Opposite results between one-side confidence interval and one-side hypothesis test for one proportion each other?

Question

I'm trying an example in PennState STAT 415 (Example 9-2)

• Method 1: a right-tailed hypothesis test

In short, the observed sample proportion is $\hat{p}=\dfrac{104}{590}=0.176$. The null hypothesis is $H_0 \colon p_0 = 0.14$, and the alternative hypothesis is $H_A \colon p > 0.14$. The significance level is $\alpha = 0.01$, so the critical region is $ > 2.326$. After calculating the test statistic $Z$, which is $2.52$ ($=\dfrac{0.176-0.14}{\sqrt{\dfrac{0.14(0.86)}{590}}}$), the null hypotheis is rejected since $2.52 > z_{0.01}$. It means that the true proportion is greater than $0.14$.

• Method 2: one-side confidence interval for the sample proportion $\hat{p}$

The formula for the one-side confidence interval (lower bound to infinity) of unknown true $p$ (not necessily $p_0$ in Method 1) is: $[\hat{p}-z_{\alpha} \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}, +\infty)$, which is: $p \geq 0.176 - 2.326\times\sqrt{\dfrac{0.176(0.824)}{590}} = 0.1395$. Because the assumed true $p$ ($0.14$) is in the interval of $[0.1395, +\infty)$, we cannot say that the sample proportion means a greater true proportion. This is the opposite conclusion with the Method 1.

Why do the conclusions by two method contradict each other?

Answer 1

$\bullet$ As PBulls noted, OP used two different things in the hope of those two yielding the same conclusion.

OP used the Wald confidence interval which was different of what one gets inverting the acceptance region of the corresponding test above.

Retaining $p$ and solving for it would lead to the Wilson confidence interval, which is better than the Wald score interval in many ways but more importantly it is boundary-respecting.

$\bullet$ Broadening the discussion, in discrete cases, to obtain a given confidence coefficient, randomized confidence sets are used, and are obtained from a corresponding randomized test procedure by incorporating a uniform variable on $[0, 1]$ independent of the sample. An illustration on binomial proportion has been discussed in Shao's Mathematical Statistics book.

$\bullet$ Finally much has been written about the confidence intervals of binomial proportion. The Wikipedia article could be a good place to start.

Answer 2

As @PBulls suggests, the method 1 or 2 uses different $p$ ($p_0$ or $\hat p$) in the formula, which makes them answer two different questions. There is a disccusion about using $p_0$ or $\hat p$ in the formula of the test statistic $Z$ at the end of Section 9.3 in the same STAT415 lesson.

In short, if we use $\hat p$ in the test statistic formula, then the results of the confidence interval will be consistent with the hypothesis test decision.

Take the example in the question, when using $\hat p$ in the formula, the $Z$ value is: $\dfrac{0.176-0.14}{\sqrt{\dfrac{0.176\times 0.824}{590}}} = 2.296$. Since $2.296 < z_{0.01}$, we cannot reject the null hypothesis. This conclusion is now consistent with the conclusion by confidence interval.