Lesson 20: Hypothesis Testing

Discussion prompt

Before we open Lesson 20: Hypothesis Testing: without looking back, what was the main idea of PCA Implementation, and what could you do by the end of it that you could not do before?

Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.

Comparison matrix

From The running example: refill the accuracy column from what you know. The rest of the table is as it appeared.

Discussion prompt

Hypothesis testing works like a trial. The defendant is presumed innocent — that's H₀ (no effect). You never prove innocence; you either gather enough evidence to convict, or you don't.

That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.

Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.

Comparison matrix

From Null, alternative, and the two errors: every row here is a choice with a cost. Fill the H₀ true column, then say which row you would actually pick and what you give up for it.

Discussion prompt

Explain What 'surprising' means by analogy to something with no Machine Learning in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.

Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.

Discussion prompt

Explain α is a pre-commitment, not a dial to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.

Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.

Predict first

A student writes this, and it looks reasonable:

We got p = 0.03, so there is a 3% chance the null hypothesis is true.

It is wrong. Say what breaks — and say it before you turn the page.

Correct: This flips the conditioning. The p-value is computed by ASSUMING H₀ and measuring the data's tail — it is P(data | H₀).

p = 0.03 means: if H₀ were true, data this extreme would occur only 3% of the time.

Why: This flips the conditioning. The p-value is computed by ASSUMING H₀ and measuring the data's tail — it is P(data | H₀).

Predict first

The 10 accuracies sum, then divide by n = 10. Add them in order — no shortcut:

Commit before you compute: what does Step 1 — the sample mean, by hand come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.

Correct: x̄ = 8.25 / 10 = 0.825

Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The sample mean sits 0.025 above the claimed μ₀ = 0.80.

Step through it

Step through Step 1 — the sample mean, by hand one row at a time. What is driving the change, and what would the row after the last one be?

1. Step 1: quantity is Σ xᵢ
2. Step 2: quantity is n
3. Step 3: quantity is x̄ = Σxᵢ / n

Discussion prompt

Spread uses the sample standard deviation with ddof = 1 (divide by n−1, not n). First every deviation dᵢ = xᵢ − x̄ and its square:

What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.

Hint: Anything you would have to invent to get started is a thing the problem must supply.

Step through it

Step through Step 2 — the sample std, deviation by deviation and watch the numbers move. Now imagine the input ten times bigger: which column is the one that stops this being practical?

1. Step 1: xᵢ is 0.82
2. Step 2: xᵢ is 0.79
3. Step 3: xᵢ is 0.85
4. Step 4: xᵢ is 0.83
5. Step 5: xᵢ is 0.78
6. Step 6: xᵢ is 0.86
7. Step 7: xᵢ is 0.81
8. Step 8: xᵢ is 0.84
9. Step 9: xᵢ is 0.80
10. Step 10: xᵢ is 0.87

Predict first

The table runs: x̄ − μ₀ | 0.025 · SE = s/√n | 0.0095743

In Step 3 — standard error, then the t-statistic, given the rows so far: what is the next one — the row where quantity is t = (x̄−μ₀)/SE?

Correct: t = (x̄−μ₀)/SE | 2.6112

Why: The relationship between the columns, not the individual numbers, is what generates the next row. Averaging 10 folds cuts the noise by √10 — that √n is why bigger samples resolve smaller effects.

Annotate

From Step 3 — standard error, then the t-statistic — read this one piece at a time. What is each part doing?

On: \( t = \frac{\bar x - \mu_0}{s/\sqrt{n}} = \frac{0.825 - 0.80}{0.0095743} = \frac{0.025}{0.0095743} \)

• Averaging 10 folds cuts the noise by √10 — that √n is why bigger samples resolve smaller effects.
• The sample mean is 2.61 standard errors above the claimed value. Now we ask how often noise alone reaches |t| ≥ 2.61 — that's the p-value.

Fill in the blanks

From Step 3 in code — the t-statistic — one line has had its right-hand side removed. Put it back.

import numpy as np
# 10 held-out accuracy scores of a model claimed to hit 0.80
data = np.array([0.82,0.79,0.85,0.83,0.78,0.86,0.81,0.84,0.80,0.87])
mu0 = 0.80
n = len(data)
xbar = data.mean() # sample mean
s = data.std(ddof=1) # SAMPLE sd (ddof=1)
se = s / np.sqrt(n) # standard error
t = (xbar - mu0) / se # t statistic
print(round(xbar,4), round(s,4), round(t,4))

Why: s is what everything below it consumes, so the wrong expression here fails later and somewhere else. Drop ddof=1 and std divides by n, shrinking s, inflating t and faking significance.

