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Translate a claim into a one-sided or two-sided alternative, choose the correct tail and reverse direction when the counted event is complemented.
Before you startParameters, null hypotheses and binomial success counts.
01 / Direction belongs to the claim
H₁ chooses the tail.
An increase in the defined probability gives p>p₀; a decrease gives p<p₀; a change in either direction gives p≠p₀.
The word success labels the event being counted. It does not mean a desirable outcome. If X counts faults, a lower value may be evidence of improvement. If X counts working items, improvement points upwards.
02 / An increase uses the upper tail
H₀: p=0.2; H₁: p>0.2
The claim is an increase in faults, so this is a one-tailed test.
Let X be the number faulty among n tested components
Under suitable assumptions, H₀ gives X~B(n,0.2).
For observed x, consider P(X≥x)
Include the observed count and all larger counts in the evidence tail.
An old support proportion is 0.35. A campaign is predicted to increase support. Define p and write H₁.
State the probability being investigated.
Let p be the current proportion of the relevant population supporting the campaign. H₁: p>0.35; use an upper tail for the support count.
If eight faulty components are observed, which upper-tail probability is needed: P(X>8) or P(X≥8)?
Include the observed result.
P(X≥8). Using P(X>8) wrongly excludes the actual observed count.
03 / A decrease uses the lower tail
H₀: p=0.9; H₁: p<0.9
The event counted is on-time delivery.
Let X count on-time deliveries in a fixed sample
A low count points in the direction of the alternative.
For observed x, consider P(X≤x)
This includes the observed count and all smaller values.
A new process aims to reduce a fault probability of 0.08. With X counting faults, write H₁ and identify the tail.
Improvement need not mean an upper tail.
H₁: p<0.08, where p is the new fault probability. Use the lower tail of the fault count.
For that reduction claim, the observed fault count is unusually high. Should the alternative be switched to p>0.08 after seeing it?
Was an increase the question chosen before sampling?
No. Retain the planned lower-tailed test. An unexpectedly high count does not justify choosing a different one-tailed test after inspection. A separate investigation may be appropriate.
04 / A change uses both tails
For a claim of any change, use H₁: p≠p₀. With the equal-tail convention used in these lessons, allocate half the stated significance level to each tail. A 5% two-tailed test uses a 2.5% threshold at each end; it does not use 5% at both ends.
A coin is being tested for any bias. Define p and state both hypotheses.
Heads and tails bias must both be allowed.
Let p be the probability of heads. H₀: p=0.5; H₁: p≠0.5. This is two-tailed.
What tail allocation is used for a 10% equal-tail test?
Split the total in two.
5% in each tail. Discrete binomial boundaries can make the actual total smaller; later lessons calculate it.
For a two-tailed test, does observing a high count turn H₁ into p>p₀?
The preselected alternative remains fixed.
No. H₁ remains p≠p₀. The observation tells us which tail contains the observed evidence, not which alternative to invent.
05 / Changing the counted event reverses direction
Pause, replay or seek freely. The notes explain the same idea and stay in view.
p = probability faulty; H₁: p<0.08
A lower-tailed test for the number faulty.
q = probability working = 1−p
The baseline is q₀=0.92.
H₁: q>0.92
For a fixed sample n, working count Y=n−X. A small fault count is a large working count.
If p>0.3 and q=1−p, what statement about q is equivalent?
Subtracting reverses the direction.
q<0.7. An increase in the probability of one event means a decrease in its complement.
Among 20 items, X counts faults and Y=20−X counts working items. Rewrite X≤2 using Y.
Subtract X from 20.
Y≥18. The events are identical, so their probabilities are equal.
Rewrite p≠0.08 using q=1−p.
A difference stays a difference.
q≠0.92. The test remains two-tailed.
06 / Read the event and the claim together
“Better” can mean a lower failure probability or a higher pass probability. “Biased towards blue” specifies a direction; “biased” without a specified direction normally allows both. “The new process is different” calls for both tails unless the question gives a directional claim.
Why is “better means upper-tailed” unreliable?
What is X counting?
The direction depends on the success definition. Fewer defects is a lower tail for defect counts but an upper tail for non-defective counts.
A spinner’s baseline blue probability is 0.25. A preselected claim says it is biased towards blue. State H₁.
Towards blue means more blue.
H₁: p>0.25, with p defined as the probability of blue.
07 / Keep the plan separate from the evidence
A student checks both one-sided probabilities at 5% and reports whichever rejects as “a 5% one-tailed test”. Explain the problem.
They allowed evidence in both directions.
Their decision rule effectively checks two tails, so it is not the preplanned single-tail procedure claimed. Use an appropriate two-tailed allocation if either direction is being investigated.
Under H₀, X~B(50,0.02), and x=4 is observed. Is four below or above the null mean?
Compare with np, not with the sample size.
The null mean is 50×0.02=1, so four is above it. In a two-tailed test this observation lies on the upper side; being a small absolute count does not make it lower-tail evidence.
08 / Define success, then translate the claim
Write the alternative about the defined population probability. Choose it before examining the sample. Include the observed value in the evidence tail. If you complement the event, reverse a one-sided direction and translate the count. Two-tailed tests retain their two-sided alternative whichever side the observation falls on.
Section 1 of 8 · Direction belongs to the claim