Measures of central tendency and dispersion, probability distributions, confidence intervals for a single mean, regression and curve fitting, and hypothesis testing.
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Exam areas
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B. Probability distributions
Probability Distributions
Model environmental counts and concentrations with the binomial, Poisson, and normal distributions, and use the central limit theorem to normalize sample means.
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C. Estimation for a single mean
Descriptive Statistics and Confidence Intervals
Summarize a sample with mean, median, mode, and the n−1 standard deviation, then bound the true mean with a z or t confidence interval.
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D. Regression (linear, multiple), curve fitting, and goodness of fit
Regression, Curve Fitting, and Goodness of Fit
Fit a least-squares line, judge it with the correlation coefficient, R-squared, and standard error, and linearize power and exponential data.
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E. Hypothesis testing
Hypothesis Testing
Frame null and alternative hypotheses, run one- and two-tailed z and t tests, control Type I and II error, and decide with critical values and p-values.
Compliance decisions are hypothesis tests in disguise: is the measured effluent mean truly above a permit limit, or could the exceedance be sampling noise? Hypothesis testing gives a disciplined yes/no answer with a stated risk of being wrong. The FE exam reduces this to a repeatable procedure — state two hypotheses, compute one test statistic, compare it to a critical value or a p-value — and rewards candidates who keep the error types and the tails straight. The framework, statistics, and rejection criteria below follow the FE Reference Handbook — Engineering Probability and Statistics.
Null and alternative hypotheses
The null hypothesis H0 is the status-quo claim you assume true until the data overturn it — typically that the mean equals a standard, μ=μ0. The alternative H1 is what you are trying to demonstrate. Its form sets the tails: μ=μ0 is two-tailed, while μ>μ0 or μ<μ0 is one-tailed. Choose the alternative from the engineering question before you look at the data — “does the effluent exceed the limit?” is one-tailed upper; “is the calibration off in either direction?” is two-tailed.
H0:μ=μ0vs.H1:μ=μ0,μ>μ0,orμ<μ0
The test statistic: z when σ is known, t when it is not
The test statistic measures how many standard errors the sample mean sits from the hypothesized mean. When the population standard deviation σ is known you get a standard-normal Z0; when you must estimate it with the sample s, you get a Student's t0 with n−1 degrees of freedom. The structure is identical — (observed minus hypothesized) over the standard error — and the choice between them mirrors the confidence-interval logic exactly.
Z0=σ/nXˉ−μ0(σknown),t0=s/nXˉ−μ0(σunknown)
Significance level, critical values, and rejection regions
The significance level α is the risk you accept of wrongly rejecting a true H0; it fixes the critical value that bounds the rejection region. A two-tailed test splits α between both tails and rejects when ∣Z0∣>zα/2 (e.g. 1.960 at α=0.05); a one-tailed test puts the whole α in one tail and rejects when Z0>zα or Z0<−zα (1.645 at α=0.05). This is why a one-tailed test is more powerful in its chosen direction — a smaller critical value — but blind to deviations the other way.
reject H0 if ∣Z0∣>zα/2(two-tailed)orZ0>zα(one-tailed upper)
Type I and Type II error
Two ways to be wrong. A Type I error rejects a true H0 — a false alarm, with probability α (you flag compliant water as a violation). A Type II error fails to reject a false H0 — a missed detection, with probability β (you pass truly contaminated water). The two trade off: shrinking α enlarges β unless you raise n. The test's power, 1−β, is its ability to catch a real effect; for a fixed α, more samples or a larger true effect both increase power.
α=P(reject H0∣H0 true),β=P(fail to reject H0∣H0 false),power=1−β
P-values: the strength of the evidence
The p-value is the probability, computed assuming H0 is true, of getting a test statistic at least as extreme as the one observed. It converts the statistic into a direct measure of evidence: reject H0 when p≤α, fail to reject when p>α. For a two-tailed test you double the one-tail area. A small p-value says the data would be surprising if H0 held — but it never proves H1 is true, and “fail to reject” never proves H0; it only means the evidence is insufficient.
p=P(∣Z∣≥∣Z0∣∣H0)(two-tailed);reject H0 if p≤α
Comparing a sample mean to a standard — the workflow
Almost every FE problem here compares one sample mean to a fixed standard μ0. The recipe never changes: write H0 and H1 (the engineering question picks the tail); pick z or t by whether σ is known; compute the statistic; compare to the critical value at α (or compare p to α); state the decision in plain language. The same machinery extends to comparing two means with a pooled variance, but the single-mean test is the exam's bread and butter.
Lock down four things in order: the tail (from H1), the statistic (z if σ given, t with n−1 df if only s), the critical value at α, and the decision. Halve α for two-tailed critical values — the most common slip. Remember the directions: large positive Z0 rejects an upper-tail H1; a negative Z0 cannot reject μ>μ0. A result can be statistically significant yet operationally trivial, so read what the decision means for the system. And never write that the test “proves” a hypothesis — you reject or fail to reject, with a known error rate.
Key equations
z test statistic (σ known)Z0=σ/nXˉ−μ0
Standardized distance of the sample mean from μ0 when the population σ is known.
t test statistic (σ unknown)t0=s/nXˉ−μ0
Same form using the sample s; compare to tα,n−1 or tα/2,n−1.