Step through it

Step through Step 3 in code — the t-statistic one row at a time. What is driving the change, and what would the row after the last one be?

1. Step 1: printed is xbar
2. Step 2: printed is s
3. Step 3: printed is t

Fill in the blanks

Step 4 — tail area to p-value, with the scaffolding fading: two lines are gone now — fill both.

import numpy as np
from scipy import stats
data = np.array([0.82,0.79,0.85,0.83,0.78,0.86,0.81,0.84,0.80,0.87])
mu0 = 0.80
n = len(data)
t = (data.mean() - mu0) / (data.std(ddof=1)/np.sqrt(n))
df = n - 1
one_tail = stats.t.sf(abs(t), df=df) # upper-tail area
p = **2 * one_tail** # two-sided
print(round(t,4), df, round(one_tail,4), round(p,4))

Why: Reproducing these unaided, rather than reading them, is what tells you the method has transferred. By symmetry of t₉, the lower tail beyond −|t| is also 0.0141.

Discussion prompt

Work backwards. The example finished here:

one tail = 0.0141, so p = 2 × 0.0141 = 0.0282

What was it asked to do, and what must it have been given? Reconstruct the problem from its answer.

Hint: Every quantity in the result had to enter somewhere. Account for each one.

Annotate

Walk the callouts on The same decision by critical value. Each one is a place this is easy to get subtly wrong.

• ppf is the inverse CDF: the value with 97.5% of t₉ below it, so 2.5% above. Two tails → 5% total = α.
• |t| lands in the rejection region beyond the cutoff, so p < α automatically. The p-value route and the critical-value route ALWAYS agree.

Fill in the blanks

From Compute the 95% CI — one line has had its right-hand side removed. Put it back.

import numpy as np
from scipy import stats
data = np.array([0.82,0.79,0.85,0.83,0.78,0.86,0.81,0.84,0.80,0.87])
n = len(data)
xbar = data.mean()
se = data.std(ddof=1) / np.sqrt(n)
tc = stats.t.ppf(0.975, df=n-1) # 2.5% in each tail
lo, hi = xbar - tcse, xbar + tcse
print(round(lo,4), round(hi,4))
print('0.80 inside?', lo <= 0.80 <= hi)

Why: se is what everything below it consumes, so the wrong expression here fails later and somewhere else. The plausible range for the true accuracy.

Draw it

Draw what Compute the 95% CI just did — the shape of it, not the line-by-line working. One picture, labels only where you need them. Then check it against the steps: anything you could not draw is a step you followed rather than understood.

Predict first

Our from-scratch t and p must match the library's ttest_1samp exactly. Self-contained:

Commit before you compute: what does Verify the whole test against scipy come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.

Correct: ttest_1samp returns (statistic, pvalue) — both match

Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. np.isclose confirms agreement to floating-point tolerance.

Predict first

A student writes this, and it looks reasonable:

Use data.std() — NumPy's default — for the standard deviation in t.

It is wrong. Say what breaks — and say it before you turn the page.

Correct: NumPy's default divides by n, giving the POPULATION sd.

Use data.std(ddof=1) — the sample sd — because we estimated the mean from the same data.

Why: NumPy's default divides by n, giving the POPULATION sd. On this data that's 0.02872, smaller than the sample sd 0.030277.

Discussion prompt

The rule this trap just fixed:

Dividing by n−1 gives an unbiased variance estimate: s = 0.030277, matching the t₉ theory.

Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?

Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.

Fill in the blanks

A chi-squared goodness-of-fit test, with the scaffolding fading: two lines are gone now — fill both.

import numpy as np
from scipy import stats
observed = np.array([22,17,20,18,12,11]) # 100 die rolls, faces 1..6
expected = np.array([100/6]*6) # fair-die expectation
chi2 = np.sum((observed-expected)2/expected)
p = stats.chi2.sf(chi2, df=5)** # df = k-1 = 5
print(round(chi2,4), round(p,4))
print(stats.chisquare(observed).pvalue.round(4))

Why: Reproducing these unaided, rather than reading them, is what tells you the method has transferred. Six faces → 5 degrees of freedom (the counts must sum to 100, costing one).

Draw it

Draw what A chi-squared goodness-of-fit test just did — the shape of it, not the line-by-line working. One picture, labels only where you need them. Then check it against the steps: anything you could not draw is a step you followed rather than understood.

Comparison matrix

From A two-sample t-test (Welch): refill the value (verified) column from what you know. The rest of the table is as it appeared.

