L-INF-6: Hypothesis Tests for a Mean (σ Unknown) and for a Proportion
Module 4 · Statistical Inference
Section 1: Introduction
▾
A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
One survey, three p-values. (The scenario is hypothetical; the agency and the watchdog are invented.) A government agency says 15% of households in a region report food insecurity. A consumer watchdog surveys a random sample of 80 households and finds 18 that do: . After seeing that number, the watchdog’s press release asks whether the true rate is higher than 15%. Three analysts send the newspaper a report on the same data.
Analyst
Tails counted
Standard error built from
p-value
Headline at
1
one (higher)
the claim,
“The rate is higher than claimed”
2
two
the claim,
“Not enough evidence of a difference”
3
two
the sample,
“Not enough evidence of a difference”
Make a prediction before reading on. Which number should the newspaper print: 0.030, 0.060 or 0.108? Nobody did the arithmetic wrong; the analysts made different choices.
Pause and write. Write your choice and one sentence of reasoning on paper or in your notes. Core Concepts C5 and C6 settle it, and the Boss Fight brings it back.
This lesson runs two tests with the five-step reasoning you used in L-INF-5. The t-test checks a claim about a mean when is unknown. The one-proportion z-test checks a claim about a share. Along the way you will meet the two choices that separate the three analysts.
After this lesson, you will be able to:
Choose between a t-test, a z-test for a mean, a one-proportion z-test, “stop” and “estimate instead” from what a study tells you.
Compute and bound its p-value from the t-table with .
Check and , then compute , with in the standard error.
Fix the tail before seeing the data, and state a conclusion that separates statistical evidence from practical importance and never “accepts .”
Section 2: Prerequisites
▾
A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
What you need coming in, and why it matters today:
The five-step test (L-INF-5): hypotheses, conditions, test statistic, p-value, conclusion in context. Every test today follows it.
The decision rule (L-INF-5): reject when ; otherwise fail to reject. Never “accept .”
The t-distribution and (L-INF-3): when is estimated by , the reference curve is , read in the row . Recall Gosset’s bet: with , captured only about 85.5% of the time.
The proportion interval (L-INF-4): an interval estimates with a standard error built from . A test checks a claimed , which changes the standard error.
Quick check — can you recall these?
What are the degrees of freedom for a one-sample -test?
Bridge to today: In L-INF-5 you knew , so the test statistic was and a z-table gave the p-value. Today is usually unknown: the statistic uses , it follows a t-distribution, and the t-table can only bracket the p-value (for example ), which is enough to decide. Sample size does not change this: is used because is estimated, at any . For a proportion the statistic is still , but its standard error comes from the claimed , not from as in L-INF-4’s interval.
Section 3: Core Concepts
▾
A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
How this section is organized:
C1–C3: the t-test: when to use it, the statistic, and bounding its p-value from the t-table.
C4–C5: the proportion test: its conditions, and why its standard error uses (Analyst 3’s 0.108).
C6: one tail or two, fixed before the data (Analyst 1’s 0.030 against Analyst 2’s 0.060).
C7: statistically significant is not the same as important.
C1 — When to Use the t-Test
When is unknown, a test of a mean estimates it with , and the test statistic then follows a t-distribution with , not the normal curve. Judging it against the normal’s 1.96 makes the test reject true claims too often, most of all in small samples; is the right reference at every sample size.
Gosset’s bet, test edition. An analyst tests a claim that is actually true, using 4 measurements from a normal population. Not knowing , she computes and rejects whenever it lies beyond , as if were .How often will she wrongly reject the true claim: about 5%, 10%, 15% or 25% of the time?
Pause and write. Write your estimate on paper or in your notes, then check it with the figure.
Predict first: with n = 4 and a true claim, what share of tests will the z rule (|t| > 1.96, using s) reject? Then test.
