the right tool.

3.1 The Mean

The arithmetic mean is the sum of the values divided by how many there are:

The mean uses every data value, which makes it the most common measure — but also makes it sensitive to outliers: one wild value can drag the mean far from the bulk of the data.

3.2 The Median and the Mode

The median is the middle value when the data are sorted. If n is odd, it is the single middle value; if n is even, it is the average of the two middle values. The median is resistant to outliers, which makes it the preferred center for skewed data such as incomes, lifetimes, and repair costs.

The mode is the most frequently occurring value. A data set may have one mode, several modes, or none. The mode is the only measure of center that works for qualitative data (the most common failure type, the most used material).

3.3 The Weighted Mean

When values carry different importance, weight each value wᵢ and compute:

Course grades, composite material properties, and blended costs are all weighted means.

3.4 Mean of a Frequency Distribution

When only grouped data are available, use each class midpoint m as the representative value of its class:

This is an approximation — exact values inside each class are unknown — but it is usually very close.

Worked examples

Try each example with pencil and calculator first, then open the solution and check yourself line by line.

Practice problems

Each problem mirrors the matching worked example above. Reveal an answer only after you have attempted the problem.

1. 3.1 Seven flow-rate readings (L/min): 8, 10, 9, 12, 10, 11, 10. Find the mean, median, and mode.

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2. 3.2 Five repair invoices: $45, $47, $48, $50, $190. Compute the mean and the median, and state which is the better “typical” value and why.

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3. 3.3 A composite score weights a design project 50%, a midterm 30%, and quizzes 20%. A student scores 85, 90, and 60 respectively. Find the weighted mean.

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4. 3.4 Estimate the mean of the downtime distribution from Problem 2.1 (midpoints 8.5, 16.5, 24.5, 32.5, 40.5; frequencies 5, 5, 5, 3, 2).

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5. 3.5 The mean of six pressure readings is 15.0 psi. Five of them are 12, 14, 16, 18, and 10. Find the sixth reading.

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Chapter 4

Measures of Variability and Position

Two processes can have the same average and behave completely differently: one holds tolerance, the other scatters parts everywhere. Variability is often the engineer’s real enemy, and this chapter gives you the tools to measure it — range, variance, standard deviation, coefficient of variation — plus measures of position: z-scores, percentiles, quartiles, and the outlier fences used in boxplots.

4.1 Range, Variance, and Standard Deviation

The range = maximum − minimum is quick but uses only two values. The workhorse measures are the variance and standard deviation, built from the deviations (x − x̄) of every value from the mean:

The sample formula divides by n − 1 (the degrees of freedom), which corrects the tendency of small samples to underestimate spread. The standard deviation has the same units as the data, which is why it is quoted far more often than the variance.

A computational shortcut avoids computing each deviation:

4.2 Coefficient of Variation

To compare the spread of variables with different units or very different means, use the coefficient of variation — the standard deviation as a percentage of the mean:

4.3 The Empirical Rule and Chebyshev’s Theorem

For bell-shaped distributions, the Empirical Rule says approximately:

• 68% of the data lie within 1 standard deviation of the mean (μ ± σ),
• 95% within 2 standard deviations (μ ± 2σ),
• 99.7% within 3 standard deviations (μ ± 3σ).

For ANY distribution, bell-shaped or not, Chebyshev’s theorem guarantees that at least 1 − 1/k² of the data lie within k standard deviations of the mean (k > 1): at least 75% within 2σ, at least 88.9% within 3σ.

4.4 Measures of Position: z-Scores, Percentiles, Quartiles

The z-score converts a value into “standard deviations from the mean”:

A z-score near 0 is typical; values beyond ±2 are unusual; beyond ±3, rare. Z-scores also let you compare apples to oranges — a measurement’s relative standing in its own distribution.

The pth percentile is the value below which about p% of the data fall. The quartiles Q1, Q2 (the median), and Q3 split the data into four quarters, and the interquartile range IQR = Q3 − Q1 measures the spread of the middle 50%.

Outlier fences: values below Q1 − 1.5·IQR or above Q3 + 1.5·IQR are flagged as outliers. The five-number summary (min, Q1, median, Q3, max) is drawn as a boxplot: a box from Q1 to Q3 with a line at the median and whiskers to the most extreme non-outlier values.

Worked examples

Try each example with pencil and calculator first, then open the solution and check yourself line by line.

Practice problems

Each problem mirrors the matching worked example above. Reveal an answer only after you have attempted the problem.

