Grade 7-12 Math Review · Grade 11-12 · Statistics and Probability
Testing a Hypothesis About a Mean
Someone makes a claim. You take a sample. The test asks one question: is what you saw too far from the claim to be luck?
Which pair of hypotheses is correct?
A supplier claims sacks average 50 kg. A buyer suspects they are lighter.
The null is always the claim of no difference, μ = 50, and the alternative carries the suspicion, which here points one way: μ < 50. B has them the wrong way round — the equals sign always belongs to H₀, because that is the value the test computes with. C is two-tailed and would test “not 50”, wasting half the evidence on a direction nobody suspects. D writes x̄; hypotheses are always about the population μ.
1
The null is the boring claim; the alternative is the suspicion.
H₀μ = the claimed valuealways has the equals sign
H₁ two-tailedμ ≠ value“different from”
H₁ left-tailedμ < value“less than”, “lighter”
H₁ right-tailedμ > value“more than”, “improved”
H₀ gets the equals sign because it is the only one specific enough to compute with.
The whole test assumes H₀ is true, works out how surprising the sample would be, and
abandons it only if the surprise is large enough.
2
The critical region is where the surprise lives.
Test
α = 0.05
α = 0.01
Left-tailed
z < −1.645
z < −2.326
Right-tailed
z > 1.645
z > 2.326
Two-tailed
|z| > 1.96
|z| > 2.576
A two-tailed test splits α between both ends, which is why its critical value is
further out. α is the risk you accept of rejecting a true H₀ — chosen
before the data are seen, never after.
3
The five steps, worked through.
1. H₀: μ = 50, H₁: μ < 50
“lighter” is one-tailed
2. α = 0.05, so reject if z < −1.645
left tail only
3. σ/√n = 2.17 = 0.3
the standard error
z = 49.2 − 500.3 = −2.67
the test statistic
4. −2.67 < −1.645, so reject H₀
inside the critical region
5. Evidence the sacks are underweight
answer in the question’s own words
Step five is a mark of its own and the one most often dropped. “Reject
H₀” is algebra; “there is evidence at the 5% level that the sacks weigh
less than 50 kg” is the answer somebody asked for.
4
What a test can and cannot conclude.
Failing to reject is not proving H₀ true
A test that does not reach the critical value says only that the evidence was not strong
enough. The sacks might genuinely average 50 kg, or they might be underweight and the
sample too small to show it. Write “there is insufficient evidence
that…”, never “the sacks weigh 50 kg”. A court that
acquits has not proved innocence.
Type I errorreject a true H₀probability α
Type II errorretain a false H₀missed a real effect
Lowering α from 0.05 to 0.01 makes a Type I error rarer and a Type II error more
likely. You cannot shrink both at once on the same data — only a larger sample does
that.
5
The p-value says the same thing, differently.
p < αreject H₀the sample was surprising
p ≥ αretain H₀the sample was ordinary
The p-value is the probability of a sample at least this extreme if H₀ were
true. For our z = −2.67, p ≈ 0.0038, well under 0.05 — the same
verdict the critical value gave, which is the point: the two routes never disagree.
6
Variations: a two-tailed retain, a right tail at 1%, one z with two verdicts.
The same five steps; what changes is the tail, the level and the verdict.
Batteries are claimed to last 300 hours, σ = 20. A sample of 64 averages 304. Test at 5% whether the mean differs.
H₀: μ = 300, H₁: μ ≠ 300; reject if |z| > 1.96
“differs” is two-tailed
z = 304 − 3002.5 = 1.6
SE = 20√64 = 2.5
1.6 < 1.96, so retain H₀
p = 2(1 − 0.9452) ≈ 0.11 > 0.05
Insufficient evidence that the mean life differs from 300 h
not “it is 300”
A two-tailed p-value doubles the one-tail area.
Reading scores averaged 75, σ = 10. After a programme, 100 pupils average 77.5. Test at 1% whether scores improved.
H₀: μ = 75, H₁: μ > 75; reject if z > 2.326
“improved” is right-tailed
z = 77.5 − 751 = 2.5
SE = 10√100 = 1
2.5 > 2.326, so reject H₀
evidence at the 1% level that scores rose
A stricter α moves the cut-off out to 2.326, and this sample still clears it.
A two-tailed test gives z = 2.2. Decide at α = 0.05 and at α = 0.01.
α = 0.05: 2.2 > 1.96 → reject
inside the 5% region
α = 0.01: 2.2 < 2.576 → retain
short of the 1% cut-off
The same data give opposite verdicts at different levels, which is why α is fixed before the sample is seen.
On the exam
Items ask you to write H₀ and H₁ from a claim, pick the critical value for the tail and α, compute z = x̄ − μσ/√n, and state the decision in words. Let the wording pick the tail (“differs” two, “less” left, “more” right), then compare z with the table. The 10-second check: the sign of z must match the direction of H₁ before a one-tailed rejection can happen.
Now you try
Practice problem
Reach the conclusion
A two-tailed test at α = 0.05 gives a test statistic of z = 1.72.
A two-tailed test at 5% rejects only when |z| > 1.96, and 1.72 does not reach it, so H₀ is retained. B uses the one-tailed cut-off 1.645, which 1.72 does clear — the tail count changes the verdict, which is why it is fixed before the data. C states the conclusion too strongly: retaining H₀ never proves it, it only reports that the evidence fell short.
Hint: check the number of tails first, because it sets the critical value.
Then choose the wording carefully — a retained null is never a proved one.