Fill in the blanks

From An F-test: three models at once (ANOVA) — one line has had its right-hand side removed. Put it back.

import numpy as np
from scipy import stats
g1 = np.array([0.82,0.79,0.85]) # three models,
g2 = np.array([0.80,0.75,0.79]) # three folds each
g3 = np.array([0.88,0.90,0.86])
F, p = stats.f_oneway(g1, g2, g3) # one-way ANOVA
print(round(F,4), round(p,4))

Why: g2 is what everything below it consumes, so the wrong expression here fails later and somewhere else. Between-group spread is 11.4× the within-group noise — far more than chance would produce.

Predict first

The table runs: 10 | 2.108 | 0.5589 · 30 | 3.651 | 0.9546

In Power climbs with n, given the rows so far: what is the next one — the row where n is 100?

Correct: 100 | 6.667 | 1.0000

Why: The relationship between the columns, not the individual numbers, is what generates the next row. With only 10 folds we'd miss this real effect 44% of the time.

Discussion prompt

A two-proportion z-test on the 2-point gap, swept over sample size per group. Pooled proportion p̂ = (0.85+0.83)/2 = 0.84:

What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.

Hint: Anything you would have to invent to get started is a thing the problem must supply.

Discussion prompt

Work backwards. The example finished here:

n=100: z=0.386, p=0.70 (noise). n=10000: z=3.858, p=0.0001 (decisive)

What was it asked to do, and what must it have been given? Reconstruct the problem from its answer.

Hint: Every quantity in the result had to enter somewhere. Account for each one.

Predict first

A student writes this, and it looks reasonable:

Model A scored higher on the test set, so A is the better model — ship it.

It is wrong. Say what breaks — and say it before you turn the page.

Correct: Ignores sampling noise. At n=100 the 2-point gap has p=0.70 — squarely inside what chance produces.

Test the difference; report significance (or a confidence interval), not just the raw numbers.

Why: Ignores sampling noise. At n=100 the 2-point gap has p=0.70 — squarely inside what chance produces.

Step through it

Step through False positives pile up one row at a time. What is driving the change, and what would the row after the last one be?

1. Step 1: m tests is 1
2. Step 2: m tests is 5
3. Step 3: m tests is 10
4. Step 4: m tests is 20

Predict first

Run 20 one-sample t-tests on completely random data — no real effect anywhere — and count the 'significant' ones. Seeded so you see these exact results:

Commit before you compute: what does p-hacking on pure noise come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.

Correct: 0 survive Bonferroni's 0.05/20 = 0.0025

Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The smallest p (0.0111) exceeds 0.0025, so the corrected test correctly reports nothing significant — which is the truth.

Annotate

Walk the callouts on p-hacking on pure noise. Each one is a place this is easy to get subtly wrong.

• Seed 7 produced 4 p-values below 0.05 (min p = 0.0111) on noise. You EXPECT ~1 by chance (20×0.05); this draw gave 4.
• The smallest p (0.0111) exceeds 0.0025, so the corrected test correctly reports nothing significant — which is the truth.

Discussion prompt

Explain When Bonferroni is too strict: FDR to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.

Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.

Predict first

A student writes this, and it looks reasonable:

Tried 20 features; feature 7 came out significant (p = 0.011), so report that discovery.

It is wrong. Say what breaks — and say it before you turn the page.

Correct: With 20 tests at α=0.05 you EXPECT a false positive — P(≥1) = 0.64.

Correct for the number of tests before claiming significance.

Why: With 20 tests at α=0.05 you EXPECT a false positive — P(≥1) = 0.64. The 'winner' is very likely noise.

Sort into buckets

Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.

Holds up

Key consequence: 'fail to reject H₀' means 'not enough evidence to convict' — not 'proven innocent'. Absence of evidence isn't evidence of absence.; Crucially this is a statement about the data, computed under the assumption H₀ holds. It is not the probability that H₀ is true.; The p-value is the probability, assuming H₀, of a test statistic at least as extreme as the one observed:

Breaks

We got p = 0.03, so there is a 3% chance the null hypothesis is true.; Use data.std() — NumPy's default — for the standard deviation in t.

sound

These are stated as this lesson states them — each one survives the edge cases Lesson 20: Hypothesis Testing puts it through.

flawed

Each of these is lifted from a trap in this deck: reasonable-sounding, and wrong in a way that only shows up once you rely on it.

Put in order

These are the steps of The hypothesis-testing recipe, scrambled. Put them back in order before the next slide shows you.