1. 4.1 Five thickness readings (mm): 10, 12, 14, 16, 18. Find the range, sample variance, and sample standard deviation using the definition formula.

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2. 4.2 For the data 3, 5, 5, 7, 10, compute the sample variance and standard deviation using the computational formula (Σx and Σx² first).

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3. 4.3 Process A: mean 120 units/hr, s = 6. Process B: mean 80 units/hr, s = 5. Use the coefficient of variation to decide which process is relatively more consistent.

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4. 4.4 Resistor values are bell-shaped with μ = 60 Ω and σ = 4 Ω. (a) Between what two values do about 68% of resistors fall? (b) Find the z-score of a 70 Ω resistor. (c) About what percent exceed 68 Ω?

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5. 4.5 Nine response times (ms): 3, 6, 7, 8, 10, 12, 13, 15, 40. Find Q1, the median, Q3, the IQR, the outlier fences, and identify any outliers.

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Chapter 5

Probability Fundamentals

Probability is the language of uncertainty. Will this relay fail? Is the defect from machine A or machine B? Probability assigns a number between 0 (impossible) and 1 (certain) to events, and gives us rules for combining those numbers. Everything in inferential statistics — confidence intervals, hypothesis tests, control charts — is built on the rules in this chapter.

5.1 Experiments, Sample Spaces, and Events

A probability experiment is any process with an uncertain outcome (test one part, roll a die, measure a voltage). The sample space S is the set of all possible outcomes, and an event is any subset of the sample space. Three ways to assign probability:

• Classical: if all outcomes are equally likely, P(E) = (number of outcomes in E) / (number in S).
• Empirical (relative frequency): P(E) ≈ (times E occurred) / (number of trials).
• Subjective: an expert judgment when neither of the above applies.

The complement of E, written E′, is “E does not happen,” and:

5.2 The Addition Rule

For any two events A and B:

The subtraction removes the double-counted overlap. If A and B are mutually exclusive (cannot happen together), P(A and B) = 0 and the rule reduces to P(A or B) = P(A) + P(B).

5.3 Conditional Probability and Independence

The conditional probability of A given that B has occurred is:

Events A and B are independent if knowing one tells you nothing about the other: P(A | B) = P(A). For independent events the multiplication rule simplifies:

For dependent events, use the general multiplication rule P(A and B) = P(A) · P(B | A). Independence is the key assumption behind series-system reliability calculations: a chain of independent components works only if every component works.

5.4 Total Probability and Bayes’ Theorem

When an outcome can arise from several sources (machines, vendors, production lines), the law of total probability adds the weighted contributions, and Bayes’ theorem “reverses” the conditioning to find where an observed outcome most likely came from:

Worked examples

Try each example with pencil and calculator first, then open the solution and check yourself line by line.

Practice problems

Each problem mirrors the matching worked example above. Reveal an answer only after you have attempted the problem.

1. 5.1 A tray holds 50 machined parts, 4 of which are out of tolerance. One part is picked at random. Find the probability it is out of tolerance, and the probability it is in tolerance.

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2. 5.2 For castings, P(porosity) = 0.25, P(crack) = 0.15, and P(both) = 0.05. Find the probability that a casting has porosity or a crack (or both).

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3. 5.3 Vendor X supplied 200 sensors with 8 defective; vendor Y supplied 100 sensors with 12 defective. One sensor is chosen at random from the 300. Find P(defective) and P(defective | vendor Y). Are “defective” and “vendor Y” independent?

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4. 5.4 Three independent smoke detectors each work with probability 0.95. Find the probability that (a) all three work, and (b) at least one works.

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5. 5.5 Line 1 makes 70% of production with a 3% defect rate; line 2 makes 30% with an 8% defect rate. A defective unit is found. What is the probability it came from line 2?

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Chapter 6

Counting Methods: Permutations and Combinations

Classical probability requires counting outcomes, and in real problems the counts are far too large to list. Three tools — the fundamental counting principle, permutations, and combinations — let you count millions of possibilities in one line. They also feed directly into the binomial distribution of Chapter 7 and acceptance-sampling calculations used throughout industry.

6.1 The Fundamental Counting Principle

If a task consists of a sequence of steps, and step 1 can be done in n₁ ways, step 2 in n₂ ways, and so on, the whole task can be done in n₁ · n₂ · n₃ ··· ways. This one idea handles part numbers, passwords, menu configurations, and test plans.

