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In the simple scenarios you're probably considering, it's a logical equivalence: the point defining the null hypothesis is inside a confidence interval with confidence level $\gamma=1-\alpha$ if and only if the observed value of the test statistic is outside the critical region of a test with size $\alpha$.

Consider the case of the usual $Z$ test. You have a random sample $X_1,\dots,X_n$ from a normal distribution with unknown mean $\mu$ and known variance $\sigma_0^2$. You want to test the null hypothesis $H_0:\mu=\mu_0$ against the alternative $H_A:\mu\ne\mu_0$. The test statistic is $Z=(\bar{X}-\mu_0)/(\sigma_0/\sqrt{n})$ and, under the null hypothesis $H_0$, $Z$ has a standard normal distribution. The critical region is $$ \mathscr{C}_\alpha = ( -\infty; -z_{1-\alpha/2}] \;\cup\; [z_{1-\alpha/2};\infty). $$ On the other hand, a confidence interval for $\mu$, with confidence level $\gamma=1-\alpha$, is given by $$ \text{CI}[\mu;\gamma = 1-\alpha] = \left( \bar{x} - z_{1-\alpha/2} \frac{\sigma_0}{\sqrt{n}} \;;\; \bar{x} + z_{1-\alpha/2} \frac{\sigma_0}{\sqrt{n}} \right). $$ It follows by simple algebra that $$ \mu_0\in \text{CI}[\mu;\gamma = 1-\alpha] \;\;\Leftrightarrow\;\; z_{\text{obs}}=(\bar{x}-\mu_0)/(\sigma_0/\sqrt{n})\notin\mathscr{C}_\alpha. $$ The connection with the usual definition of a $p$-value for this problem is also immediate.

In the simple scenarios you're probably considering, it's a logical equivalence: the point defining the null hypothesis is inside a confidence interval with confidence level $\gamma=1-\alpha$ if and only if the observed value of the test statistic is outside the critical region of a test with size $\alpha$.

Consider the case of the usual $Z$ test. You have a random sample $X_1,\dots,X_n$ from a normal distribution with unknown mean $\mu$ and known variance $\sigma_0^2$. Suppose you want to perform a two tailed hypothesis test with null hypothesis $H_0:\mu=\mu_0$ and alternative $H_A:\mu\ne\mu_0$. The test statistic is $Z=(\bar{X}-\mu_0)/(\sigma_0/\sqrt{n})$ and, under the null hypothesis $H_0$, $Z$ has a standard normal distribution. The critical region is $$ \mathscr{C}_\alpha = ( -\infty, -z_{\alpha/2}] \;\cup\; [z_{\alpha/2} ,\infty). $$ On the other hand, a confidence interval for $\mu$, with confidence level $\gamma=1-\alpha$, is given by $$ \left( \bar{x} - z_{\alpha/2} \frac{\sigma_0}{\sqrt{n}} , \bar{x} + z_{\alpha/2} \frac{\sigma_0}{\sqrt{n}} \right). $$ It follows by simple algebra that $$ \mu_0\in \left( \bar{x} - z_{\alpha/2} \frac{\sigma_0}{\sqrt{n}} , \bar{x} + z_{\alpha/2} \frac{\sigma_0}{\sqrt{n}} \right) \;\;\Leftrightarrow\;\; z_{\text{obs}}=(\bar{x}-\mu_0)/(\sigma_0/\sqrt{n})\notin\mathscr{C}_\alpha. $$ The connection with the usual definition of a $p$-value for this problem is also immediate.

In the simple scenarios you're probably considering, it's a logical equivalence: the point defining the null hypothesis is inside a confidence interval with confidence level $\gamma=1-\alpha$ if and only if the observed value of the test statistic is outside the critical region of a test with size $\alpha$.

Consider the case of the usual $Z$ test. You have a random sample $X_1,\dots,X_n$ from a normal distribution with unknown mean $\mu$ and known variance $\sigma_0^2$. You want to test the null hypothesis $H_0:\mu=\mu_0$ against the alternative $H_A:\mu\ne\mu_0$. The test statistic is $Z=(\bar{X}-\mu_0)/(\sigma_0/\sqrt{n})$ and, under the hull hypothesis $H_0$, $Z$ has a standard normal distribution. The critical region is $$ \mathscr{C}_\alpha = ( -\infty; -z_{1-\alpha/2}] \;\cup\; [z_{1-\alpha/2};\infty). $$ On the other hand, a confidence interval for $\mu$, with confidence level $\gamma=1-\alpha$, is given by $$ \text{CI}[\mu;\gamma = 1-\alpha] = \left( \bar{x} - z_{1-\alpha/2} \frac{\sigma_0}{\sqrt{n}} \;;\; \bar{x} + z_{1-\alpha/2} \frac{\sigma_0}{\sqrt{n}} \right). $$ It follows by simple algebra that $$ \mu_0\in \text{CI}[\mu;\gamma = 1-\alpha] \;\;\Leftrightarrow\;\; z_{\text{obs}}=(\bar{x}-\mu_0)/(\sigma_0/\sqrt{n})\notin\mathscr{C}_\alpha. $$ The connection with the usual definition of a $p$-value for this problem is also immediate.

