8.6 Relationship Between Confidence Intervals and Hypothesis Tests
Confidence intervals (CI) and hypothesis tests should give consistent results: we should not reject [latex]H_0[/latex] at the significance level [latex]\alpha[/latex] if the corresponding [latex](1  \alpha) \times 100\%[/latex] confidence interval contains the hypothesized value [latex]\mu_0[/latex]. Twosided confidence intervals correspond to twotailed tests, uppertailed confidence intervals correspond to righttailed tests, and lowertailed confidence intervals correspond to lefttailed tests.
A [latex](1  \alpha) \times 100\%[/latex] twosided [latex]t[/latex] confidence interval is given in the form [latex](\bar{x}  t_{\alpha / 2} \frac{s}{\sqrt{n}}, \bar{x} + t_{\alpha / 2} \frac{s}{\sqrt{n}})[/latex]. A [latex](1  \alpha) \times 100\%[/latex] uppertailed t confidence interval is given by [latex](\bar{x}  t_{\alpha} \frac{s}{\sqrt{n}}, \infty)[/latex] and the number [latex]\bar{x}  t_{\alpha} \frac{s}{\sqrt{n}}[/latex] is called the lower bound of the interval. A [latex](1  \alpha) \times 100\%[/latex] lowertailed t confidence interval is given by [latex]( \infty, \bar{x} + t_{\alpha} \frac{s}{\sqrt{n}})[/latex] and the number [latex]\bar{x} + t_{\alpha} \frac{s}{\sqrt{n}}[/latex] is called the upper bound of the interval. We can also use confidence intervals to make conclusions about hypothesis tests: reject the null hypothesis [latex]H_0[/latex] at the significance level [latex]\alpha[/latex] if the corresponding [latex](1  \alpha) \times 100\%[/latex] confidence interval does not contain the hypothesized value [latex]\mu_0[/latex]. The relationship is summarized in the following table.
Table 8.3: Relationship Between Confidence Interval and Hypothesis Test
Null hypothesis  [latex]H_0: \mu = \mu_0[/latex]  [latex]H_0: \mu \leq \mu_0[/latex]  [latex]H_0: \mu \geq \mu_0[/latex] 

Alternative  [latex]H_a: \mu \neq \mu_0[/latex]  [latex]H_a: \mu \: \gt \: \mu_0[/latex]  [latex]H_a: \mu < \mu_0[/latex] 
[latex](1  \alpha) \times 100\%[/latex] CI  [latex](\bar{x}  t_{\alpha / 2} \frac{s}{\sqrt{n}}, \bar{x} + t_{\alpha / 2} \frac{s}{\sqrt{n}})[/latex]  [latex](\bar{x}  t_{\alpha} \frac{s}{\sqrt{n}}, \infty)[/latex]  [latex]( \infty, \bar{x} + t_{\alpha} \frac{s}{\sqrt{n}})[/latex] 
Decision 
Reject [latex]H_0[/latex] if [latex]\mu_0[/latex] is outside the interval

Here is the reason we should reject [latex]H_0[/latex] if [latex]\mu_0[/latex] is outside the corresponding confidence interval.
Take the righttailed test for example, we should reject [latex]H_0[/latex] if the observed test statistic [latex]t_o[/latex] falls in the rejection region, that is if [latex]t_o \geq t_{\alpha}[/latex]. This implies [latex]t_o = \frac{\bar{x}  \mu_0}{s / \sqrt{n}} \geq t_{\alpha} \Longrightarrow \mu_0 \leq \bar{x}  t_{\alpha} \frac{s}{\sqrt{n}}.[/latex] Given that the uppertailed confidence interval for a righttailed test is [latex](\bar{x}  t_{\alpha / 2} \frac{s}{\sqrt{n}}, \infty)[/latex], [latex]\mu_0 \leq \bar{x}  t_{\alpha} \frac{s}{\sqrt{n}}[/latex] means the value of [latex]\mu_0[/latex] is outside the confidence interval. The same rationale applies to twotailed and lefttailed tests. Therefore, we can reject [latex]H_0[/latex] at the significance level [latex]\alpha[/latex] if [latex]\mu_0[/latex] is outside the corresponding (1–[latex]\alpha[/latex])×100% confidence interval.
Example: Relationship Between Confidence Intervals and Hypothesis Tests
The anklebrachial index (ABI) compares the blood pressure of a patient’s arm to the blood pressure of the patient’s leg. The ABI can be an indicator of different diseases, including arterial diseases. A healthy (or normal) ABI is 0.9 or greater. Researchers obtained the ABI of 100 women with peripheral arterial disease and obtained a mean ABI of 0.64 with a standard deviation of 0.15.
 At the 5% significance level, do the data provide sufficient evidence that, on average, women with peripheral arterial disease have an unhealthy ABI?
Steps: Set up the hypotheses: [latex]H_0: \mu \geq 0.9[/latex] versus [latex]H_a: \mu < 0.9[/latex].
 The significance level is [latex]\alpha = 0.05[/latex].
 Compute the value of the test statistic: [latex]t_o = \frac{\bar{x}  \mu_0}{s / \sqrt{n}} = \frac{0.64  0.9}{0.15 / \sqrt{100}} = \frac{0.26}{0.015} = 17.333[/latex] with [latex]df = n1 = 100 1 = 99[/latex] (not given in Table IV, use 95, the closest one smaller than 99).
 Find the Pvalue. For a lefttailed test, the Pvalue is the area to the left of the observed test statistic [latex]t_o[/latex]. [latex]\mbox{Pvalue} = P(t \leq t_o) = P(t \leq 17.333) = P(t \geq 17.333) < 0.005,[/latex] since [latex]17.333> 2.629(t_{0.005})[/latex].
 Decision: Since the P value [latex]< 0.005 < 0.05(\alpha)[/latex], we should reject the null hypothesis [latex]H_0[/latex].
 Conclusion: At the 5% significance level, the data provide sufficient evidence that, on average, women with peripheral arterial disease have an unhealthy ABI.
 Obtain a confidence interval corresponding to the test in part a).
For a lefttailed test at the significance level [latex]\alpha = 0.05[/latex], we should obtain a [latex](1  \alpha) \times 100\% = 95\%[/latex] lowertailed interval. For [latex]df = 99[/latex], not given in Table IV, use [latex]df=95, t_{\alpha} = t_{0.05} = 1.661[/latex][latex]\left(  \infty, \bar{x} + t_{\alpha} \frac{s}{\sqrt{n}} \right)= \left(  \infty, 0.64 + 1.661 \times \frac{0.15}{\sqrt{100}} \right) = ( \infty , 0.665)[/latex].
Interpretation: We are 95% confident that women with peripheral arterial disease have an average ABI below 0.665.
 Does the interval in part b) support the conclusion in part a)?
In part a), we reject [latex]H_0[/latex] and claim that the mean ABI is below 0.9 for women with peripheral arterial disease. In part b), we are 95% confident that the mean ABI is less than 0.9 since the entire confidence interval is below 0.9. In other words, the hypothesized value 0.9 is outside the corresponding confidence interval, we should reject the null. Therefore, the results obtained in parts a) and b) are consistent.