6.2 Factorials and Permutations

n! (“n factorial”) = n(n−1)(n−2)···(2)(1), with 0! = 1. A permutation is an ordered arrangement. The number of ways to arrange r items chosen from n distinct items, where order matters, is:

Use permutations when the positions are different: first/second/third place, which part goes to which test rig, orderings of a startup sequence.

6.3 Combinations

A combination is a selection where order does NOT matter — a handful, a committee, a sample pulled for inspection:

6.4 Counting in Probability: Sampling Without Replacement

Combinations turn “what is the chance my random sample contains k defectives?” into a ratio of counts:

where the lot has N items, d defective, and n are sampled. This hypergeometric-style counting is the mathematical core of acceptance sampling.

Worked examples

Try each example with pencil and calculator first, then open the solution and check yourself line by line.

Practice problems

Each problem mirrors the matching worked example above. Reveal an answer only after you have attempted the problem.

1. 6.1 A serial number has two letters followed by three digits (repeats allowed everywhere). How many serial numbers are possible?

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2. 6.2 Eight instruments are available and two must be assigned, one to the inlet station and one to the outlet station. How many assignments are possible?

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3. 6.3 From a lot of 12 gearboxes, 4 will be selected for endurance testing. How many different samples are possible?

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4. 6.4 A shipment of 15 valves contains 4 defective ones. Three valves are drawn at random without replacement. Find the probability that none of the three is defective.

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5. 6.5 A project team needs exactly 2 electrical engineers from a pool of 5 and exactly 3 mechanical engineers from a pool of 7. How many teams are possible?

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Chapter 7

Discrete Random Variables: Binomial and Poisson Distributions

A random variable attaches a number to each outcome of an experiment — the number of defects on a board, the number of failures in a fleet. This chapter shows how to describe a discrete random variable with a probability distribution, how to compute its mean and standard deviation, and how to use the two most important discrete models in engineering: the binomial and the Poisson.

7.1 Probability Distributions of Discrete Random Variables

A discrete probability distribution lists every possible value x and its probability P(x). Two requirements: every P(x) is between 0 and 1, and ΣP(x) = 1. Probabilities of events are found by adding the P(x) values involved.

7.2 Mean and Variance of a Random Variable

The mean (expected value) and variance of a discrete random variable are weighted by probability:

The expected value is the long-run average per trial — the basis of expected-cost and expected-life calculations.

7.3 The Binomial Distribution

Use the binomial when an experiment consists of n identical, independent trials, each ending in “success” or “failure,” with the same success probability p on every trial. Then X = number of successes has:

“Success” is just the outcome being counted — often a defective part! Inspecting n parts from a large lot with defect rate p is the classic engineering binomial.

7.4 The Poisson Distribution

Use the Poisson to count events scattered randomly over an interval of time, length, or area at a constant average rate λ per interval — flaws per panel, calls per minute, failures per year:

Worked examples

Try each example with pencil and calculator first, then open the solution and check yourself line by line.

Practice problems

Each problem mirrors the matching worked example above. Reveal an answer only after you have attempted the problem.

1. 7.1 The number of hydraulic leaks X per machine has distribution P(0) = 0.60, P(1) = 0.25, P(2) = 0.10, P(3) = 0.05. Verify the distribution is valid and find P(X ≥ 1).

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2. 7.2 For the distribution in Problem 7.1, find μ, σ², and σ.

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3. 7.3 A lot is 10% defective. Eight items are sampled at random. Find (a) P(no defectives), (b) P(exactly one), (c) P(at most one).

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4. 7.4 Circuit boards are 4% defective. For a batch of 250 boards, find the expected number of defectives and the standard deviation.

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5. 7.5 A call center receives an average of 2 alarm calls per minute (Poisson). Find (a) P(no calls in a minute), (b) P(exactly 3 calls), (c) P(at least one call).

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Chapter 8

Continuous Distributions and the Normal Curve

Measurements — lengths, voltages, times, strengths — are continuous, and their probabilities come from areas under a curve called a probability density function. This chapter introduces the uniform distribution, then the most important curve in all of statistics: the normal (bell) curve, along with the standard normal table, z-scores, and inverse (“work backward”) problems.

8.1 Continuous Random Variables

For a continuous random variable, P(X = one exact value) = 0; only intervals carry probability, computed as area under the density curve. The total area under any density curve is 1. Consequently P(X ≤ a) and P(X < a) are the same for continuous variables.