Sample size n
False alarm: a true claim rejected
Claim kept
Your batches so far
Figure 1: Each batch draws 40 samples of size n from a normal population whose mean
really equals the claimed μ₀, so H₀ is true and every rejection is a false alarm. Each sample's t
= (x̄ − μ₀)/(s/√n) is judged twice: by the z rule (reject when |t| > 1.96, as if s were σ) and by
the t rule (reject when |t| > t* from the df = n − 1 row). After the first batch, bars show the
z rule's long-run rate at six sample sizes beside the rates your batches produced.
With the z rule rejects a true claim about 14.5% of the time, not 5%. That is L-INF-3’s 85.5% coverage seen from the test side: a test rejects exactly when the matching interval misses . Samples whose comes out smaller than make the statistic too large. Using with brings the false-alarm rate back to 5% at every sample size.
When to Use the t-Test for a Mean
Use the one-sample t-test when:
is unknown and estimated by the sample standard deviation ;
the observations are independent (a random sample; when sampling without replacement, less than 10% of the population);
the population shape suits the sample size: with a small sample, approximately normal with no strong skew or extreme outliers. Larger samples tolerate more, but no sample size switches the check off.
When is known from outside the sample (records, a specification), use the z-test from L-INF-5, with the same independence and shape conditions.
Decision rule in plain English: first make sure the design and the shape license a test of a mean. Then, if is given → z-test; if only is available → t-test. Sample size never switches one into the other.
Before the next figure, try one: a clinic wants to know its mean wait time. It has 25 recorded waits and no claimed value. Where does it end up in the map below?
Bold border: this lesson's boxes
Gold bar, “new”: a question no earlier lesson asked
① Estimate or test a claim? newis there a claimed value to test?
estimate →Interval (L-INF-2 to L-INF-4), no H₀, SE from the data
test a claim →go to ②
② Mean or proportion? measured amount vs. yes/no count
mean μ →go to ③ for a mean
proportion p →go to ③ for a proportion
For a mean ③ Is σ known? not “is n small?”
σ known →z-test for μ, SE = σ/√n
s only →t-test for μ, SE = s/√n, df = n − 1
either test →then ④
For a proportion ③ np₀ ≥ 10 and n(1 − p₀) ≥ 10? counts from the claim p₀, not p̂
yes →z-test for p, SE = √(p₀(1 − p₀)/n)
no →Stop: too few expected counts
Both mean tests ④ Shape OK for this n? newlook for strong skew or outliers when n is small
OK →run the test
not OK →Stop: collect more data or use another method
Figure 2:
Which procedure? Four questions, asked in order, lead from a research question to an
interval, a test, or a stop. Each box names what to check; a stop means the methods of
this course do not apply as they stand.
The clinic has no claimed value, so the first question sends it to an interval (L-INF-3), not a test. A test always starts from a claim.
A very common mistake is using the z-test when only is given, reasoning “n is large enough.” The choice between z and t is about whether is known. With a large the t-distribution is close to the normal, so the numbers barely differ, but is still an estimate, not a known .
C2 — The t Test Statistic
The formula has the same shape as L-INF-5’s z statistic. Replace with , and with .
t Test Statistic for a Population Mean
where is the value claimed by , is the sample standard deviation and is the sample size. counts how many standard errors sits from the claim: a large is strong evidence against .
Mini-example: Suppose ,,,.
: the p-value is read along the row of the t-table.
Colour key for every figure in this lesson:green is the claim (,,);blue is what the sample shows (,, the test statistic); orange is the p-value area, the one thing to read; a faint dashed line or curve is a wrong path, drawn so you can see what a mistaken analysis would do.
The most common arithmetic slip is instead of . For ,, not 9. The wrong row gives the wrong critical values and can change your decision.
C3 — Reading p-Values from the t-Table
The t-table gives critical values for a few tail areas. To find the p-value for a computed , find your row and the two critical values your falls between. That gives a bound: ” is between 0.02 and 0.05.”
Bounding the p-Value from the t-Table
For a computed with known :
Find the row.
Find the two critical values that bracket .
Read their column headings: the one-tail area (1) for a one-sided , the two-tail area (2) for with .
