The shortest section in the chapter, and the book concedes at the outset that the situation it describes is not likely, since knowing the population standard deviations rarely happens. Everything from section 10.1 carries over unchanged except two things. The two population standard deviations replace the two sample standard deviations in the standard error, so the variances added are the true ones rather than estimates, and the sampling distribution of the difference becomes normal rather than a Student t. That second change removes the Aspin-Welch degrees of freedom entirely, which is the whole practical simplification. Both populations must be normal, the samples must be independent, and the test statistic is a z-score formed exactly as before: the difference of the sample means divided by the standard error of that difference.
Subject: Statistics · 65 slides · symbolic lesson
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Title
Statistics · Chapter 10 — Hypothesis Testing with Two Samples
Two Population Means with Known Standard Deviations
Objectives
Five outcomes, and the first two are the only differences from section 10.1.
OpenStax Introductory Statistics 2e, §10.2 Two Population Means with Known Standard Deviations §10.2, pp. 520-522 — the section these objectives are drawn from
Warm-up
Section 10.1 built a two-sample test from a standard error and Aspin-Welch degrees of freedom.
Discussion prompt
Suppose both population standard deviations were handed to you rather than estimated from the samples. Which parts of section 10.1's procedure would still be needed?
Hint: What were the degrees of freedom there for?
Answer:
The standard error is still needed, with the same structure — each variance over its own sample size, added, then rooted. What changes is that the variances are now the true ones rather than estimates.
The degrees of freedom are not needed at all. They existed to describe the extra uncertainty in estimating two standard deviations from data, and with the standard deviations given there is no such uncertainty to describe.
So the distribution becomes a plain normal. That is the whole content of this section, which is why the book covers it in three pages and opens by admitting the situation is unlikely.
Concept
Even though this situation is not likely, since knowing the population standard deviations is not likely, the same comparison can be made when they are known. The sampling distribution for the difference between the means is normal, and both populations must be normal.
the two-sample z test — A test of a difference of means when both population standard deviations are known. The statistic is a z-score, so no degrees of freedom are involved.
\[ z = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \]
The requirement that both populations be normal is worth noticing, because section 10.1 relaxed it for large samples. Here it is stated flatly, and the reason is that this test is used precisely where sigma is known from long experience — typically manufacturing or measurement settings, where the population's normality is usually known from the same long experience.
Figure (svg): Two columns contrasting the unknown-sigma test with the known-sigma test
OpenStax Introductory Statistics 2e, §10.2 Two Population Means with Known Standard Deviations §10.2, p. 520
Section
Section 1
Concept
The sample standard deviations are replaced by the population ones, and the Student t is replaced by the normal. Everything else — the hypotheses, the tail, the decision rule, the conclusion — is exactly as in section 10.1.
what carries over — The parameter is still a difference of means, the point estimate is still the difference of sample means, and the variances are still added. Only the source of the spreads and the resulting distribution differ.
\[ s_i \;\to\; \sigma_i, \qquad t_{\text{df}} \;\to\; N(0,1) \]
It is worth being explicit that the simplification runs one way only. Knowing sigma makes the test sharper, because the normal's multiplier is smaller than any t's, and it removes a computation. It does not change any conclusion's meaning, and it cannot be assumed into existence — a sigma taken from the sample is an estimate whatever it is called.
Figure (svg): Two columns contrasting the unknown-sigma test with the known-sigma test
OpenStax Introductory Statistics 2e, §10.2 Two Population Means with Known Standard Deviations §10.2, p. 520 — the section's opening and the normal distribution
Picture it
Six rows, of which two are genuine changes.
Figure (svg): Two columns contrasting the unknown-sigma test with the known-sigma test
The right column's fifth row is the book's own assessment, and it is unusually blunt for a textbook. The section exists to complete the pattern — every combination of one or two samples with known or unknown spreads — rather than because the case arises often.
Worked example
Example 10.6's setup.
\[ \text{two floor waxes; population standard deviations } 0.33 \text{ and } 0.36 \]
Count the samples
Why: Two waxes.
Identify the parameter
Why: Mean months lasting.
Check the spreads
Why: Given as population values.
Choose the distribution
Why: Nothing is estimated.
Figure (svg): The solution to Worked example identifying the case shown as a ladder of expressions, one row per legal move
\[ z \sim N(0, 1) \]
Verify: confirm the word population is doing the work
Why: The table's column heading reads population standard deviation rather than sample standard deviation, and that single word selects the whole method. Had it read sample, section 10.1's t and its Aspin-Welch degrees of freedom would apply instead. Reading table headings carefully is the practical version of section 9.3's rule about where the spread came from.