8.2 The Uniform Distribution

If X is equally likely anywhere between a and b, its density is a flat rectangle of height 1/(b−a). Probability of an interval is simply its length divided by (b−a):

8.3 The Normal Distribution

The normal distribution is a symmetric bell curve completely determined by its mean μ (the center) and standard deviation σ (the spread). It describes measurement error, part dimensions, and — by the Central Limit Theorem of Chapter 9 — sample means of almost anything. The Empirical Rule (68–95–99.7) of Chapter 4 is exact for the normal curve.

8.4 The Standard Normal and z-Table

The standard normal Z has μ = 0 and σ = 1. Any normal X is converted to Z by standardizing:

A standard normal table (or calculator) gives the cumulative area Φ(z) = P(Z ≤ z). Then:

• P(X ≤ x) = Φ(z)
• P(X > x) = 1 − Φ(z)
• P(a < X < b) = Φ(z_b) − Φ(z_a)

8.5 Inverse Normal Problems

Sometimes the probability is given and the cutoff value is wanted (a warranty life, a spec limit, a percentile). Work backward: find the z-value whose cumulative area matches the given probability, then un-standardize:

Useful landmarks: z = 1.28 cuts off the top 10%, z = 1.645 the top 5%, z = 1.96 the top 2.5%, z = 2.33 the top 1%.

Worked examples

Try each example with pencil and calculator first, then open the solution and check yourself line by line.

Practice problems

Each problem mirrors the matching worked example above. Reveal an answer only after you have attempted the problem.

1. 8.1 Mixing time is uniformly distributed between 4 and 10 minutes. Find (a) P(X < 5.5), (b) the mean, and (c) the standard deviation.

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2. 8.2 Find (a) P(Z < 0.75), (b) P(Z > 2.33), (c) P(−1.50 < Z < 0.50).

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3. 8.3 Fill volumes are normal with μ = 250 mL and σ = 4 mL. Find P(244 < X < 258).

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4. 8.4 For the filler in Problem 8.3, the specification limits are 242 to 258 mL. What percentage of containers is out of spec?

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5. 8.5 Lamp lifetimes are normal with μ = 1,200 h and σ = 100 h. Find the lifetime that only 10% of lamps fail to reach (the 10th percentile).

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Chapter 9

Sampling Distributions and the Central Limit Theorem

Every sample gives a slightly different mean. If you repeatedly drew samples of n parts and computed x̄ each time, those means would themselves form a distribution — the sampling distribution of the mean. Understanding how sample means behave is the bridge from descriptive statistics to inference: it is what makes confidence intervals and hypothesis tests possible.

9.1 The Sampling Distribution of the Sample Mean

If a population has mean μ and standard deviation σ, then the collection of all possible sample means x̄ (samples of size n) has:

Sample means center on the true mean, and they scatter LESS than individual values — averaging cancels out variation. Quadrupling the sample size halves the standard error.

9.2 The Central Limit Theorem (CLT)

The Central Limit Theorem: for a sufficiently large sample size (n ≥ 30 is the usual rule of thumb), the sampling distribution of x̄ is approximately NORMAL — regardless of the shape of the population. If the population itself is normal, x̄ is exactly normal for any n.

To find probabilities about a sample mean, standardize with the standard error:

9.3 The Sampling Distribution of a Proportion

A sample proportion p̂ = x/n from a population with true proportion p has:

and is approximately normal when np ≥ 10 and n(1−p) ≥ 10. Standardize with:

Worked examples

Try each example with pencil and calculator first, then open the solution and check yourself line by line.

Practice problems

Each problem mirrors the matching worked example above. Reveal an answer only after you have attempted the problem.

1. 9.1 Cable tensile strengths have μ = 50 kN and σ = 8 kN. For samples of n = 16, find the mean and standard error of x̄.

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2. 9.2 The population in Problem 9.1 is normal. Find P(x̄ > 53) for n = 16.

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3. 9.3 Cure times have μ = 200 s and σ = 30 s. For samples of n = 36, find P(195 < x̄ < 210).

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4. 9.4 Repair costs are right-skewed with μ = $75 and σ = $20. For samples of n = 100 invoices, find P(x̄ < 71) and state why the normal curve may be used.

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5. 9.5 A process runs at p = 0.10 nonconforming. For samples of n = 225, find P(p̂ > 0.13).

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Chapter 10

Confidence Intervals

A sample mean is a single guess — a point estimate — and it is almost certainly not exactly equal to the true population mean. A confidence interval turns the point estimate into an honest range: estimate ± margin of error, together with a confidence level (usually 90%, 95%, or 99%) describing how reliable the procedure is. This chapter builds intervals for a mean (σ known and unknown) and for a proportion, and shows how to choose the sample size needed for a target precision.

