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8.6 Relationship Between Confidence Intervals and Hypothesis Tests

Confidence intervals (CI) and hypothesis tests should give consistent results: we should not reject H0 at the significance level α if the corresponding (1α)×100% confidence interval contains the hypothesized value μ0. Two-sided confidence intervals correspond to two-tailed tests, upper-tailed confidence intervals correspond to right-tailed tests, and lower-tailed confidence intervals correspond to left-tailed tests.

A (1α)×100% two-sided t confidence interval is given in the form (x¯tα/2sn,x¯+tα/2sn). A (1α)×100% upper-tailed t confidence interval is given by (x¯tαsn,) and the number x¯tαsn is called the lower bound of the interval. A (1α)×100% lower-tailed t confidence interval is given by (,x¯+tαsn) and the number x¯+tαsn is called the upper bound of the interval. We can also use confidence intervals to make conclusions about hypothesis tests: reject the null hypothesis H0 at the significance level α if the corresponding (1α)×100% confidence interval does not contain the hypothesized value μ0. The relationship is summarized in the following table.

Table 8.3: Relationship Between Confidence Interval and Hypothesis Test

Null hypothesis H0:μ=μ0 H0:μμ0 H0:μμ0
Alternative Ha:μμ0 Ha:μ>μ0 Ha:μ<μ0
(1α)×100% CI (x¯tα/2sn,x¯+tα/2sn) (x¯tαsn,) (,x¯+tαsn)
Decision
Reject H0 if μ0 is outside the interval

Here is the reason we should reject H0 if μ0 is outside the corresponding confidence interval.

Take the right-tailed test for example, we should reject H0 if the observed test statistic to falls in the rejection region, that is if totα. This implies to=x¯μ0s/ntαμ0x¯tαsn. Given that the upper-tailed confidence interval for a right-tailed test is (x¯tα/2sn,), μ0x¯tαsn means the value of μ0 is outside the confidence interval. The same rationale applies to two-tailed and left-tailed tests. Therefore, we can reject H0 at the significance level α if μ0 is outside the corresponding (1–α)×100% confidence interval.

 

Example: Relationship Between Confidence Intervals and Hypothesis Tests

The ankle-brachial 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.

  1. At the 5% significance level, do the data provide sufficient evidence that, on average, women with peripheral arterial disease have an unhealthy ABI?

    Steps:

    1. Set up the hypotheses: H0:μ0.9 versus Ha:μ<0.9.
    2. The significance level is α=0.05.
    3. Compute the value of the test statistic: to=x¯μ0s/n=0.640.90.15/100=0.260.015=17.333 with df=n1=1001=99 (not given in Table IV, use 95, the closest one smaller than 99).
    4. Find the P-value. For a left-tailed test, the P-value is the area to the left of the observed test statistic to. P-value=P(tto)=P(t17.333)=P(t17.333)<0.005, since 17.333>2.629(t0.005).
    5. Decision: Since the P- value <0.005<0.05(α), we should reject the null hypothesis H0.
    6. Conclusion: At the 5% significance level, the data provide sufficient evidence that, on average, women with peripheral arterial disease have an unhealthy ABI.
  2. Obtain a confidence interval corresponding to the test in part a).
    For a left-tailed test at the significance level α=0.05, we should obtain a (1α)×100%=95% lower-tailed interval. For df=99, not given in Table IV, use df=95,tα=t0.05=1.661

    (,x¯+tαsn)=(,0.64+1.661×0.15100)=(,0.665).

    Interpretation: We are 95% confident that women with peripheral arterial disease have an average ABI below 0.665.

  3. Does the interval in part b) support the conclusion in part a)?
    In part a), we reject H0 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.