1. State H₀ and H₁; fix α before looking at the data
2. Pick the test: t (means), χ² (categorical counts), F (variance)
3. Compute the statistic — e.g. t=(x̄−μ₀)/(s/√n) with ddof=1 — and its p-value under H₀
4. Decide: reject if p < α (or |t| > t_crit); report the effect size, not just 'significant'
5. Correct for multiple tests with α/m (Bonferroni) or FDR if you ran many

Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.

Discussion prompt

The hypothesis-testing recipe works on the cases you have just seen. Push it to the edge: what is the most degenerate input it still handles — empty, zero, one item, everything equal — and what is the first case where it stops being true? Name the case, not just "it breaks".

Hint: Try the smallest legal input, then the largest, then the one where two things collide. Methods are specified at their edges; the middle takes care of itself.

Eliminate the wrong options

Our t-test gave p = 0.0282. Which statement is correct?

3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.

• A. If μ were 0.80, a sample this far from 0.80 would occur about 2.8% of the time
• B. There is a 2.8% probability that H₀ (μ = 0.80) is true
• C. There is a 97.2% probability the model's accuracy exceeds 0.80
• D. The effect size is 2.8%

Survives elimination: A

Why: A p-value is P(data this extreme | H₀). Assuming μ = 0.80, a sample mean at least 2.61 SE away arises ~2.8% of the time — rare, so we reject H₀. It never states P(H₀).

Check your understanding

Our t-test gave p = 0.0282. Which statement is correct?

• A. If μ were 0.80, a sample this far from 0.80 would occur about 2.8% of the time (correct)
• B. There is a 2.8% probability that H₀ (μ = 0.80) is true
• C. There is a 97.2% probability the model's accuracy exceeds 0.80
• D. The effect size is 2.8%

Answer: A

Why: A p-value is P(data this extreme | H₀). Assuming μ = 0.80, a sample mean at least 2.61 SE away arises ~2.8% of the time — rare, so we reject H₀. It never states P(H₀).

Why B tempts people

That's P(H₀ | data), the reversed conditioning. Getting a posterior needs a prior and Bayes' rule — the p-value is not a posterior.

Why C tempts people

Also a posterior claim (about H₁). 1 − p is not P(H₁ true); the p-value can't deliver the probability an effect is real.

Why D tempts people

The p-value measures surprise under H₀, not effect size. Here the effect is x̄−μ₀ = 0.025, and a large n could make a tiny effect have p = 0.028.

Predict first

Computing t = (x̄ − μ₀)/(s/√n) in NumPy, what should s be?

Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.

Correct: data.std(ddof=1) — divide by n−1 (sample sd)

Why: The mean was estimated from the same data, so the unbiased variance divides by n−1: ddof=1. That gives s = 0.030277, t = 2.6112, matching scipy and the t₉ theory.

Check your understanding

Computing t = (x̄ − μ₀)/(s/√n) in NumPy, what should s be?

• A. data.std(ddof=1) — divide by n−1 (sample sd) (correct)
• B. data.std() — NumPy's default (divide by n)
• C. data.var() — the variance, not the sd
• D. data.std(ddof=1) / n — divide by n once more

Answer: A

Why: The mean was estimated from the same data, so the unbiased variance divides by n−1: ddof=1. That gives s = 0.030277, t = 2.6112, matching scipy and the t₉ theory.

Why B tempts people

ddof=0 is the population sd (divide by n). It underestimates spread here (0.02872), inflating t to 2.752 and faking a stronger result.

Why C tempts people

var is s², not s. The t formula needs the standard deviation; using the variance is dimensionally wrong.

Why D tempts people

The √n is already in the SE denominator (s/√n). Dividing s by n as well double-counts the sample size and collapses t toward 0.

Predict first

Model A 85% vs B 83% is statistically significant when…

Commit to an answer, then rate it — certain, fairly sure, or guessing — and write the rating down before you turn the page.

Correct: the sample is large enough (e.g. ~10000/group), not at ~100/group

Why: The two-proportion test gives p = 0.70 at n=100 (noise) but p = 0.0001 at n=10000. Significance of a fixed gap depends on the sample size behind it.

The rating matters as much as the answer: confident-and-wrong is the combination that survives revision, because nothing about it feels like it needs revisiting.

Check your understanding

Model A 85% vs B 83% is statistically significant when…

• A. the sample is large enough (e.g. ~10000/group), not at ~100/group (correct)
• B. always — 85 > 83
• C. never — 2 points is too small to matter
• D. only if both models use the same architecture

Answer: A

Why: The two-proportion test gives p = 0.70 at n=100 (noise) but p = 0.0001 at n=10000. Significance of a fixed gap depends on the sample size behind it.