OpenStax Introductory Statistics 2e, §10.2 Two Population Means with Known Standard Deviations §10.2, p. 520
Sorting
Ask where each spread came from.
Sort into buckets
Sort each two-sample comparison.
Item (d) is the one that catches people: a pilot study's estimate is still an estimate, however carefully it was made. Only a genuinely known population value licenses the normal.
Worked example
The same numbers treated as sample standard deviations.
\[ \text{if } 0.33 \text{ and } 0.36 \text{ were } s_1 \text{ and } s_2 \]
The standard error
Why: Identical formula.
\[ \text{still } 0.1092 \]
The statistic
Why: Identical arithmetic.
\[ \text{still } 0.9157 \]
The distribution
Why: A t instead.
The p-value
Why: Slightly larger.
\[ \text{about } 0.183 \]
Figure (svg): The solution to Worked example what would have changed shown as a ladder of expressions, one row per legal move
\[ p_z = 0.1799 \quad\text{against}\quad p_t \approx 0.183 \]
Verify: confirm the direction and size of the difference
Why: The t p-value is larger, as it always is, since the t has heavier tails — so treating estimates as known values would overstate the evidence. Here the difference is about 0.003 and changes nothing, because the samples are large enough that the t has nearly 38 degrees of freedom. On samples of five the same substitution would matter a great deal.
OpenStax Introductory Statistics 2e, §10.2 Two Population Means with Known Standard Deviations §10.2, pp. 520-521
Trap
\[ s_1 = 0.33 \text{ from the data, used as if it were } \sigma_1 \]
Use the number given, whatever it is called
Why: Both are standard deviations.
\[ \text{a normal is used where a t belongs} \]
The p-value comes out too small, since the normal has lighter tails than any t.
\[ \text{check whether the spread was GIVEN or COMPUTED} \]
Let the source decide, as in section 8.2
Why: Estimated spreads need a t.
This is the same decision section 8.2 settled for one-sample intervals and section 9.3 restated for one-sample tests, and it is settled the same way here. The rule refers to where the number came from, never to how large the samples are.
Fill the middle
The practical simplification of knowing sigma.
Fill in the blanks
\textdegrees of freedom ___ \text___
Why: Degrees of freedom. They described the uncertainty in estimating the spreads, and with the spreads given there is none to describe — so the normal applies at any sample size.
Two truths and a lie
All three concern what changes.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. The hypotheses concern a difference of population means and are written identically whether or not the spreads are known — what is known about the spreads affects the distribution used, not the claim being tested.
Prediction
Commit before reasoning.
Predict first
On identical data, which gives the smaller p-value: the z test or the t test?
Correct: The z.
Why: The normal has lighter tails than any t, so less area lies beyond a given statistic and the p-value is smaller. That is why knowing sigma produces a sharper test — and why using a z when sigma was only estimated always overstates the evidence.
Section
Section 2
Concept
The standard deviation of the difference is the square root of the first population's variance over the first sample size plus the second population's variance over the second sample size. The test statistic is the difference of sample means divided by it.
the known-sigma standard error — Identical in form to section 10.1's, with sigma-squared where s-squared stood. Because the sigmas are exact, this standard error is exact rather than estimated.
\[ \text{SE} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \]
The variances are added for the same reason as before: the two samples are independent, so their uncertainties combine rather than cancel. Nothing about that argument depended on whether the spreads were known — it depended only on the samples being independent, which both sections assume.
Figure (svg): A card giving the standard error of a difference when both population standard deviations are known
OpenStax Introductory Statistics 2e, §10.2 Two Population Means with Known Standard Deviations §10.2, p. 520 — the standard deviation and the test statistic
Picture it
The formula, and the statistic built from it.
Figure (svg): A card giving the standard error of a difference when both population standard deviations are known
It is worth noticing what is NOT here: no degrees of freedom line, and no warning about pooling. Pooling was a question about how to combine two estimated spreads, and with both given exactly there is nothing to combine.