The p-value lies between those two areas.
Try it before the figure: for with and a two-sided test, which two columns of the row will light up, and what bound do they give?
Test
t-table row for df = 11: critical values t* by two-tail area α
two-tail α
0.20
0.10
0.05
0.02
0.01
0.001
df = 11
1.363
1.796
2.201
2.718
3.106
4.437
0.02 < p < 0.05
2.201 < |t| = 2.45 < 2.718, so the shaded area lies between the area beyond 2.718
(0.02) and the area beyond 2.201 (0.05).
Shaded: the p-value (the tails continue past the axis)
|t| from the sample
Table cutoff: the area beyond it is its α
Figure 3: One row of the t-table above the t curve with the same degrees of freedom. Each
dashed line is a table cutoff, and the area beyond it is that column's α. The shaded area beyond
|t| is the p-value. Slide |t| and read which two cutoffs it falls between: the table can only
bracket the shaded area, never measure it exactly.
, so the shaded area beyond lies between the two-tail areas 0.02 and 0.05: . Software gives , inside the bound.
Why bounds are enough: if and , then : reject . If , then : fail to reject. The bound tells you which side of the p-value falls on, and that is all the decision needs. The full t-table is in Section 10.
For a two-sided test, read the two-tail (2) headings. If uses , the p-value is . Reading the one-tail headings instead halves the p-value and can lead you to reject when you should not.
C4 — Conditions for the Proportion Test
When the question is about a population proportion , the test uses the normal curve, because is approximately normal when the null model predicts enough of each outcome.
Conditions for the One-Sample Proportion Test
Before testing , check:
: at least 10 expected successes under ;
: at least 10 expected failures under ;
Independent observations: a random sample, less than 10% of the population when sampling without replacement.
If a count falls below 10, stop: there is no trustworthy p-value from the normal curve. Gather a larger sample, or use an exact method (beyond this course).
Check the conditions with , not . They ask whether the null model predicts enough of each outcome. Surveying 40 households against a claim of 15% gives : stop, even if 12 of the 40 answered yes.
C5 — The Proportion Test Statistic
The statistic has the same shape as L-INF-5’s z: what the sample shows () minus the claim (), divided by the spread would have if were true, which is built from .
z Test Statistic for a Population Proportion
where is the sample proportion, is the value claimed by , and is the sample size. The numerator is what the sample shows minus the claim; the denominator is the spread predicts. For a two-sided test, .
Back to the watchdog: and , so the test may run. Predict before the figure: will the watchdog’s two-tailed p-value clear 0.05?
Survey
Alternative Hₐ
The claim, H₀
What the sample shows, z
The p-value area (tails continue past the axis)
✗ Wrong path: SE from p̂
Figure 4: The one-sample z test for a proportion, opening on the watchdog's survey. The
readout takes the statistic apart: the numerator is what the sample shows minus the claim, and
the denominator is the spread the null hypothesis predicts, built from p₀. The shaded area on
the curve is the p-value for the chosen alternative. Before any curve is drawn, the null
expected counts n·p₀ and n(1 − p₀) must both reach 10; a valid test also needs random sampling
or assignment and independent observations. The “What if” switch shows, faint and dashed, the
path a student takes when the SE is built from p̂ instead.
Why in the denominator: a test asks “if were true, how surprising is this sample?” So every part of the yardstick must come from : if the true proportion were , would vary with standard error . Using would assume the sample is exactly right in order to judge whether the sample is surprising.
That settles Analyst 3. With : and . With , the standard error grows to about 0.0467, shrinks to about 1.61, and the two-tailed p-value becomes about 0.108. That number answers a question about an interval (L-INF-4), not about the agency’s claim. Analyst 3’s report is out.
The most common error in proportion tests is putting in the denominator. The numerator measures the gap between what was observed () and what claims (); the denominator measures the variability expected if were true, so it uses .
C6 — One Tail or Two: Decided Before the Data
The rules are the ones from L-INF-5. Choose from the research question before looking at the data.
