Justifying a Claim from an Interval · Justificar una afirmación a partir de un intervalo
Does a poll support a half-support claim?
- A checked 95% interval for population support is $(0.52,0.61)$. The value 0.50 is outside this interval, whereas 0.55 is inside.
- Use the interval to assess plausible parameter values. It does not show that 95% of people have responses between 0.52 and 0.61.
Confidence refers to a repeated method
- If we repeatedly used the same valid sampling and interval method, its capture rate would be approximately the stated confidence level.
- The population proportion is fixed. A particular interval either captures it or misses it; we do not assign that interval a 95% probability for $p$ under this method.
The correct interpretation of a 95% confidence level is...
Confidence describes the method over repeated sampling.
Name the population and measured outcome
- Write: we are 95% confident that 52% to 61% of users in the sampled library population support later opening.
- A value inside the interval is not proved true. A value outside is excluded by this interval procedure, subject to the study assumptions.
A 95% interval of (0.52, 0.61) gives evidence against the claim that p = 0.50.
0.50 is outside the interval, so it's implausible.
If a claimed value lies inside the confidence interval, you can rule it out.
Inside the interval → plausible, cannot be ruled out.
Match the claim to its interval assessment.
Compare population-parameter values with endpoints; do not confuse a parameter interval with individual data.
An interval runs from 0.52 to 0.61. What is its margin of error?
Half its width is (0.61 - 0.52)/2 = 0.045.
Keep an interval distinct from a test
- Exact interval-test agreement requires an interval constructed by inverting that same test. Do not promise it for every pair of formulas.
- The usual proportion interval uses an estimated standard error; the one-proportion test uses the null proportion. Borderline decisions can differ.
The interval (0.52, 0.61) excludes 0.50 but includes 0.55. This is evidence against 0.50 using this interval method, not proof of exactly 0.55.
Inspect a borderline example
- For · A favor $n=100$ and · y $\hat p=0.45$, the 95% normal interval is about $(0.3525,0.5475)$, containing 0.355.
- A two-sided null test of $p_0=0.355$ gives $z\approx1.985$ and p-value about 0.047. It rejects at 0.05 despite that Wald interval containing the null value.
A Wald interval and a null-standard-error test can disagree near a boundary; exact agreement needs a matching interval/test pair.
Increasing the sample size, holding confidence fixed, makes the margin of error... Assume the observed proportion stays fixed.
Larger n → smaller SE → smaller margin of error.
To get a narrower interval AND higher confidence, you need a larger sample size. Assume the observed proportion stays fixed.
Bigger n overcomes the precision-vs-confidence trade-off.
Make a bounded claim
- For · A favor $(0.52,0.61)$, explain that 0.50 is outside the interval and name the target population. To report a test decision, calculate the specified test.
- Precision also depends on design and assumptions. An apparently narrow interval from a biased survey does not establish an accurate population estimate.
For · A favor $(0.52,0.61)$, explain that 0.50 is outside the interval and name the target population. To report a test decision, calculate the specified test.
A Wald interval contains p0 but a null-standard-error test rejects p0. Which explanations are valid?
An interval obtained by inverting the same test agrees with that test. A Wald interval and a null-SE z-test are not an identical pair.