Worked example
Example 10.6, with twenty floors per wax.
\[ \sigma_1 = 0.33, \; \sigma_2 = 0.36, \; n_1 = n_2 = 20 \]
Wax 1's term
Why: 0.33 squared over 20.
\[ 0.005445 \]
Wax 2's term
Why: 0.36 squared over 20.
\[ 0.006480 \]
Add
Why: The variance of the difference.
\[ 0.011925 \]
Take the root
Why: The standard error.
\[ 0.1092 \]
Figure (svg): The solution to Worked example the floor wax standard error shown as a ladder of expressions, one row per legal move
\[ \text{SE} = \sqrt{\frac{0.33^2}{20} + \frac{0.36^2}{20}} \approx 0.1092 \]
Verify: confirm the result exceeds each wax's own standard error
Why: Wax 1's mean has a standard error of 0.33 over root 20, about 0.0738, and wax 2's is about 0.0805. The difference's 0.1092 exceeds both, as combining independent uncertainties must. That check works identically here and in section 10.1, since it follows from independence rather than from what is known.
OpenStax Introductory Statistics 2e, §10.2 Two Population Means with Known Standard Deviations §10.2, pp. 520-521
Faded example
Two engines, thirty each, with population standard deviations 1.5 and 1.2.
Fill in the blanks
\text0.123 = \sqrt0.351___ + \frac______} = \sqrt___} \approx ___
Why: 0.075 plus 0.048 gives 0.123, whose square root is about 0.351 — the typical size of a difference of sample means under a null of equality.
Worked example
Standardising the observed difference.
\[ \bar{x}_1 = 3, \; \bar{x}_2 = 2.9 \]
Compute the difference
Why: Three minus 2.9.
\[ 0.1\text{ months} \]
Recall the null's claim
Why: Mu-1 at most mu-2.
\[ \text{difference at most } 0 \]
Divide by the standard error
Why: 0.1 over 0.1092.
\[ 0.9157 \]
Interpret
Why: Under one standard error.
Figure (svg): The solution to Worked example the test statistic shown as a ladder of expressions, one row per legal move
\[ z = \frac{0.1}{0.1092} \approx 0.92 \]
Verify: confirm why the curve is centred at zero
Why: The book notes that since mu-1 is at most mu-2, the difference mu-1 minus mu-2 is at most zero, and the mean for the normal distribution is zero. The boundary value of the null is what the distribution is centred on — the same rule chapter 9 gave, applied to a difference rather than to a single mean.
OpenStax Introductory Statistics 2e, §10.2 Two Population Means with Known Standard Deviations §10.2, p. 521
Error analysis
The correct value is about 0.1092.
Annotate
On: \( \begin{aligned} &(1)\; \sqrt{\tfrac{0.33^2 + 0.36^2}{20}} \approx 0.1092 \\ &(2)\; \tfrac{0.33 + 0.36}{\sqrt{20}} \approx 0.1543 \\ &(3)\; \sqrt{\tfrac{0.33^2}{20} - \tfrac{0.36^2}{20}} \;\text{undefined} \\ &(4)\; \sqrt{\tfrac{0.33^2}{20} + \tfrac{0.36^2}{20}} \approx 0.1092 \end{aligned} \)
Error (1) is the instructive one. It gives the right answer on this problem and the wrong one on almost any other, which is exactly the kind of coincidence that makes a wrong method survive — worth remembering when checking work against a single example.
Two truths and a lie
All three concern the standard error.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. Knowing sigma does not shrink the standard error — the formula and the numbers are identical. What knowing sigma buys is a smaller MULTIPLIER, because a normal is used instead of a t, and that is where the sharper test comes from.
Estimation
A gap of 0.1 with a standard error of 0.1092.
Predict first
Will the test reject at any conventional level?
Correct: No: the gap is under one standard error.
Why: A z under 1 leaves a one-tailed area above 0.15, which no conventional alpha would reject. Comparing the gap with the standard error before computing gives an immediate read on the outcome, exactly as in section 10.1.
Prediction
Commit before reasoning.
Predict first
For a null that mu-1 is at most mu-2, where is the sampling distribution of the difference centred?
Correct: At zero.
Why: The book says so directly: since mu-1 is at most mu-2, the difference is at most zero and the mean for the normal distribution is zero. As throughout chapter 9, the distribution is centred on the value the null names, and for a one-sided null that is the boundary.
Section
Section 3
Concept
With the standard error computed and the distribution known to be normal, the test proceeds exactly as every test since chapter 9: read the tail from the alternative, find the p-value, compare with alpha, decide, and write the conclusion in context.
the unchanged procedure — Hypotheses, tail, distribution, p-value, decision, conclusion. Only step three's answer differs from section 10.1, and only because sigma is known.
\[ z = 0.92 \;\to\; p = 0.1799 \;\to\; \text{do not reject} \]
The book's conclusion for this example is worth reading for its care about direction: at the 5 percent level there is not sufficient evidence to conclude that the mean time wax 1 lasts is longer, meaning that wax 1 is more effective, than the mean time wax 2 lasts. It names the level, the direction and the practical claim, and it says not sufficient evidence rather than that the waxes are equal.