Choosing the Tail
Two-tailed( or ): the question asks whether the parameter differs, in either direction. The p-value counts both tails.
One-tailed(,, or ): the question named one direction before the data were collected. The p-value counts that tail only.
Now the last disagreement. The watchdog wrote “higher than 15%” only after it saw . Which p-value is honest: 0.030, 0.060 or 0.108? Flip the first switch in the figure and watch which tails are shaded.
Watchdog survey: 18 of 80 households, p̂ = 0.225. Claim H₀: p = 0.15 (np₀ = 12 and n(1 − p₀) =
68, both at least 10).
Analyst 1one tail · SE from p₀p ≈ 0.030 → rejectCurrent setting
Analyst 2two tails · SE from p₀p ≈ 0.060 → fail to reject
Analyst 3two tails · SE from p̂p ≈ 0.108 · ✗ wrong SE
Hₐ written
SE from
The claim, H₀
The sample's z
p-value area
✗ a wrong path
Statistic
z ≈ 1.88, with SE from p₀ = 0.15: √(0.15 × 0.85 / 80) ≈ 0.0399
p-value
p ≈ 0.030 (one tail)
Decision
Analyst 1: reject H₀ at α = 0.05
Why
Legitimate only because Hₐ: p > 0.15 was written before the survey. The watchdog wrote it after seeing p̂ = 0.225: flip the first switch.
Figure 5: One survey, three analysts. The curve is the standard normal model for z when
H₀: p = 0.15 is true; the orange area is the p-value. The first switch says when the alternative
hypothesis was written, the second where the standard error comes from. A faint dashed mark is a
path an analyst could take that the test does not allow.
Analyst 1’s 0.030 would be honest only if the direction had been fixed before the survey. Here it was chosen after the data pointed up, so the one-tailed test gives the data two chances to look surprising: choosing the tail afterwards doubles the false-alarm rate. The honest report is Analyst 2’s: , fail to reject at . The data are suggestive, not conclusive.
Choose the tail from the research question, not from the data. Selecting a one-tailed test after seeing which way the sample went is data snooping: it inflates the real false-alarm rate to about . If no direction was stated before the data, use a two-tailed test.
C7 — Practical vs. Statistical Significance
A statistically significant result () says the data are hard to reconcile with at the chosen threshold. It does not say the difference is large, meaningful, or caused by anything in particular.
Predict before the figure: a filler’s true mean is 0.1 mL above the claimed 500 mL, and mL is known. At roughly what sample size does a two-tailed z-test first call this 0.1 mL “significant” at 0.05?
Jump to
Figure 6: A bottle filler whose true mean is 0.1 mL above the claimed 500 mL, with known
σ = 2 mL, tested by a two-tailed z test at α = 0.05. The top bar is the difference, measured
against a 5 mL change that would matter in practice; the bottom bar is the p-value, with α
marked. Change n with the slider, type it, or step it by one, and watch which bar moves.
From bottles on, passes 1.96: the same tiny difference becomes “significant” because the sample is huge, not because 0.1 mL matters.
With a very large sample, even a tiny difference can give . Always report the estimated difference ( or ), preferably with a confidence interval, beside the p-value, and ask: “Does this difference matter in the real world?” Statistical significance alone cannot answer that.
Section 4: Worked Examples
▾
A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
Example 1 — Fully Worked: Caloric Intake (Two-Tailed t-Test)
A nutritionist claims adults in a certain region consume a mean of kcal/day. A random sample of adults—less than 10% of the region’s adult population—yields kcal and kcal. Caloric intake in the target population is approximately normally distributed. Test at .
Student's t-Distribution Table
Critical values (t*) for given degrees of freedom (df) and tail area.