10.1 Confidence Interval for a Mean, σ Known (z-Interval)

The critical values are z = 1.645 for 90%, 1.96 for 95%, and 2.576 for 99% confidence. The quantity after the ± sign is the margin of error E.

Interpretation: “95% confident” means the procedure captures the true mean in 95% of all samples — not that there is a 95% probability μ is inside this one interval. μ is a fixed number; it is the interval that varies from sample to sample.

10.2 Confidence Interval for a Mean, σ Unknown (t-Interval)

In practice σ is rarely known; we estimate it with s, which adds uncertainty. The correction is Student’s t distribution — bell-shaped but slightly wider than z, with df = n − 1 degrees of freedom:

As n grows, t approaches z. Requirements: a random sample, and either a roughly normal population or a large n.

10.3 Confidence Interval for a Proportion

valid when np̂ ≥ 10 and n(1−p̂) ≥ 10 (at least 10 “successes” and 10 “failures” in the sample).

10.4 Choosing the Sample Size

To achieve a desired margin of error E, solve for n and always round UP:

If no prior estimate of the proportion exists, use p̂ = 0.5, which gives the largest (safest) sample size.

Worked examples

Try each example with pencil and calculator first, then open the solution and check yourself line by line.

Practice problems

Each problem mirrors the matching worked example above. Reveal an answer only after you have attempted the problem.

1. 10.1 A process has known σ = 7. A sample of n = 49 gives x̄ = 118. Construct a 95% confidence interval for μ.

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2. 10.2 Using the data of Problem 10.1, construct a 99% confidence interval and state how it compares to the 95% interval.

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3. 10.3 Twenty-five oil samples have x̄ = 8.35 (viscosity index) with s = 0.90. Build a 95% t-interval for the true mean.

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4. 10.4 In a sample of 240 molded parts, 60 show flash defects. Build a 95% confidence interval for the true defect proportion.

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5. 10.5 (a) With σ ≈ 12, what sample size estimates a mean within E = 3 at 95% confidence? (b) What sample size estimates a proportion within E = 0.04 at 95% confidence with no prior estimate?

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Chapter 11

Hypothesis Testing: One Sample

A hypothesis test answers a yes-or-no question about a population: Has the mean shifted from its target? Is the defect rate above the contract limit? The logic is proof by contradiction: assume the status quo (the null hypothesis), compute how surprising your sample would be under that assumption, and reject the assumption only when the data are too surprising to believe.

11.1 Hypotheses, Errors, and Significance Level

The null hypothesis H₀ states the status quo with equality (μ = μ₀ or p = p₀). The alternative H₁ is what you seek evidence for: two-tailed (≠), left-tailed (<), or right-tailed (>).

Two errors are possible. A Type I error rejects a true H₀ (a false alarm); its probability is the significance level α, chosen in advance (commonly 0.05). A Type II error fails to reject a false H₀ (a missed detection); its probability is β. Decreasing α raises β unless the sample size increases.

11.2 The Test Procedure

• Step 1. State H₀ and H₁, and choose α.
• Step 2. Compute the test statistic from the sample.
• Step 3. Find the critical value(s) for α, or compute the p-value.
• Step 4. Decide: reject H₀ if the statistic falls in the rejection region (equivalently, if p-value ≤ α).
• Step 5. State the conclusion in plain language, in context.

The p-value is the probability, assuming H₀ is true, of a sample result at least as extreme as the one observed. Small p-value → the data are inconsistent with H₀.

11.3 Test for a Mean

11.4 Test for a Proportion

valid when np₀ ≥ 10 and n(1−p₀) ≥ 10. Note that the standard error uses the hypothesized p₀, not p̂ — under H₀ we must assume p₀ is the truth.

Worked examples

Try each example with pencil and calculator first, then open the solution and check yourself line by line.

Practice problems

Each problem mirrors the matching worked example above. Reveal an answer only after you have attempted the problem.

1. 11.1 Write H₀ and H₁: (a) a rod diameter must equal 25 mm, either direction bad; (b) a new coating is claimed to make corrosion resistance exceed 900 h; (c) a firm suspects its scrap rate has fallen below 8%.

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2. 11.2 A beverage filler targets 12.00 oz with known σ = 0.18 oz. A sample of 36 bottles averages 11.92 oz. Test H₀: μ = 12 vs. H₁: μ ≠ 12 at α = 0.05, and give the p-value.