Figure (svg): A normal curve centred at zero with the right tail beyond nought point one shaded
OpenStax Introductory Statistics 2e, §10.2 Two Population Means with Known Standard Deviations §10.2, p. 521 — Example 10.6 worked in full
Picture it
Example 10.6: does wax 1 last longer?
Figure (svg): A normal curve centred at zero with the right tail beyond nought point one shaded
The shaded area is large — nearly a fifth of the curve — which is what a statistic under one standard error looks like. Nothing about the observed 0.1-month advantage is hard to explain if the two waxes are in truth equally durable.
Worked example
Example 10.6, at the 5 percent level.
\[ \bar{x}_1 = 3, \; \bar{x}_2 = 2.9; \; \alpha = 0.05 \]
Set the hypotheses
Why: Wax 1 more effective means longer.
\[ H 0: \mu 1 \le \mu 2, H a: \mu 1 > \mu 2 \]
Read the tail
Why: Longer is a greater-than.
Compute the statistic
Why: 0.1 over 0.1092.
\[ z = 0.92 \]
Find the p-value and decide
Why: Right tail; alpha below it.
\[ 0.1799;\text{ do not reject} \]
Figure (svg): The solution to Worked example the floor wax test in full shown as a ladder of expressions, one row per legal move
\[ p = 0.1799 > 0.05 \]
Verify: confirm the conclusion does not claim the waxes are equal
Why: The book's wording is not sufficient evidence to conclude that wax 1 lasts longer, which leaves open that it might. Concluding the two waxes are equally durable would be accepting the null, which section 9.1 forbids — and here it would be a commercially consequential overstatement, since a real advantage could easily have gone undetected with twenty floors per wax.
OpenStax Introductory Statistics 2e, §10.2 Two Population Means with Known Standard Deviations §10.2, p. 521
Faded example
A right-tailed z test with a statistic of 0.9157.
Fill in the blanks
p\text0.1799 = 1 - \Phi(0.9157) \approx do not reject, \text___ \alpha = 0.05 \text___ ___
Why: The right-tail area beyond 0.9157 is about 0.1799, comfortably above 0.05, so the null survives and the evidence is insufficient.
Worked example
Assessing the study's sensitivity after the fact.
\[ \text{SE} = 0.1092; \; \text{a right-tailed test at } 5 \text{ percent} \]
Find the critical z
Why: For a 5 percent right tail.
\[ 1.645 \]
Convert to a difference
Why: 1.645 times 0.1092.
\[ 0.18\text{ months} \]
Compare with the observation
Why: 0.1 months.
State the implication
Why: Smaller effects go undetected.
\[ \text{under about } 0.18 \]
Figure (svg): The solution to Worked example what the test could have detected shown as a ladder of expressions, one row per legal move
\[ 1.645 \times 0.1092 \approx 0.18 \text{ months} \]
Verify: confirm this is a statement about the design rather than the result
Why: The calculation uses only the standard error and the significance level, both of which were fixed before the data arrived — so it describes what the study was capable of, which is section 9.2's power question. Reporting it alongside a non-significant result is what turns an uninformative conclusion into a useful one: a difference of a fortnight in a wax's lifetime may or may not matter commercially, and the study simply could not see it.
OpenStax Introductory Statistics 2e, §10.2 Two Population Means with Known Standard Deviations §10.2, pp. 521-522
Trap
\[ p = 0.1799 \;\Rightarrow\; \text{the two waxes last equally long} \]
Read a failure to reject as showing no difference
Why: The test found no evidence of one.
\[ \text{a true gap of } 0.15 \text{ months would also have given } p > 0.05 \]
The data are consistent with equality and with a range of real differences, and cannot distinguish them.
\[ \text{not sufficient evidence that wax 1 lasts longer} \]
Report the absence of evidence, and say what the study could detect
Why: Section 9.1's rule, unchanged.
With twenty floors per wax and a standard error of 0.109 months, the study could only reliably detect advantages above about 0.18 months. Stating that alongside the p-value tells a reader whether the non-result is informative, which the p-value alone does not.
Two truths and a lie
All three concern running the test.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. Any test can miss a real difference — that is a Type II error, and its probability depends on the sample size and the size of the true gap. Here a real advantage under about 0.18 months would usually be missed.