Why B tempts people

Raw ordering ignores sampling noise; at n=100 the gap (z=0.386) is indistinguishable from chance.

Why C tempts people

A 2-point gap CAN be significant — with enough data it's decisive (p=0.0001 at n=10000).

Why D tempts people

Architecture is irrelevant to whether the accuracy difference is statistically distinguishable; the test needs only the counts and n.

Predict first

You run 20 independent tests at α = 0.05 on data with no real effects. Roughly how many come out 'significant', and how do you control it?

Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.

Correct: About 1 by chance; Bonferroni's α/20 = 0.0025 suppresses it

Why: Each test has a 5% false-positive rate, so the expected count is 20 × 0.05 = 1 (our seed happened to give 4). Bonferroni lowers the threshold to 0.05/20 = 0.0025 to hold the family-wise rate at 5%.

Check your understanding

You run 20 independent tests at α = 0.05 on data with no real effects. Roughly how many come out 'significant', and how do you control it?

• A. About 1 by chance; Bonferroni's α/20 = 0.0025 suppresses it (correct)
• B. 0 — there are no real effects
• C. All 20
• D. Exactly 5

Answer: A

Why: Each test has a 5% false-positive rate, so the expected count is 20 × 0.05 = 1 (our seed happened to give 4). Bonferroni lowers the threshold to 0.05/20 = 0.0025 to hold the family-wise rate at 5%.

Why B tempts people

Even with no real effect, the 5% Type I rate produces false positives — that IS the multiple-testing problem. P(≥1) = 0.64 at m=20.

Why C tempts people

Each test independently has only a 5% false-positive chance; all 20 firing by chance is astronomically unlikely (0.05²⁰).

Why D tempts people

5 would be a 25% rate. The expected count is 20 × 0.05 = 1, not 5.

Eliminate the wrong options

Our 95% CI for the true accuracy is [0.8033, 0.8467]. What does that tell you about testing H₀: μ = 0.80 at α = 0.05?

3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.

• A. Reject H₀ — 0.80 lies outside the interval
• B. Fail to reject H₀ — the interval is wide
• C. There's a 95% probability the true mean is exactly 0.825
• D. The p-value must be greater than 0.05

Survives elimination: A

Why: A 95% CI is exactly the set of μ₀ values a two-sided test at α=0.05 would NOT reject. Since 0.80 is outside [0.8033, 0.8467], the test rejects — matching p = 0.0282 < 0.05.

Check your understanding

Our 95% CI for the true accuracy is [0.8033, 0.8467]. What does that tell you about testing H₀: μ = 0.80 at α = 0.05?

• A. Reject H₀ — 0.80 lies outside the interval (correct)
• B. Fail to reject H₀ — the interval is wide
• C. There's a 95% probability the true mean is exactly 0.825
• D. The p-value must be greater than 0.05

Answer: A

Why: A 95% CI is exactly the set of μ₀ values a two-sided test at α=0.05 would NOT reject. Since 0.80 is outside [0.8033, 0.8467], the test rejects — matching p = 0.0282 < 0.05.

Why B tempts people

Interval width doesn't decide the test; what matters is whether μ₀ is inside. 0.80 is outside, so we reject.

Why C tempts people

The CI is a range of plausible means, not a probability spike on 0.825. The interval either contains the fixed true μ or not; 95% refers to the procedure's long-run coverage.

Why D tempts people

The opposite: because 0.80 is outside the 95% CI, p must be BELOW 0.05. CI-exclusion and p < α are equivalent.

Discussion prompt

Explain Project: one-sample t-test from scratch by analogy to something with no Machine Learning in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.

Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.

Step through it

Step through Milestone 2 — the p-value one row at a time. What is driving the change, and what would the row after the last one be?

1. Step 1: quantity is df
2. Step 2: quantity is p-value
3. Step 3: quantity is reject H₀?

Comparison matrix

From Milestone 3 — verify vs scipy: every row here is a choice with a cost. Fill the t column, then say which row you would actually pick and what you give up for it.

Comparison matrix

From The full program: refill the value column from what you know. The rest of the table is as it appeared.

Discussion prompt

Slides closed, out loud: explain (1) why a p-value is P(data | H₀) and not P(H₀), (2) why ddof=1 and not ddof=0, and (3) why the same 85-vs-83 gap flips significance with n.

That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.

Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.

Draw it

One page, no notation unless you need it: draw how these connect — The framework · Build the t-test · The test family · Power & sample size · Multiple testing · Recipe & checks. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.