Triola Elementary Statistics · 11th Edition
Chapter 8: Hypothesis Testing
Every key formula from Triola Elementary Statistics Chapter 8, in one searchable page. Click a card to study it — worked examples included.
The null hypothesis H0
H0: parameter = claimed value
Assume H0 true, then test it: reject or fail to reject.
The alternative hypothesis H1
H1: uses only ≠, <, >
The tail follows H1
≠ two-tailed · < left-tailed · > right-tailed
Rare event rule: if the sample result is extremely unlikely under H0, the assumption is probably wrong.
WORKED EXAMPLE
Claim: XSORT raises P(girl) above 0.5:
H0: p = 0.5, H1: p > 0.5 — right-tailed.
H0: p = 0.5, H1: p > 0.5 — right-tailed.
KEY NOTES
- To support your claim, word it as H1 — you can never “prove” H0.
Test statistic
Measures disagreement between the sample and H0. The critical region (bounded by critical values) holds the extreme outcomes.
Significance level α
α = P(statistic in critical region | H0 true)
Common choice: α = 0.05. Note: α = P(Type I error).
P-value method
P-value ≤ α ⇒ reject H0 · P-value > α ⇒ fail to reject H0
WORKED EXAMPLE
P-value = 0.03, α = 0.05:
0.03 ≤ 0.05 ⇒ reject H0 — sufficient evidence to support the claim.
0.03 ≤ 0.05 ⇒ reject H0 — sufficient evidence to support the claim.
KEY NOTES
- Always say “fail to reject H0“ — NEVER “accept H0“.
8.3 · TYPE I/II ERRORS & POWER
Type I error
Reject a true H0 · P(Type I) = α
Mnemonic: RTN = Reject True Null.
Type II error
Fail to reject a false H0 · P(Type II) = β
Mnemonic: FRFN = Fail to Reject a False Null.
Power
Power = 1 − β = P(rejecting a false H0)
WORKED EXAMPLE
XSORT test (H0: p = 0.5):
Type I = conclude the method works when it doesn’t;
Type II = miss a method that really works.
Type I = conclude the method works when it doesn’t;
Type II = miss a method that really works.
8.4 · CLAIM: PROPORTION
z = p̂ − p√(pq/n)
Use the claim’s p in the SE, not p̂. No continuity correction.
Requirements: binomial conditions; np ≥ 5, nq ≥ 5.
Requirements: binomial conditions; np ≥ 5, nq ≥ 5.
WORKED EXAMPLE
n = 100, x = 60, claim p = 0.5, α = 0.05, right-tailed:
p̂ = 0.6; z = (0.6 − 0.5)/√(0.25/100) = 2.00;
P-value = 0.0228 < 0.05 ⇒ reject H0.
p̂ = 0.6; z = (0.6 − 0.5)/√(0.25/100) = 2.00;
P-value = 0.0228 < 0.05 ⇒ reject H0.
8.5 · CLAIM ABOUT A MEAN
σ known
z = x̄ − μσ/√n
σ unknown
t = x̄ − μs/√n (df = n − 1)
Robust vs mild non-normality — check outliers and the histogram.
WORKED EXAMPLE
n = 25, x̄ = 103, s = 12, claim μ = 100, two-tailed, α = 0.05:
t = (103 − 100)/(12/5) = 1.25; df = 24; critical ±2.064;
|1.25| < 2.064 ⇒ fail to reject H0.
t = (103 − 100)/(12/5) = 1.25; df = 24; critical ±2.064;
|1.25| < 2.064 ⇒ fail to reject H0.