Estimation
A standard error of 0.1092 and a right-tailed test at 5 percent.
Predict first
Roughly what is the smallest difference this study would usually detect?
Correct: About 0.18.
Why: The critical value of 1.645 times the standard error of 0.1092 gives about 0.18. Anything below that produces a statistic under 1.645 and so a p-value above 0.05 — which is what makes reporting this figure alongside a non-result worthwhile.
Explain it
A classmate writes: p was 0.18, so the two waxes are equally effective.
Discussion prompt
In two sentences or fewer, correct them.
Hint: Ask what other true differences would have produced the same result.
Answer:
A real advantage of a tenth of a month, or even 0.15, would also have given a p-value above 0.05 with only twenty floors per wax — so the data cannot tell equality apart from a modest real difference.
The honest conclusion is that there is not sufficient evidence that wax 1 lasts longer, and that the study could only have detected an advantage above about 0.18 months.
Section
Section 4
Concept
Section 8.2 introduced the t because the sample standard deviation varies from sample to sample, adding a second source of uncertainty. When the population standard deviations are known there is no such variation to account for, so the normal applies directly.
the price of estimating — The t's extra tail width relative to the normal. It grows as degrees of freedom shrink, and it vanishes entirely when nothing is estimated.
\[ t_{\text{df}} \;\to\; N(0,1) \;\text{ as }\; \text{df} \to \infty \]
This is the cleanest illustration in the book of what the t is for. Two tests can be identical in every respect — same data, same statistic, same hypotheses — and differ only in whether the spreads were given or estimated. The whole of the difference between their p-values is the price of that estimation, and it is exactly what the t distribution charges.
Figure (svg): A normal curve drawn over a t curve with few degrees of freedom, showing the t sitting lower in the centre and higher in the tails
OpenStax Introductory Statistics 2e, §10.2 Two Population Means with Known Standard Deviations §10.2, p. 520 — the distribution for the difference
Picture it
A normal against a t with few degrees of freedom.
Figure (svg): A normal curve drawn over a t curve with few degrees of freedom, showing the t sitting lower in the centre and higher in the tails
The gap is in the tails, which is where p-values live — so the choice matters most exactly where it is used. With sigma known, the narrower curve is the correct one, and the test is sharper as a result rather than merely more convenient.
Worked example
The 95 percent two-tailed value, by what is known.
\[ \text{sigma known; and } t \text{ at } 38, 18 \text{ and } 6 \text{ df} \]
Sigma known
Why: The normal.
\[ 1.960 \]
t with 38 df
Why: Large samples.
\[ 2.024 \]
t with 18 df
Why: Moderate samples.
\[ 2.101 \]
t with 6 df
Why: Small samples.
\[ 2.447 \]
Figure (svg): The solution to Worked example the multiplier across the cases shown as a ladder of expressions, one row per legal move
\[ 1.960 \to 2.024 \to 2.101 \to 2.447 \]
Verify: confirm the ordering can never reverse
Why: Every t exceeds the normal at every finite degrees of freedom, so the known-sigma test always gives the smallest multiplier and the sharpest result. A t multiplier below 1.96 at a 95 percent level would be impossible, which makes the ordering a reliable check on any table lookup.
OpenStax Introductory Statistics 2e, §10.2 Two Population Means with Known Standard Deviations §10.2, pp. 520-521
Two truths and a lie
All three concern why the t exists.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false, and it is the same misconception section 8.2 corrected. The deciding question is whether the standard deviations were estimated. A tiny sample with genuinely known sigmas uses a normal; a huge sample with estimated ones uses a t.
Worked example
Comparing the two treatments on Example 10.6's data.
\[ n_1 = n_2 = 20, \text{ so a } t \text{ would have about } 38 \text{ df} \]
The z p-value
Why: From the normal.
\[ 0.1799 \]
The t p-value
Why: About 38 df.
\[ \text{about } 0.183 \]
Compare
Why: The difference.
\[ \text{about } 0.003 \]
Assess
Why: No decision changes.
Figure (svg): The solution to Worked example when the difference stops mattering shown as a ladder of expressions, one row per legal move
\[ 0.1799 \quad\text{against}\quad 0.183 \]
Verify: confirm this does not license using a z when sigma is unknown
Why: The agreement is a property of these sample sizes, not of the method. With five floors per wax the t would have about 8 degrees of freedom and the two p-values would differ noticeably — and the error would run in the direction of overstating the evidence. Section 8.2's rule stands regardless of how close the numbers happen to be in a particular case.