Two-tailed rejection (z)reject H0 if ∣Z0∣>zα/2
H1:μ=μ0. At α=0.05, zα/2=1.960.
One-tailed rejection (z)Z0>zαorZ0<−zα
Upper- or lower-tail H1. At α=0.05, zα=1.645.
Type I and Type II errorα=P(reject H0∣H0true),β=P(accept H0∣H0false)
Probability of correctly rejecting a false H0; rises with n and effect size.
P-value decision rulereject H0⟺p≤α
p = probability of a statistic at least as extreme as observed, given H0. Double the one-tail area for two-tailed tests.
Pooled variance (two means)sp2=n1+n2−2(n1−1)s12+(n2−1)s22
Combined variance estimate for comparing two means with equal variances; df =n1+n2−2.
Worked examples
One-tailed z test: does effluent TSS exceed the limit?
Problem. An NPDES limit sets the mean effluent TSS at μ0=30mg/L; the instrument's standard deviation is known, σ=8mg/L. Thirty-six samples give Xˉ=33mg/L. At α=0.05, is the mean significantly above the limit? Also estimate the p-value, and the chance of missing a true mean of 34mg/L.
Solution. Engineering question is one-sided upper: H0:μ=30 vs. H1:μ>30. σ known → use z.
Z0=8/3633−30=8/63=1.3333=2.25.
Critical value z0.05=1.645. Since Z0=2.25>1.645, reject H0: the mean TSS is significantly above 30mg/L.
P-value =R(2.25)=1−F(2.25)=1−0.9878=0.012. Because 0.012<0.05, the same reject decision follows.
Type II illustration: the critical sample mean is Xˉ∗=30+1.645(1.333)=32.19. If the true mean is 34, β=P(Xˉ<32.19)=F(1.33332.19−34)=F(−1.36)=0.087, so power =1−0.087=0.913.
Sanity check: a measured mean 2.25 standard errors above the limit is convincingly high, p≈1%, and the test has ∼91% power to catch a 4mg/L true exceedance — all internally consistent.
Z0=8/3633−30=2.25>1.645⇒reject H0,p=0.012
One-tailed t test: is lead below the MCL?
Problem. A drinking-water standard is μ0=15μg/L for lead. Ten samples give Xˉ=14.2μg/L with s=2.5μg/L (σ unknown). At α=0.05, can you conclude the mean is below the standard?
Solution. To demonstrate compliance below the limit: H0:μ=15 vs. H1:μ<15. σ unknown → use t with n−1=9 df.
Standard error =s/n=2.5/10=0.791.
t0=0.79114.2−15=0.791−0.8=−1.01.
Critical value t0.05,9=1.833, so the lower-tail rejection requires t0<−1.833. Since −1.01>−1.833, fail to reject H0.
P-value =P(T9<−1.01)≈0.17>0.05, confirming the decision.
Sanity check: the sample mean is only about one standard error below the limit — well inside ordinary sampling noise — so the data cannot establish that the true mean is below 15μg/L. “Fail to reject” is not proof of compliance; it means more data are needed.
t0=2.5/1014.2−15=−1.01,t0.05,9=1.833⇒fail to reject H0
Two-tailed t test: is a temperature sensor calibrated?
Problem. A stream-temperature probe is checked against a 20.0∘C reference bath. Twelve readings give Xˉ=20.4∘C with s=0.6∘C. At α=0.05, is the probe out of calibration (in either direction)?
Solution. Bias in either direction → two-tailed: H0:μ=20.0 vs. H1:μ=20.0. σ unknown → t with n−1=11 df.
Standard error =0.6/12=0.1732.
t0=0.173220.4−20.0=0.17320.4=2.31.
Two-tailed critical value t0.025,11=2.201. Since ∣2.31∣>2.201, reject H0: the probe shows a statistically significant bias.
P-value (two-tailed) =2P(T11>2.31)≈0.041<0.05, consistent with rejection.
Sanity check: the 0.4∘C offset is just over two standard errors, landing barely inside the rejection region — a marginal but significant result, exactly what a p≈0.04 implies. Note that a one-tailed test at the same α would use t0.05,11=1.796 and reject more easily; the two-tailed form is the honest choice when bias direction is not specified in advance.
•Using zα for a two-tailed test. Two-tailed critical values use α/2 in each tail (1.960 at α=0.05, not 1.645).
•Choosing z when only s is known. If the standard deviation is estimated from the sample, use t with n−1 degrees of freedom.
•Treating “fail to reject H0” as proof that H0 is true. It only means the evidence is insufficient to reject — absence of evidence, not evidence of absence.
•Confusing Type I and Type II error: α is a false alarm (reject a true H0); β is a missed detection (keep a false H0).
•Comparing the p-value the wrong way: reject when p≤α. A large p-value supports the null, not the alternative.
•Getting the tail direction backward: a sample mean below μ0 gives a negative statistic and can never reject an upper-tail H1:μ>μ0.
•Putting σ (or s) in the statistic instead of the standard error σ/n — the n is what makes the test sensitive.
References
NCEES FE Reference Handbook — Engineering Probability and Statistics
Montgomery & Runger, Applied Statistics and Probability for Engineers — tests on means of a normal distribution
Davis & Cornwell, Introduction to Environmental Engineering — statistical compliance decisions