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3. 11.3 Sixteen specimens of an alloy have x̄ = 102.3 ksi and s = 4.0 ksi. Test whether the mean strength differs from 100 ksi at α = 0.05.

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4. 11.4 A contract allows at most 5% defective. In a sample of 400 units, 28 are defective. Test H₀: p = 0.05 vs. H₁: p > 0.05 at α = 0.05.

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5. 11.5 A test yields p-value = 0.12 with α = 0.05. State the decision, and explain which type of error is now possible and what it would mean in practice.

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Chapter 12

Hypothesis Testing: Two Samples

Engineering questions are usually comparisons: Is vendor A stronger than vendor B? Did the process change reduce defects? Does the new fixture cut cycle time on the same machines? This chapter compares two means from independent samples, two means from paired (before/after) data, and two proportions.

12.1 Independent Samples vs. Paired Data

Two samples are independent when the individuals in one sample have no connection to those in the other (parts from vendor A vs. parts from vendor B). Data are paired when each observation in one sample is naturally matched to one in the other — the same machine before and after an upgrade, the same specimen measured by two gauges. Pairing removes unit-to-unit variation and usually gives a more sensitive test. Choosing the wrong analysis is one of the most common statistical mistakes.

12.2 Comparing Two Means: Independent Samples (Pooled t)

When the two populations are roughly normal with similar variances, pool the two sample variances:

A confidence interval for the difference of means is (x̄₁ − x̄₂) ± t(α/2)·sp·√(1/n₁ + 1/n₂). If the interval contains 0, the data are consistent with “no difference.”

12.3 Comparing Two Means: Paired Data

Compute the difference d for each pair, then run a one-sample t-test on the differences:

where n is the number of pairs, d̄ the mean difference, and s_d the standard deviation of the differences.

12.4 Comparing Two Proportions

To test H₀: p₁ = p₂, pool the successes to estimate the common proportion p̄ = (x₁ + x₂)/(n₁ + n₂):

Worked examples

Try each example with pencil and calculator first, then open the solution and check yourself line by line.

Practice problems

Each problem mirrors the matching worked example above. Reveal an answer only after you have attempted the problem.

1. 12.1 Eight parts from process 1: x̄₁ = 150, s₁ = 6. Eight parts from process 2: x̄₂ = 144, s₂ = 8. Test H₀: μ₁ = μ₂ vs. H₁: μ₁ ≠ μ₂ at α = 0.05 using the pooled t-test.

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2. 12.2 Using the data of Problem 12.1, construct a 95% confidence interval for μ₁ − μ₂ and interpret it.

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3. 12.3 Surface roughness was measured on five shafts before and after a polishing change; improvements are d = 3, 5, 2, 6, 4. Test at α = 0.05 whether the change reduces roughness (H₁: μ_d > 0).

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4. 12.4 Line 1 had 24 defectives in 300 units; line 2 had 42 defectives in 300 units. Test at α = 0.05 whether the defect rates differ.

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5. 12.5 Classify the correct analysis (independent t, paired t, or two-proportion z): (a) breaking strength of 12 welds by robot vs. 12 welds by hand; (b) the same 10 circuit boards tested at 25 °C and again at 85 °C; (c) proportion of passed inspections for shift A vs. shift B.

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Chapter 13

Correlation and Simple Linear Regression

Many engineering questions involve two variables: does higher curing temperature increase strength? Does spindle speed affect surface finish? Correlation measures how strongly two quantitative variables move together; regression fits the best straight line through the data so you can predict one variable from the other.

13.1 Scatter Plots and the Correlation Coefficient

Always start with a scatter plot of the pairs (x, y): x is the explanatory (independent) variable, y the response (dependent) variable. Look for direction (positive or negative), form (linear or curved), strength, and outliers.

The linear correlation coefficient r measures the strength and direction of a LINEAR relationship, with −1 ≤ r ≤ +1. Values near ±1 indicate a tight line; values near 0 indicate no linear relationship. A convenient computing route uses three sums of squares:

13.2 The Least-Squares Regression Line

The least-squares line ŷ = b₀ + b₁x minimizes the total squared vertical error between the data points and the line:

The slope b₁ is the predicted change in y for each one-unit increase in x. The intercept b₀ is the prediction at x = 0 (meaningful only if x = 0 makes physical sense).