OpenStax Introductory Statistics 2e, §10.2 Two Population Means with Known Standard Deviations §10.2, pp. 520-521
Trap
\[ 0.1799 \text{ and } 0.183 \text{ are close, so either will do} \]
Conclude the distinction is academic
Why: The numbers agree to three decimals.
\[ \text{but that agreement depends on } n = 20 \text{ per group} \]
At five per group the gap would be substantial, and always in the direction of overstating significance.
\[ \text{use the distribution the situation calls for} \]
Let the source of the spread decide, whatever the sample size
Why: The rule has no threshold.
A rule with no exceptions is easier to apply correctly than one with a case analysis, and here it costs nothing — a calculator computes either p-value in the same keystroke. The agreement at large samples is a fact about the t approaching the normal, not a licence to choose freely.
Faded example
At a 95 percent level with 6 degrees of freedom.
Fill in the blanks
t = 2.447 \text1.96 z = 0.49, \text___ ___
Why: About 0.49 — a 25 percent wider multiplier, which is what estimating two spreads from very small samples costs.
Prediction
Commit before reasoning.
Predict first
Why does a known-sigma test have no degrees of freedom?
Correct: Because nothing is estimated except the means.
Why: Degrees of freedom count the constraints imposed by estimating parameters from the data. With both spreads given, only the two means come from the samples — and the normal distribution of a sample mean needs no degrees-of-freedom index. Normality of the populations is assumed by both tests and settles nothing here.
Estimation
Two tests on the same data, one z and one t with 8 degrees of freedom.
Predict first
Which comparison shows the larger discrepancy: 8 df or 38 df?
Correct: 8 df, by a wide margin.
Why: The t's excess width shrinks as degrees of freedom grow, so the two distributions differ most when the samples are small. At 38 degrees of freedom the p-values agreed to three thousandths; at 8 the difference would be many times larger and could change a decision.
Section
Section 5
Concept
The book opens the section by conceding that knowing the population standard deviations is not likely. The cases where it genuinely happens share a feature: the spread was established from a large body of prior experience rather than from the data being analysed.
a genuinely known sigma — A population standard deviation established independently of the current samples — from a long-running process, a manufacturer's specification, or a published standard. A value computed from the data at hand is an estimate, whatever it is called.
\[ \sigma \text{ from prior knowledge}, \quad s \text{ from these data} \]
Distinguishing the two is a reading skill rather than a computation, and it is the same one section 8.2 required. A table headed population standard deviation licenses a normal; one headed sample standard deviation does not. Where a problem is ambiguous, treating the spread as estimated is the conservative choice, since the t only ever widens the result.
Figure (svg): When the population standard deviations are genuinely known
OpenStax Introductory Statistics 2e, §10.2 Two Population Means with Known Standard Deviations §10.2, p. 520 — the book's opening concession
Picture it
Where a population standard deviation might genuinely be known.
Figure (svg): When the population standard deviations are genuinely known
The last line is the book's own assessment and the honest summary. This section exists to complete the pattern of one or two samples against known or unknown spreads, and a student is far more likely to meet section 10.1's case in practice.
Worked example
Deciding which method each calls for.
\[ \text{four descriptions of where a spread came from} \]
A decades-old filling process
Why: Measured over years.
A pilot study of twelve
Why: Computed from data.
A manufacturer's stated tolerance
Why: Independent of these data.
The samples being analysed
Why: Computed here.
Figure (svg): The solution to Worked example classifying four settings shown as a ladder of expressions, one row per legal move
\[ \text{prior knowledge} \to z; \quad \text{computed here} \to t \]
Verify: confirm the pilot study is genuinely an estimate
Why: A pilot of twelve gives a sample standard deviation whose own uncertainty is considerable — section 8.2's t multiplier at 11 degrees of freedom is 2.2 rather than 1.96, which is a measure of exactly that. Treating it as known would ignore a real source of variability, and the resulting p-values would be too small.
OpenStax Introductory Statistics 2e, §10.2 Two Population Means with Known Standard Deviations §10.2, p. 520
Sorting
Ask where the number came from.
Sort into buckets
Sort each source of a standard deviation.
Item (e) is the only case where sigma is known with complete certainty, since the simulation's author chose it. Everything on the left is known well enough for practical purposes; everything on the right needs a t.
Worked example
What to do when a problem is ambiguous.
\[ \text{a problem gives } 0.33 \text{ and } 0.36 \text{ without saying which} \]
Assume known
Why: Use a normal.
