An original synthetic study observes 99 successes among 150 independent trials. The reference claim is p0=0.60 and the prespecified alternative is p>0.60. The observed proportion is 0.66, but its difference from 0.60 must be judged relative to the sampling variation predicted by the null.
Predict
Predict whether the standard-error formula should use the observed 0.66 or the proposed 0.60. A test asks how the statistic behaves under the null model, so its denominator uses p0. An interval instead estimates uncertainty without imposing that particular null value.
Experiment
The null standard error is √[0.60·0.40/150]=0.04. Thus z=(0.66−0.60)/0.04=1.5. A supplied normal table gives upper-tail probability 0.0668. At α=0.05, fail to reject: these data do not provide sufficient evidence for p>0.60 under the model.
Expected null successes and failures are 90 and 60. Both satisfy the common large-count check. Write those expected counts beside the observed counts 99 and 51 to make clear which model supplies the reference uncertainty. Do not call the observed improvement zero merely because the test does not reject.
Understand
For n independent binary observations with a common null success probability p0 strictly between zero and one, the usual statistic is z=(p̂−p0)/√[p0(1−p0)/n]. The numerator has a sign; it indicates whether the observed proportion is above or below the null. Tail selection follows the prespecified alternative, not that sign alone.
Check a credible random sampling or experimental design, independence or an appropriate approximation, and adequate expected null counts np0 and n(1−p0), commonly at least 10 each. The null count check differs from the observed-count check used for the basic Wald interval. Neither numerical check can validate biased selection or dependent observations.
The normal distribution is an approximation to a discrete binomial reference. If expected counts are small, an exact binomial calculation or another justified method may be more suitable. The existing exact-test investigation sums binomial tail probabilities for a small model; it should not be mislabeled as this z approximation. Exactness concerns the probability calculation under the model, not the truth of the model’s independence assumption.
A batch of tightly related items may have more variability than independent Bernoulli trials. Treating every item as an independent unit can underestimate uncertainty. Repeated testing, selective reporting or stopping after a favorable result also changes the operating characteristics unless the procedure accounts for those actions.
A conclusion names the target, direction and evidence level. “Insufficient evidence that the success probability exceeds 0.60” does not establish that it equals 0.60. Conversely, rejection does not automatically imply a practically meaningful improvement or generalization beyond the sampled setting. Pair the test with an estimate, suitable uncertainty and a context-specific threshold where possible.
Apply
Recompute the example after changing the observed count while holding n and p0 fixed. Predict the statistic’s direction before calculating. Then write a complete answer for the practice lower-tail case, including null expected counts, reference tail and a qualified population conclusion.
Continue
Next consider what correct and incorrect decisions mean across repeated studies. Error rates and power require a specified truth and alternative effect.
Apply · explain · revise
Practice and reasoning
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In 100 independent trials, p̂=0.6. Test p0=0.5. Find the null-standardized z statistic.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
For an upper-tailed test with z=2, the supplied normal upper-tail area is 0.0228. What is the p-value?
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
The standard error for testing H0:p=p0 uses:
n=20 and p0=0.02 give expected null successes 0.4. A routine normal z approximation is:
In a representative independent sample of 200, 70 successes occur. Test H0:p=0.4 versus p<0.4. Give z and a conclusion using supplied lower-tail p≈0.0745 for z≈−1.443; α=0.05.
Structure your argument and preserve revisions
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
A machine produces 200 items in one tightly coupled batch. Why may a binomial z test overstate evidence even with large success and failure counts?
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Reflect and continue
Before marking this lesson complete, explain one key idea without the worked solution and identify a condition that matters. Marking complete records participation, not independent mastery.