In the simple scenarios you're probably considering, it's a logical equivalence: the point defining the null hypothesis is inside a confidence interval with confidence level $\gamma=1-\alpha$ if and only if the observed value of the test statistic is outside the critical region of a test with size $\alpha$.

Consider the case of the usual $Z$ test. You have a random sample $X_1,\dots,X_n$ from a normal distribution with unknown mean $\mu$ and known variance $\sigma_0^2$. You want to test the null hypothesis $H_0:\mu=\mu_0$ against the alternative $H_A:\mu\ne\mu_0$. The test statistic is $Z=(\bar{X}-\mu_0)/(\sigma_0/\sqrt{n})$ and, under the null hypothesis $H_0$, $Z$ has a standard normal distribution. The critical region is $$ \mathscr{C}_\alpha = ( -\infty; -z_{1-\alpha/2}] \;\cup\; [z_{1-\alpha/2};\infty). $$ On the other hand, a confidence interval for $\mu$, with confidence level $\gamma=1-\alpha$, is given by $$ \text{CI}[\mu;\gamma = 1-\alpha] = \left( \bar{x} - z_{1-\alpha/2} \frac{\sigma_0}{\sqrt{n}} \;;\; \bar{x} + z_{1-\alpha/2} \frac{\sigma_0}{\sqrt{n}} \right). $$ It follows by simple algebra that $$ \mu_0\in \text{CI}[\mu;\gamma = 1-\alpha] \;\;\Leftrightarrow\;\; z_{\text{obs}}=(\bar{x}-\mu_0)/(\sigma_0/\sqrt{n})\notin\mathscr{C}_\alpha. $$ The connection with the usual definition of a $p$-value for this problem is also immediate.

In the simple scenarios you're probably considering, it's a logical equivalence: the point defining the null hypothesis is inside a confidence interval with confidence level $\gamma=1-\alpha$ if and only if the observed value of the test statistic is outside the critical region of a test with size $\alpha$.

Consider the case of the usual $Z$ test. You have a random sample $X_1,\dots,X_n$ from a normal distribution with unknown mean $\mu$ and known variance $\sigma_0^2$. You want to test the null hypothesis $H_0:\mu=\mu_0$ against the alternative $H_A:\mu\ne\mu_0$. The test statistic is $Z=(\bar{X}-\mu_0)/(\sigma_0/\sqrt{n})$ and, under the hull hypothesis $H_0$, $Z$ has a standard normal distribution. The critical region is $$ \mathscr{C}_\alpha = ( -\infty; -z_{1-\alpha/2}] \;\cup\; [z_{1-\alpha/2};\infty). $$ On the other hand, a confidence interval for $\mu$, with confidence level $\gamma=1-\alpha$, is given by $$ \text{CI}[\mu;\gamma = 1-\alpha] = \left( \bar{x} - z_{1-\alpha/2} \frac{\sigma_0}{\sqrt{n}} \;;\; \bar{x} + z_{1-\alpha/2} \frac{\sigma_0}{\sqrt{n}} \right). $$ It follows by simple algebra that $$ \mu_0\in \text{CI}[\mu;\gamma = 1-\alpha] \;\;\Leftrightarrow\;\; z_{\text{obs}}=(\bar{x}-\mu_0)/(\sigma_0/\sqrt{n})\notin\mathscr{C}_\alpha. $$

In the simple scenarios you're probably considering, it's a logical equivalence: the point defining the null hypothesis is inside a confidence interval with confidence level $\gamma=1-\alpha$ if and only if the observed value of the test statistic is outside the critical region of a test with size $\alpha$.

Consider the case of the usual $Z$ test. You have a random sample $X_1,\dots,X_n$ from a normal distribution with unknown mean $\mu$ and known variance $\sigma_0^2$. You want to test the null hypothesis $H_0:\mu=\mu_0$ against the alternative $H_A:\mu\ne\mu_0$. The test statistic is $Z=(\bar{X}-\mu_0)/(\sigma_0/\sqrt{n})$ and, under the hull hypothesis $H_0$, $Z$ has a standard normal distribution. The critical region is $$ \mathscr{C}_\alpha = ( -\infty; -z_{1-\alpha/2}] \;\cup\; [z_{1-\alpha/2};\infty). $$ On the other hand, a confidence interval for $\mu$, with confidence level $\gamma=1-\alpha$, is given by $$ \text{CI}[\mu;\gamma = 1-\alpha] = \left( \bar{x} - z_{1-\alpha/2} \frac{\sigma_0}{\sqrt{n}} \;;\; \bar{x} + z_{1-\alpha/2} \frac{\sigma_0}{\sqrt{n}} \right). $$ It follows by simple algebra that $$ \mu_0\in \text{CI}[\mu;\gamma = 1-\alpha] \;\;\Leftrightarrow\;\; z_{\text{obs}}=(\bar{x}-\mu_0)/(\sigma_0/\sqrt{n})\notin\mathscr{C}_\alpha. $$ The connection with the usual definition of a $p$-value for this problem is also immediate.