Student's t-Distribution Table
df
Confidence
80%
90%
95%
98%
99%
99.9%
0.10 (1)0.20 (2)
0.05 (1)0.10 (2)
0.025 (1)0.05 (2)
0.01 (1)0.02 (2)
0.005 (1)0.01 (2)
0.0005 (1)0.001 (2)
1
3.078
6.314
12.706
31.821
63.657
636.619
2
1.886
2.920
4.303
6.965
9.925
31.599
3
1.638
2.353
3.182
4.541
5.841
12.924
4
1.533
2.132
2.776
3.747
4.604
8.610
5
1.476
2.015
2.571
3.365
4.032
6.869
6
1.440
1.943
2.447
3.143
3.707
5.959
7
1.415
1.895
2.365
2.998
3.499
5.408
8
1.397
1.860
2.306
2.896
3.355
5.041
9
1.383
1.833
2.262
2.821
3.250
4.781
10
1.372
1.812
2.228
2.764
3.169
4.587
11
1.363
1.796
2.201
2.718
3.106
4.437
12
1.356
1.782
2.179
2.681
3.055
4.318
13
1.350
1.771
2.160
2.650
3.012
4.221
14
1.345
1.761
2.145
2.624
2.977
4.140
15
1.341
1.753
2.131
2.602
2.947
4.073
16
1.337
1.746
2.120
2.583
2.921
4.015
17
1.333
1.740
2.110
2.567
2.898
3.965
18
1.330
1.734
2.101
2.552
2.878
3.922
19
1.328
1.729
2.093
2.539
2.861
3.883
20
1.325
1.725
2.086
2.528
2.845
3.850
21
1.323
1.721
2.080
2.518
2.831
3.819
22
1.321
1.717
2.074
2.508
2.819
3.792
23
1.319
1.714
2.069
2.500
2.807
3.768
24
1.318
1.711
2.064
2.492
2.797
3.745
25
1.316
1.708
2.060
2.485
2.787
3.725
26
1.315
1.706
2.056
2.479
2.779
3.707
27
1.314
1.703
2.052
2.473
2.771
3.690
28
1.313
1.701
2.048
2.467
2.763
3.674
29
1.311
1.699
2.045
2.462
2.756
3.659
30
1.310
1.697
2.042
2.457
2.750
3.646
40
1.303
1.684
2.021
2.423
2.704
3.551
50
1.299
1.676
2.009
2.403
2.678
3.496
60
1.296
1.671
2.000
2.390
2.660
3.460
80
1.292
1.664
1.990
2.374
2.639
3.416
100
1.290
1.660
1.984
2.364
2.626
3.390
∞
1.282
1.645
1.960
2.326
2.576
3.291
Step 1: State and
kcal kcal (two-tailed: the question is whether the mean differs in either direction)
Step 2: Check conditions and choose the procedure
I notice only the sample’s is given: is unknown and must be estimated, so I use the t-distribution. (The sample size is not the reason; is.) Random sample ✓, less than 10% ✓, approximately normal population ✓. Use t with .
Step 3: Compute the test statistic
kcal.
Step 4: Bound the p-value from the t-table
In the row:
has two-tail area 0.10;
has two-tail area 0.05.
Since , the two-tailed p-value satisfies .
Step 5: Conclusion
, so we fail to reject . There is insufficient evidence at the 5% significance level that the mean caloric intake in this region differs from 2,000 kcal/day. This does not show that the mean equals 2,000 kcal/day.
Example 2 — Predict, Then Check: Tire Durability
A tire manufacturer claims its tires last km. A random sample of tires—less than 10% of the production run—gives km and km. Tire lifetimes in the target population are approximately normally distributed with no strong outliers. Test at .
Before opening the solution, state and , justify the procedure and degrees of freedom, compute the test statistic, bound the one-tailed p-value, and decide whether the evidence supports a mean below 50,000 km. In particular, predict whether the p-value will fall below .
Pause and write. Record your answer and reasoning on paper or in your notes before continuing. Even if you are unsure, committing your thinking helps you see what you understood and what you need to revisit.
Show Solution
Step 1: State the hypotheses
km
km
This is left-tailed: the concern, stated before testing, is that the tires fall short of the claimed lower bound.
