신뢰 구간에 속하지 않은 가설과 동일한 p- 값을 사용하여 가설을 기각합니까?

공식적으로 추정치의 신뢰 구간을 도출하면서 $p$ 값이 계산 되는 방식과 매우 유사한 공식으로 끝났습니다 .

따라서 질문은 공식적으로 동등합니까? 즉,이 가설은 거부된다 $H_0 = 0$ 임계 값 $\alpha$ 에 해당 $0$ 임계 값과의 신뢰 구간에 속하지 않는 $\alpha$ ?

예, 아니오

먼저 "예"

관찰 한 것은 검정과 신뢰 구간이 동일한 통계량을 기반으로 할 때 둘 사이에 동등성이 있다는 것입니다. 값을 매개 변수의 null 값이 갖는 α 의 가장 작은 값으로 해석 할 수 있습니다. 에 포함되는 1 - α의 신뢰 구간.$p$$\alpha$$1-\alpha$

하자 파라미터 공간에서 미지 파라미터 수 Θ ⊆ R 및하자 시료 (X) = ( X 1 , ... , X의 n은 ) ∈ X N ⊆ R n은 랜덤 변수의 실현 될 X = ( X 1 , ... , X의 N ) . 간단히하기 위해 신뢰 구간 I α ( X ) 를 적용 범위 확률 P θ가 되도록 임의 구간으로 정의하십시오.$\theta$$\Theta\subseteq\mathbb{R}$$\mathbf{x}=(x_1,\ldots,x_n)\in\mathcal{X}^ n\subseteq\mathbb{R}^n$$\mathbf{X}=(X_1,\ldots,X_n)$$I_\alpha(\mathbf{X})$ (커버리지 확률이 1 - α 로 제한되거나 대략 1 - α 인 일반적인 간격을 비슷하게 고려할 수있습니다. 추론은 유사합니다.)

$$P_\theta(\theta\in I_\alpha(\mathbf{X}))= 1-\alpha\qquad\mbox{for all }\alpha\in(0,1).$$ $1-\alpha$

대안 H 1 ( θ 0 ) 에 대한 점-널 가설 양측 검정 : θ ≠ θ 0을 고려하십시오 . 하자 λ ( θ 0 , X ) 테스트의 P 값을 나타낸다. 어떤 옵션 α ∈ ( 0 , 1 ) , H 0 ( θ 0 ) λ ( θ 0$H_0(\theta_0): \theta=\theta_0$$H_1(\theta_0): \theta\neq \theta_0$$\lambda(\theta_0,\mathbf{x})$$\alpha\in(0,1)$$H_0(\theta_0)$ 레벨에서 거부 경우$\alpha$ . 레벨 α 제거 영역의 집합이며 , X 의 제거로 이어질 H 0 ( θ 0 ) : R α ( θ 0 ) =을 { X ∈ R N : λ ( θ 0 , X ) ≤ α } $\lambda(\theta_0,x)\leq\alpha$$\alpha$ $\mathbf{x}$$H_0(\theta_0)$

$$R_\alpha(\theta_0)=\{\mathbf{x}\in\mathbb{R}^n: \lambda(\theta_0,\mathbf{x})\leq\alpha\}.$$

이제 θ ∈ Θ에 대해 p- 값이 인 양면 테스트 패밀리를 고려하십시오 . 이러한 가정을 위해 우리는 정의 할 수 반전 거절 영역 Q의 α를 ( X ) = { θ ∈ Θ : λ ( θ , X ) ≤ α } $\lambda(\theta,\mathbf{x})$$\theta\in\Theta$

$$Q_\alpha(\mathbf{x})=\{\theta\in\Theta: \lambda(\theta,\mathbf{x})\leq\alpha\}.$$

