Chapter 10: Q. 10.105 (page 431)

Left-Tailed Hypothesis Tests and CIs. If the assumptions for a nonpooled $t$ -interval are satisfied, the formula for a $(1 - α)$ level upper confidence bound for the difference, $μ_{1} - μ_{2}$ . between two population means is
$(f_{1} - f_{2}) + t_{0} \cdot \sqrt{(s_{1}^{2} / n_{1}) + (s_{2}^{2} / n_{2})}$
For a left-tailed hypothesis test at the significance level $α$ , the null hypothesis $H_{0} : μ_{1} = μ_{2}$ will be rejected in favor of the alternative hypothesis $H_{2 :} μ_{1} < μ_{2}$ if and only if the $(1 - α)$ -level upper confidence bound for $μ_{1} - μ_{2}$ is less than or equal to 0. In each case, illustrate the preceding relationship by obtaining the appropriate upper confidence bound and comparing the result to the conclusion of the hypothesis test in the specified exercise.
a. Exercise $10.83$
b. Exercise $10.84$

Short Answer

Expert verified

a) $95 %$ upper bound is negative.

b) $95 %$ upper bound is negative.

Step by step solution

Part(a) Step 1: Given Information

To detect a comparison between the hypothesis test result and the confidence interval.

Part (a) Step 2: Explanation

Consider the following data,

Consider the confidence interval's upper bound of $95 %$

$({\bar{x}}_{1} - {\bar{x}}_{2}) + t_{\frac{a}{2}} \cdot \sqrt{(\frac{s_{1}^{2}}{n_{1}}) + (\frac{s_{2}^{2}}{n_{2}})} = (7.36 - 10.50) + (- 2.015)$ $\sqrt{(\frac{1 . 22^{2}}{14} + \frac{4 . 59^{2}}{6})}$

$= - 6.97256$

As a result, the 95% upper bound is negative. As a result, the mean number of acute postoperative days in the hospital is lower than in the dynamic system.

Part(b) Step 1: Given Information

To determine the confidence interval and hypothesis test result comparison.

Part(b) Step 2: Explanation

Consider the following data

Consider the confidence interval's upper bound of $95 %$

$({\bar{x}}_{1} - {\bar{x}}_{2}) + t_{\frac{a}{2}} \cdot \sqrt{(\frac{s_{1}^{2}}{n_{1}}) + (\frac{s_{2}^{2}}{n_{2}})} = (67.90 - 66.81) + (- 1.83311)$