Step 2: Check conditions and choose the procedure
The sample is random, is unknown, and the population is approximately normal, so use a one-sample t-test with
Step 3: Compute the test statistic
Step 4: Bound the p-value
For :
has one-tail area 0.05.
has one-tail area 0.025.
Because ,
Step 5: Conclude
The p-value is greater than , so we fail to reject . There is insufficient evidence at the 1% significance level that mean tire life is below 50,000 km. The prediction that would not fall below is confirmed.
Example 3 — Recycling Rate (Two-Tailed Proportion Test)
A city claims 30% of households recycle regularly. A simple random survey of households—less than 10% of the city’s households—finds 54 that recycle (). Carry out the complete two-tailed test at : state the hypotheses, check the sampling and independence conditions and both success–failure counts, compute the null-based standard error and z statistic, find the p-value, and write the contextual conclusion.
Pause and write. Record your answer and reasoning on paper or in your notes before continuing. Even if you are unsure, committing your thinking helps you see what you understood and what you need to revisit.
Show Solution
Step 1:; (two-tailed).
Step 2: Check Conditions
✓
✓
Random sample ✓. Conditions met: use the z-test for a proportion.
Step 3: Compute and the Test Statistic
Step 4: Two-tailed: .
(from the z-table, rounding to 0.93).
.
Step 5:. We fail to reject .
There is insufficient evidence at the 5% level that the proportion of households that recycle regularly differs from 30%.
Example 4 — Find and Repair the Reasoning
A study tests against with a simple random sample of from a population of at least 500. The sample has (), and . The independence condition is therefore credible, while and establish the null expected-count condition. A student writes:
“Because , I accept and conclude that the true proportion equals 0.40.”
Identify both conceptual errors, recompute the null-based standard error and z statistic, and replace the conclusion with language that says exactly what the evidence supports.
Pause and write. Record your answer and reasoning on paper or in your notes before continuing. Even if you are unsure, committing your thinking helps you see what you understood and what you need to revisit.
Show Solution
Error 1 — Wrong standard error: The student used in the denominator instead of . Under , the standard error must use the null value.
The numerical decision is the same in this instance, but the student’s formula is invalid and could change a decision in another sample.
Error 2 — Unsupported conclusion language: We never “accept ,” and failing to reject does not establish equality.
The correct conclusion is: We fail to reject . There is insufficient evidence at the 5% significance level that the true proportion differs from 0.40.
Section 5: Guided Practice
▾
A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
Every step is guided, with feedback after each one. A fresh scenario appears each time you start again.
Problem 1 — A t-Test, Step by Step (C1 + C2 + C3 + C6)
The t-table excerpt has three rows; choosing the right one is part of the work.
Problem 2 — A Proportion Test, Step by Step (C4 + C5 + C6)
Check the conditions honestly: sometimes the test may not run.
Problem 3 — Which Tool? (C1 + C4)
Sort into groups
Which tool does each claim need?
Each card is a question someone wants answered. Sort it by the procedure you would run, or into “not a test yet”. Read the dot plots: shape is part of the decision.
Drag an item into its group, or select it and then choose the group. On a keyboard: select an item and press the group’s number (1–4); 0 sends it back to Unsorted. Each item also has its own “move” buttons.
Unsorted Drop here
Every item is sorted.
Section 6: Independent Practice
▾
A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
From here on you work first and get help after. Each test asks for numbers you can only reach by doing the work: the p-value or its table bound. The steps open if you ask for them. Sample sizes are no longer perfect squares, so keep the standard error unrounded.
Problem 1 — Is the Mean Off the Claim? (C1 + C2 + C3 + C6)
Read the stem carefully: sometimes is known, and the direction of the test varies.
Problem 2 — A Directional Proportion Test (C4 + C5 + C6)
The direction was fixed before each survey. Check the conditions before you compute.
Problem 3 — Diagnose the Test Workflow (C2 + C3 + C5)
Problem 4 — Same Rate, Bigger Survey (C5 + C7)
Complete the table
Same sample rate, three sample sizes
Fill in every empty cell, then check the whole table once. The cells depend on one another.
