01 · Identify the region
Is the critical region only X=2?
Hint
What other counts are at least as low?
Worked solution
No. It includes X=0, X=1 and X=2. The critical value is the boundary two.
Understand · explore · practise
Lower critical regions
Construct a lower binomial critical region, check the largest allowed cutoff and recognise when no supported count is sufficiently unusual.
Before you startCumulative binomial probabilities and significance levels.
01 / Small counts can support a decrease
The alternative determines which end of the distribution matters.
For a lower test, reject when X≤k.
Choose the largest integer k for which the null cumulative probability P(X≤k) is at most α.
Raising k adds possible counts and increases the rejection probability. We want the largest lower-tail region that meets the significance constraint. The adjacent check is therefore k+1, not k−1.
Choose a lower rejection boundaryExplore
Under H₀, X~B(12,0.5). For H₁: p<0.5, find a 5% lower critical region.
02 / Check the working cutoff and the next integer
A larger region may exceed the allowed probability.
Watch: including three crosses the limit
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Under H₀, X~B(12,0.5). Construct a lower critical region at 5%.Worked example
P(X≤2) = 79/4096 ≈ 0.01928711
This is at most 0.05.
P(X≤3) = 299/4096 ≈ 0.07299805
Including three would exceed 0.05.
Critical region: X≤2, namely {0,1,2}
Two is the largest working cutoff and is the critical value.
02 · Why not one?
P(X≤1) = 13/4096 ≈ 0.00317383. Why is k=1 not the standard 5% cutoff?
Hint
Can a larger cutoff still work?
Worked solution
Two already meets the constraint and gives a larger rejection region. One is unnecessarily restrictive.
03 / Use the cumulative endpoint directly
No complement or minus-one shift is needed for X≤k.
The cumulative function F(k) is exactly P(X≤k). For the candidate k=2, read F(2). By contrast, the strict event X<2 is F(1). Keep the inequality in the question visible while entering the calculator command.
03 · Wrong endpoint
A learner uses F(1) to calculate the probability of X≤2. What has been omitted?
Hint
Which boundary mass is missing?
Worked solution
P(X=2) has been omitted. F(2)=F(1)+P(X=2).
04 · Equivalent strict form
Rewrite the critical region X≤2 using a strict integer inequality.
Hint
The next integer is three.
Worked solution
X<3, for this integer-valued statistic. It is not X<2.
04 / Apply the boundary without moving it
The sample either falls in the predefined region or it does not.
Suppose p is the success probability for a process formerly operating at 0.5, and the alternative is a decrease. With a fixed sample of 12, the region above leads to rejection for observed x=0,1 or 2. An observed 3 is outside it.
05 · Boundary observation
Observe x=2. Give a contextual conclusion for H₁: p<0.5 at 5%.
Hint
The boundary is included.
Worked solution
Reject H₀. There is sufficient evidence at 5% that the process success probability is below 0.5, assuming the model is appropriate.
06 · Adjacent observation
Observe x=3 instead. What is the conclusion?
Hint
Three is not in the region.
Worked solution
Do not reject H₀. There is insufficient evidence at 5% of a decrease. The result does not establish that p equals 0.5.
05 / Calculate the achieved level
The nominal limit and actual probability are different quantities.
The actual significance of X≤2 is 79/4096≈0.01928711, about 1.929%. The nominal level remains 5%. The gap occurs because the next possible region jumps to about 7.300%.
07 · Nonrejection probability
Under H₀, find the probability of a count outside the region.
Hint
Complement the whole rejection region.
Worked solution
1 − 79/4096 = 4017/4096 ≈ 0.98071289.
08 · Boundary mass
Find P(X=3) from the two cumulative values.
Hint
Subtract F(2) from F(3).
Worked solution
(299−79)/4096 = 220/4096 ≈ 0.05371094. This extra mass pushes the region over 5%.
06 / Zero may still be too likely
A supported lower critical region always includes zero.
Under H₀, Y~B(10,0.1). Is there a nonempty lower critical region at 5%?Worked example
P(Y≤0) = P(Y=0) = 0.9¹⁰ ≈ 0.34867844
Even the smallest nonempty lower tail exceeds 0.05.
Every larger lower tail is at least as probable
No supported cutoff can meet the rule.
The critical region is empty
Y≤−1 represents this impossible event. Zero is not a valid rejection cutoff here.
09 · Why zero does not force rejection
Does observing no successes always provide significant evidence of a decrease?
Hint
Consider the baseline and sample size.
Worked solution
No. In this model, zero successes has probability about 0.349 under H₀, so it is quite plausible. The baseline and sample size determine how unusual zero is.
10 · Null probability versus conclusion
Does an empty lower region prove that the baseline is correct?
Hint
The test lacks a sufficiently rare supported lower outcome.
Worked solution
No. It means this sample size and test rule cannot reject for a decrease at the specified level. It does not prove H₀.
07 / Use the correct neighbouring cutoff
Monotonicity supplies the boundary proof.
11 · Wrong neighbour
A student proves F(2)≤0.05 and F(1)≤0.05. Have they shown two is the largest cutoff?
Hint
Which candidate would enlarge the region?
Worked solution
No. They must check F(3)>0.05. A smaller working cutoff does not establish maximality.
12 · Smaller significance level
For B(12,0.5), find the largest cutoff at 1%, given F(1)≈0.00317383 and F(2)≈0.01928711.
Hint
Compare with 0.01.
Worked solution
k=1. The critical region becomes {0,1}; F(1) works while F(2) fails.
13 · Alternative direction
Would this lower region test a preselected claim of an increase?
Hint
An increase is supported by high counts.
Worked solution
No. Use an upper critical region for an increase in the same defined success probability.
14 · Outside support
For a 12-trial count, what is P(X≤−1)?
Hint
Negative counts are impossible.
Worked solution
Zero. A cutoff below the support represents an empty region, not a possible observation.
08 / Largest working lower cutoff
Use F(k), check k+1 and state all rejecting counts.
Construct the region under H₀ before applying it. Check both adjacent probabilities, include the boundary and calculate the achieved level. If even zero is too likely, state an empty region instead of forcing a cutoff.
Section 1 of 8 · Small counts can support a decrease