13.3 Prediction, Residuals, and r²

A residual is (observed − predicted) = y − ŷ; the line runs through the “middle” of the residuals, which sum to zero. The coefficient of determination r² is the fraction of the variation in y explained by the linear relationship with x — an r of 0.9 means r² = 0.81, i.e., 81% of the variation is explained.

Two cautions: never extrapolate far outside the range of the x-data, and never trust a fitted line you have not plotted.

Worked examples

Try each example with pencil and calculator first, then open the solution and check yourself line by line.

Practice problems

Each problem mirrors the matching worked example above. Reveal an answer only after you have attempted the problem.

1. 13.1 Predict the direction of association (positive, negative, or none): (a) hours of machine use vs. remaining bearing life; (b) weld current vs. weld penetration depth over the normal range; (c) operator badge number vs. parts per hour.

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2. 13.2 For x = 2, 4, 6, 8 and y = 3, 7, 5, 9, compute Sxx, Syy, Sxy, and the correlation coefficient r.

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3. 13.3 Machine hours x and output y (units): (1, 3), (2, 5), (3, 6), (4, 9), (5, 11). Find the least-squares line ŷ = b₀ + b₁x.

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4. 13.4 Using the line from Problem 13.3: (a) predict output at x = 3.5 hours; (b) find the residual at x = 3, where the observed output was 6.

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5. 13.5 For the data of Problem 13.3, Syy = 40.8. Compute r and r², and state in words what r² means here.

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Chapter 14

Chi-Square Tests: Goodness of Fit and Independence

Not all data are measurements — much engineering data are counts in categories: defect types, vendors, shifts, pass/fail grades. Chi-square (χ²) tests compare observed counts with the counts expected under a hypothesis. The goodness-of-fit test checks whether one categorical variable follows a claimed distribution; the test of independence checks whether two categorical variables are related.

14.1 The Chi-Square Statistic

Both tests use the same statistic, summed over all categories or cells:

where O is the observed count and E is the expected count under H₀. Large χ² means the observed data are far from what H₀ predicts, so chi-square tests are always right-tailed. Requirement: every expected count should be at least 5.

14.2 Goodness-of-Fit Test

H₀ states the claimed category proportions. Each expected count is E = n × (claimed proportion). Degrees of freedom: df = k − 1, where k is the number of categories. Reject H₀ when χ² exceeds the critical value χ²(α, k−1).

14.3 Test of Independence

Counts are arranged in a contingency table of r rows and c columns. H₀: the row variable and column variable are independent. Expected counts come from the table margins:

Rejecting H₀ means the two variables are associated — for example, defect rate depends on which shift produced the part.

Worked examples

Try each example with pencil and calculator first, then open the solution and check yourself line by line.

Practice problems

Each problem mirrors the matching worked example above. Reveal an answer only after you have attempted the problem.

1. 14.1 Four identical machines should share breakdowns equally. Of 80 breakdowns: machine 1 had 26, machine 2 had 18, machine 3 had 22, machine 4 had 14. Test at α = 0.05 whether breakdowns are equally likely.

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2. 14.2 A supplier claims its shipments grade 60% premium, 25% standard, 15% economy. In 200 shipments you observe 105, 60, and 35. Test the claim at α = 0.05.

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3. 14.3 Material X: 12 cracked, 88 sound. Material Y: 24 cracked, 76 sound. Test at α = 0.05 whether cracking is independent of material.

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4. 14.4 A 2 × 3 table has row totals 150 and 150, and column totals 90, 120, 90 (n = 300). Compute the six expected counts.

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5. 14.5 For a 4 × 3 contingency table: (a) find the degrees of freedom; (b) state the expected-count condition that must be checked before testing.

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Chapter 15

One-Way Analysis of Variance (ANOVA)

The two-sample t-test compares two means. But engineers often compare three or more: three suppliers, four cure temperatures, five machines. Running many t-tests inflates the chance of a false alarm; one-way ANOVA tests all the means at once with a single F-test by asking a simple question — is the variation BETWEEN group means large compared with the variation WITHIN groups?

15.1 The Logic and Hypotheses of ANOVA

ANOVA splits the total variation in the data into two parts: SSB, the sum of squares between groups (how far the group means are from the grand mean), and SSW, the sum of squares within groups (ordinary scatter inside each group). If H₀ is true, both reflect only random noise and their ratio is near 1; if some means differ, SSB inflates.

Assumptions: independent random samples, roughly normal populations, and roughly equal group variances.