The gap is the same in every row of the table. What changed between the first row and the last?
Problem 5 — What May the Report Say? (C6 + C7)
Build the Conclusion
Build the conclusion
Choose a phrase for each blank so that the sentence states the correct conclusion.
Problem 6 — Same Survey, Two Questions (C5)
These two cases use the same data. Choose the standard error for each, then use the feedback to name the one fact that changes it.
Structural Difference
Interval or test: which standard error?
Identify what key word or condition separates Case A from Case B, then select the method for each.
Scenario A
18 of 80 randomly sampled households report food insecurity (hypothetical). Estimate the population rate with a 95% confidence interval.
Scenario B
18 of 80 randomly sampled households report food insecurity (hypothetical). An agency claims 15%. Test whether the rate differs from the claim.
Mixed Review — Retrieval from Earlier Lessons
These problems draw on earlier lessons. Attempting them without re-reading is the point: retrieval strengthens memory more than re-reading does.
Review Problem 1 — t-Confidence Interval (L-INF-3)
Find from the excerpt yourself; then enter the interval.
Review Problem 2 — z Hypothesis Test, Full Five Steps (L-INF-5)
Review Problem 3 — Is the t-Procedure Trustworthy? (L-INF-3)
A health economist has billing amounts for emergency-room visits at one hospital and wants to test whether the mean charge differs from a regional benchmark of $1,200. The hospital’s historical records show a strongly right-skewed distribution: a few very expensive visits stretch the upper tail. is unknown.
(a) Is a one-sample t-test trustworthy here?
(b) What are her options?
Compare with a model answer
The t-procedure is robust to mild non-normality, and that robustness improves as grows. But with strong skew is outside that comfort zone. Reasonable options:
Collect a substantially larger sample. As grows, becomes approximately normal even for skewed data.
Transform the data to reduce skew (for right-skewed money data a log transform is common), then test on the transformed scale and interpret on that scale.
Use a distribution-free method that does not assume normality (later study).
What she should not do: report the t-test p-value as if it were reliable.
Section 7: Mastery Check
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A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
Question 1 — Feynman Test
In your own words, explain why the denominator of the one-proportion test statistic uses rather than . Write as if teaching a classmate who used the confidence-interval standard error by mistake. Your explanation should connect the null hypothesis, the standard error under , and what the test statistic measures.
Pause and write. Record your answer and reasoning on paper or in your notes before continuing. Even if you are unsure, committing your thinking helps you see what you understood and what you need to revisit.
Compare with a model answer
The denominator of the test statistic is the standard error of under : the spread we would expect for the sample proportion if the null hypothesis were exactly right.
says the true proportion is . If is true, varies with standard deviation . That is the yardstick for how surprising is under .
Using instead would compute the spread as if the true proportion equalled our sample estimate: we would be assuming our sample is right in order to judge whether our sample is surprising. A confidence interval has no claimed value, which is why it uses ; a test has one, so it uses .
Question 2 — Which Procedure?
Question 3 — A Cold t-Test
Question 4 — May the Proportion Test Run?
Question 5 — Significant, or Important?
Self-Assessment
Still confusedReady for the Boss Fight
Section 8: Boss Fight
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A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
Choose one of two equally demanding missions. Neither tells you which method to use: choosing the tool is part of the fight. Each starts with your own complete attempt; open the steps only if you need to find where it went wrong. Lose, and the rematch brings a differently built case.
📰 The Fact-Checker
Back to the opening puzzle: one survey, three analysts, three p-values. Whose report do you sign?
🗞️ The Claims Desk
A newsroom gets a new claim every day. Decide what each one needs, and write the line the editor can print.
📰 The Fact-Checker
A second survey. This time the honest test is already run. Write the line the newspaper may print.
Build the Conclusion
What may the newspaper print?
Choose a phrase for each blank so that the sentence states the correct conclusion.
Then, on paper: go back to the prediction you wrote in the Introduction. Which analyst’s choices were fixed before the data, and why are the other two reports not the ones to print?