Assume estimated
Why: Use a t.
Ask which error is worse
Why: Overstating evidence.
Choose
Why: The conservative one.
Figure (svg): The solution to Worked example the conservative default shown as a ladder of expressions, one row per legal move
\[ \text{ambiguous} \;\Rightarrow\; \text{use the } t \]
Verify: confirm the direction of the risk
Why: A t always gives the larger p-value, so assuming estimation can only make a conclusion more cautious. The reverse assumption can turn a non-significant result into a significant one — an error that flatters the analyst and is invisible in the output. Where a choice must be made without evidence, the one that cannot manufacture significance is the right default.
OpenStax Introductory Statistics 2e, §10.2 Two Population Means with Known Standard Deviations §10.2, pp. 520-521
Trap
\[ \text{s from } 500 \text{ prior observations, treated as } \sigma \]
Treat a very precise estimate as exact
Why: Five hundred observations pin it down well.
\[ \text{it is still an estimate, and a t with 499 df is available} \]
The difference is tiny here, but the rule needs no exception since the t costs nothing to use.
\[ \text{use a t; at large df it is nearly the normal anyway} \]
Let the source decide, and let the df handle the precision
Why: A well-estimated spread gives a t that is almost normal.
This is the elegant part of the t's design: it does not need a threshold, because the degrees of freedom already encode how well the spread is known. A spread from five observations gets a wide correction and one from five hundred gets almost none, automatically.
Two truths and a lie
All three concern recognising the case.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. A large sample gives a precise estimate, and the t with many degrees of freedom already accounts for that precision — its multiplier is nearly the normal's. There is no sample size at which an estimate becomes a known value.
Prediction
Commit before reasoning.
Predict first
A problem does not say whether its standard deviations are population or sample values. Which assumption is safer?
Correct: Assume they are estimates.
Why: The t gives the larger p-value, so that assumption can only make a conclusion more cautious. Assuming knowledge that is not there can manufacture significance, and nothing in the output would reveal it — which makes the cautious default the right one.
Fill the middle
How the section opens.
Fill in the blanks
\textlikely ___
Why: Likely. The section exists to complete the pattern of one or two samples against known or unknown spreads, and the book is candid that a student will meet section 10.1's case far more often.
Comparison
Fill the blanks. Two rows differ and the rest are identical.
Comparison matrix
| Sigmas unknown (10.1) | Sigmas known (10.2) | |
|---|---|---|
| Spread used | s-1 and s-2, estimated | sigma-1 and sigma-2, given |
| Distribution | Student t | normal |
| Degrees of freedom | Aspin-Welch, usually fractional | none |
| Standard error formula | the same: each variance over its own n, added | the same |
The last row is the point worth carrying: the standard error's structure does not depend on what is known, because it follows from the samples being independent. Only what is substituted into it, and what distribution the result is read against, differ.
Pattern
Six steps, and the third is the only one that differs from section 10.1.
If the problem is ambiguous about whether the spreads are known, treat them as estimated and use section 10.1's t — that assumption never overstates the evidence.
OpenStax Introductory Business Statistics 2e, §10.5 Two Population Means with Known Standard Deviations §10.5 Two Population Means with Known Standard Deviations
Check
The distribution.
Check your understanding
Two population standard deviations are given. Which distribution does the test use?
Answer: A
Why: Nothing beyond the means is estimated, so no t correction is needed and no degrees of freedom arise.
Check
The standard error.
Check your understanding
For sigma = 0.33 with n = 20 and sigma = 0.36 with n = 20, what is the standard error of the difference?
Answer: A
Why: Add 0.005445 and 0.006480 to get 0.011925, then take the square root.
Check
Recognising the case.
Check your understanding
A pilot study of twelve subjects supplies a standard deviation. Which test applies?
Answer: A
Why: A value computed from a sample is an estimate however carefully it was obtained, and twelve observations pin it down loosely.
Real world
A factory runs two filling lines whose long-term standard deviations are known from years of records to be 1.8 ml and 2.4 ml. A shift supervisor samples 25 bottles from each and finds mean fills of 501.2 ml and 499.6 ml. He reports that line 1 overfills relative to line 2 and asks for line 1 to be recalibrated.
Discussion prompt
Test the claim properly, and say whether recalibration is the right response.
Hint: This is one of the rare settings where the population standard deviations really are known.