15.2 The F Statistic and the ANOVA Table

With k groups and N total observations (group sizes nⱼ, group means x̄ⱼ, grand mean x̄̄):

Reject H₀ when F exceeds the critical value F(α; k−1, N−k). The results are conventionally displayed in an ANOVA table with columns Source, SS, df, MS, F.

15.3 After ANOVA

A significant F says only that SOME difference exists. To find which means differ, examine the group means and follow up with multiple-comparison procedures (such as Tukey’s method, covered in later courses). Always report the group means alongside the ANOVA table — the sizes of the differences matter as much as their significance.

Worked examples

Try each example with pencil and calculator first, then open the solution and check yourself line by line.

Practice problems

Each problem mirrors the matching worked example above. Reveal an answer only after you have attempted the problem.

1. 15.1 State the ANOVA hypotheses for comparing the mean cycle times of four CNC machines, and explain the advantage over six pairwise t-tests.

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2. 15.2 Etch rates for three bath temperatures, four wafers each — T1: 10, 12, 11, 11; T2: 14, 13, 15, 14; T3: 12, 12, 13, 11. Run the complete ANOVA at α = 0.05.

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3. 15.3 A study with k = 3 and N = 18 gives SST = 90 and SSB = 30. Complete the ANOVA table and test at α = 0.05 [use F(0.05; 2, 15) = 3.68].

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4. 15.4 An ANOVA with k = 4 groups and N = 24 observations gives F = 2.90. Find both df values and the decision at α = 0.05 [use F(0.05; 3, 20) = 3.10].

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5. 15.5 Explain why an engineer should not simply run a t-test on the largest and smallest of five group means as a substitute for ANOVA.

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Chapter 16

Statistical Quality Control: Control Charts and Process Capability

This capstone chapter applies everything — means, ranges, the normal curve, sampling distributions, hypothesis-testing logic — to the daily work of keeping a manufacturing process healthy. Control charts detect when a process changes; capability indices measure whether a stable process can actually meet its specification limits.

16.1 Common-Cause and Special-Cause Variation

Every process varies. Common-cause variation is the steady background noise inherent to the process; a process showing only common causes is in statistical control and is predictable. Special-cause (assignable) variation comes from something specific — a tool breaks, a new material lot, a setup error. Control charts exist to flag special causes quickly, so they can be found and removed, while leaving a stable process alone.

16.2 x̄ and R Charts

Small subgroups (typically n = 4 or 5 consecutive parts) are sampled periodically. Each subgroup yields a mean x̄ and a range R. With x̄̄ = the average of the subgroup means and R̄ = the average range, the 3-sigma control limits are:

Constants by subgroup size n: n = 2: A₂ = 1.880, D₃ = 0, D₄ = 3.267; n = 3: A₂ = 1.023, D₃ = 0, D₄ = 2.574; n = 4: A₂ = 0.729, D₃ = 0, D₄ = 2.282; n = 5: A₂ = 0.577, D₃ = 0, D₄ = 2.114.

A process is judged out of control if any point falls outside the limits, or if non-random patterns appear (runs of 8+ points on one side of the centerline, steady trends, cycles). Check the R chart first — if spread is unstable, the x̄ chart limits mean nothing.

16.3 Process Capability: Cp and Cpk

Control answers “is the process stable?”; capability answers “can it meet spec?” For a stable process with standard deviation σ and specification limits LSL and USL:

Cp compares the spec width to the natural process width (6σ) but ignores centering; Cpk penalizes a process that is off-center. Cpk = Cp only when the process is perfectly centered. Common benchmarks: Cpk ≥ 1.00 is minimally capable (about 0.27% out of spec if centered); Cpk ≥ 1.33 is the usual industrial requirement.

Worked examples

Try each example with pencil and calculator first, then open the solution and check yourself line by line.

Practice problems

Each problem mirrors the matching worked example above. Reveal an answer only after you have attempted the problem.

1. 16.1 Subgroups of n = 4 give x̄̄ = 25.00 and R̄ = 3.00. Compute the centerline and control limits for the x̄ chart.

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2. 16.2 For the process in Problem 16.1, compute the R chart limits.

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3. 16.3 Using your limits from Problem 16.1, judge the process from the subgroup means 24.5, 25.2, 27.4, 25.0, 24.8, and state the correct response.

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4. 16.4 A specification runs from LSL = 48 to USL = 52, and the process σ = 0.50. Compute Cp.

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5. 16.5 The process in Problem 16.4 is centered at μ = 50.8. Compute Cpk, compare it with Cp, and recommend an action.

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