Compare with a model answer
Analyst 3 built the standard error from , which belongs to an interval: a test judges the data in the world where is true, so its yardstick comes from . Analysts 1 and 2 used the same ; they differ only in the tail. The watchdog chose “higher” after seeing , so Analyst 1’s one-tailed 0.030 lets the data pick its own test. The honest report is Analyst 2’s two-tailed : not enough evidence at 0.05, and not proof that the agency is right either. Had the watchdog filed “higher” before the survey, Analyst 1’s report would have been the one to sign.
🗞️ The Claims Desk
A second claim. The test has been run; build the line for the story.
Build the Conclusion
The line for the story
Choose a phrase for each blank so that the sentence states the correct conclusion.
Then, on paper: which question did you ask first to choose the tool, and which clue in the claim answered it?
Compare with a model answer
A strong answer names the order of questions and the clue for each. “First: is there a claimed value? A question like ‘how long is the wait?’ has none, so it needs an interval. Then: a mean or a yes/no count? For a mean, is known from records, or only ? Then the licence: a small, strongly skewed sample, or a claim that predicts fewer than 10 successes or failures, means stop. Only then compute.”
Section 9: Challenge Problems
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A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
Optional stretch. These cold-transfer problems go beyond the lesson objectives.
Challenge 1 — Paired Data: t-Test on Differences
A fresh paired-data scenario each attempt. Work it yourself, then use the optional solution to check the transfer procedure.
Challenge 2 — How Many Bottles to Catch a 5 mL Drift?
A filling line’s volumes have mL. Quality control wants an 80% chance (power) of detecting a 5 mL shift in the mean with a two-sided test at . Explore the figure first, but keep your answer to yourself until you have computed it.
True shift Δ = 5 mL · planning SD σplan = 20 mL · two-sided α = 0.05
H₀ curve: the claim, μ = 500 mL
Truth curve: μ = 505 mL
α: reject region under H₀
Power: the truth lands in the reject region
β: the shift goes undetected
Figure 7: Power for Challenge 2's design in the normal planning model. The green curve is
where the standardised sample mean lands if the claim is true; the ink curve is where it lands
if the filler really runs 5 mL high. Power is the share of the truth's curve beyond the cutoffs;
β is the share between them. Change n and watch the truth's curve move away from the claim; the
meter compares power with the 80% target. Exact t-test power is slightly lower at every n.
Follow-up, on paper: an exact calculation for the t-test (which also estimates from the sample) says the minimum is 128 bottles: power 0.7984 at 127 and 0.8015 at 128. Why does the t-test need a few more bottles than the normal formula?
Compare with a model answer
The normal formula pretends is known. A real test estimates with , so it uses , which is a little larger than 1.960, and its statistic has heavier tails. A slightly larger signal is needed to clear the higher bar with 80% probability, which costs a couple of extra observations. The gap is small here because is large; with small samples it would be bigger.
Section 10: Solutions Reference
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A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
Complete solutions for the problems in Sections 5–9 are on the solutions page, with the reasoning behind every step, the t-table lookups, and interpretation guidance.
If you’re stuck: re-read the relevant Core Concept in Section 3. For a mean test, decide whether is known; if not, use and read the t-table at . For a proportion test, check the counts with and use in the standard error. In either test, take the tail from the question as it was asked before the data.
Quick-Reference Formulas
t Test Statistic (mean, unknown):
z Test Statistic (proportion):
Proportion Conditions (check before testing):
Decision Rule (both tests):
Which procedure?
No claimed value → estimate with an interval, not a test.
Mean, known → z-test; mean, only → t-test (any sample size), once the shape suits .
Proportion → z-test with SE from , once and are both at least 10.
Bounding p from the t-table:
Find the df row.
Locate the two critical values bracketing .
Read the one-tail (1) or two-tail (2) headings, as requires → those are the bounds on .
Compare with to decide.
Student's t-Distribution Table
Critical values (t*) for given degrees of freedom (df) and tail area.