Answer:
The known sigmas make this a genuine z test. The standard error is the root of 1.8 squared over 25 plus 2.4 squared over 25, which is the root of 0.1296 plus 0.2304, or about 0.6 ml.
\[ z = \frac{501.2 - 499.6}{0.6} \approx 2.67, \qquad p_{\text{two}} \approx 0.0077 \]
The difference of 1.6 ml is about 2.67 standard errors, giving a two-tailed p-value of about 0.008. So there is strong evidence that the two lines differ in mean fill — the supervisor's observation is real and not sampling noise.
But recalibrating line 1 does not follow from the test. The test establishes that the lines differ from each other, not that either differs from the target. If the specification is 500 ml, line 1 is 1.2 ml high and line 2 is 0.4 ml low, and both deserve attention — a one-sample test of each line against 500 would answer the question actually being asked, and is the analysis the supervisor needed.
Two further points are worth making. The 2.4 ml spread on line 2 is a third larger than line 1's, and a line's variability matters as much as its centre for meeting a fill specification — that comparison is section 13.4's test of two variances rather than anything here. And because the sigmas are known from long records rather than estimated, this test is genuinely sharper than a t would be: the same data analysed with estimated spreads would give a slightly larger p-value, though at 25 per line the difference would not change the conclusion.
Commit first
Answer, then rate your confidence honestly.
Predict first
Why does a known-sigma two-sample test have no degrees of freedom?
Correct: Because nothing beyond the means is estimated.
\[ z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \sim N(0,1) \]
Why: The t distribution exists to account for the variability of an estimated standard deviation, and its degrees of freedom measure how well that estimate is pinned down. With both spreads given exactly there is nothing to account for, so the sampling distribution of the standardised difference is simply normal. Normality of the populations is assumed by both tests and does not distinguish them.
Explain it
They used a normal because their two samples were large, even though both standard deviations came from the data.
Discussion prompt
In two sentences or fewer, correct them.
Hint: Ask where their standard deviations came from.
Answer:
Their spreads were computed from the samples, so they are estimates — and the deciding question is the source of the number, never the sample size.
A t with large degrees of freedom is almost identical to a normal anyway, so using it costs nothing and keeps the rule free of exceptions.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the first, ask where the number came from rather than how large the samples are. For the second, each variance over its own n, added, then rooted. For the third, centre the curve on zero, the null's boundary difference. For the fourth, the t prices an estimated spread, and here nothing is estimated. Do five problems of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Twelve minutes — this is the chapter's shortest section.
Draw it
At the top, draw two columns headed sigmas unknown and sigmas known, and fill in four rows: the spread used, the distribution, the degrees of freedom, and the standard error formula — marking clearly that only the first three differ. Beneath, write the standard error formula with sigmas and note beside it that the variances are still added, for the same independence reason as in section 10.1. In the middle of the page, work Example 10.6 completely: compute each variance over 20, add them to get 0.011925, take the root to get 0.1092, form the statistic of about 0.92, draw a normal curve centred at ZERO with the right tail beyond 0.1 months shaded, and label the p-value 0.1799. Write the conclusion naming the level and the direction, and beneath it one sentence on why concluding the waxes are equally durable would be wrong. To the right, draw a normal and a t with six degrees of freedom on one pair of axes and write the four 95 percent multipliers — 1.96, 2.024, 2.101 and 2.447 — noting what each corresponds to. At the bottom, list three settings where sigma is genuinely known and one where it only looks known.
Check the standard error by confirming 0.1092 exceeds both 0.0738 and 0.0805, the two individual standard errors. Check the multipliers by confirming they increase as the degrees of freedom fall and that all three t values exceed 1.96, which they must at every finite df.
Recap
Five things, and only the first two are new material.
| If you see | Then |
|---|---|
| Population standard deviations given | A normal distribution, no degrees of freedom |
| Sample standard deviations computed | Section 10.1's t, with Aspin-Welch df |
| A spread from a pilot study | An estimate, however large the pilot |
| A one-sided null on a difference | Centre the curve on zero, its boundary |
| An ambiguous problem | Assume estimated, since the t never overstates |
| A non-significant result | Say what difference the study could have detected |
| Two very unequal sample sizes | The smaller group dominates the standard error |
Section 10.3 changes the parameter rather than what is known about the spread. When the two groups are compared on a proportion rather than a mean, the standard error is built from a single pooled proportion — because under a null of equality both samples are estimating the same value.
OpenStax Introductory Statistics 2e, §10.2 Two Population Means with Known Standard Deviations §10.2, pp. 520-522 — everything on these slides traces back here
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