고정 된 , x ∈ R α ( θ 0 ) 인 경우 H 0 ( θ 0 ) 은 기각되며 , 이는 θ 0 ∈ Q α ( x ) , 즉 x ∈ R α ( θ 0 ) 인 경우에만 발생합니다 ⇔ θ 0 ∈ Q α ( x ) . 검정이 완전히 연속적으로 널 분포를 완전히 지정한 검정 통계량을 기반으로하는 경우$\theta_0$$H_0(\theta_0)$$\mathbf{x}\in R_\alpha(\theta_0)$$\theta_0\in Q_\alpha(\mathbf{x})$

$$\mathbf{x}\in R_\alpha(\theta_0) \Leftrightarrow \theta_0\in Q_\alpha(\mathbf{x}).$$ $\lambda(\theta_0,\mathbf{X})\sim \mbox{U}(0,1)$$H_0(\theta_0)$ $$P_{\theta_0}(\mathbf{X}\in R_\alpha(\theta_0))=P_{\theta_0}(\lambda(\theta_0,\mathbf{X})\leq\alpha)=\alpha.$$ $\theta_0\in\Theta$ $$P_{\theta_0}(\mathbf{X}\in R_\alpha(\theta_0))=P_{\theta_0}(\theta_0\in Q_\alpha(\mathbf{X})),$$ it follows that the random set $Q_\alpha(\mathbf{x})$ always covers the true parameter $\theta_0$ with probability $\alpha$. Consequently, letting $Q_\alpha^C(\mathbf{x})$ denote the complement of $Q_\alpha(\mathbf{x})$, for all $\theta_0\in\Theta$ we have $$P_{\theta_0}(\theta_0\in Q_\alpha^C(\mathbf{X}))=1-\alpha,$$ meaning that the complement of the inverted rejection region is a $1-\alpha$ confidence interval for $\theta$.

An illustration is given below, showing rejection regions and confidence intervals corresponding to the the $z$-test for a normal mean, for different null means $\theta$ and different sample means $\bar{x}$, with $\sigma=1$. $H_0(\theta)$ is rejected if $(\bar{x},\theta)$ is in the shaded light grey region. Shown in dark grey is the rejection region $R_{0.05}(-0.9)=(-\infty,-1.52)\cup(-0.281,\infty)$ and the confidence interval $I_{0.05}(1/2)=Q_{0.05}^C(1/2)=(-0.120,1.120)$.

(Much of this is taken from my PhD thesis.)

Now for the "no"

Above I described the standard way of constructing confidence intervals. In this approach, we use some statistic related to the unknown parameter $\theta$ to construct the interval. There are also intervals based on minimization algorithms, which seek to minimize the length of the interval condition on the value of $X$. Usually, such intervals do not correspond to a test.

This phenomenon has to do with problems related to such intervals not being nested, meaning that the 94 % interval can be shorter than the 95 % interval. For more on this, see Section 2.5 of this recent paper of mine (to appear in Bernoulli).

And a second "no"

In some problems, the standard confidence interval is not based on the same statistic as the standard test (as discussed by Michael Fay in this paper). In those cases, confidence intervals and tests may not give the same results. For instance, $\theta_0=0$ may be rejected by the test even though 0 is included in the confidence interval. This does not contradict the "yes" above, as different statistics are used.

And sometimes "yes" is not a good thing

As pointed out by f coppens in a comment, sometimes intervals and tests have somewhat conflicting goals. We want short intervals and tests with high power, but the shortest interval does not always correspond to the test with the highest power. For some examples of this, see this paper (multivariate normal distribution), or this (exponential distribution), or Section 4 of my thesis.

Bayesians can also say both yes and no

Some years ago, I posted a question here about whether a test-interval-equivalence exists also in Bayesian statistics. The short answer is that using standard Bayesian hypothesis testing, the answer is "no". By reformulating the testing problem a little bit, the answer can however be "yes". (My attempts at answering my own question eventually turned into a paper!)

When looking at a single parameter, it is possible that a test about the value of the parameter and the confidence interval "mismatch" depending on how they are constructed. In particular, a hypothesis test is a level $\alpha$-test, if it rejects the null hypothesis a proportion $\leq \alpha$ of the time when the null hypothesis is true. For that reason one can e.g. use estimates of model parameters (e.g. the variance) that are only valid under the null hypothesis. If one then tried to construct a CI by inverting this test, the coverage may be not quite right under the alternative hypothesis. For that reason one would usually construct a confidence interval differently so that the coverage is also right under the alternative, which can then lead to a (usually very small) mismatch.