Question

Right-Tailed Hypothesis Tests and CIs. If the assumptions for a pooled $t$-interval are satisfied, the formula for a $(1-\alpha )$-level lower confidence bound for the difference, ${\mu }_1-{\mu }_2$, between two population means is

$\left({\overline{x}}_1-{\hat{x}}_2\right)-t_a·S_{\mathrm{p}}\sqrt{\left(1/n_1\right)+\left(1/n_2\right)}$

For a right-tailed hypothesis test at the significance level $\alpha$,

the null hypothesis $H_0:{\mu }_1={\mu }_2$will be rejected in favor of the alternative hypothesis $H_{\mathrm{a}}:{\mu }_1>{\mu }_2$if and only if the $(1-\alpha )$-level lower confidence bound for ${\mu }_1-{\mu }_2$is greater than or equal to $0$. In each case, illustrate the preceding relationship by obtaining the appropriate lower confidence bound and comparing the result to the conclusion of the hypothesis test in the specified exercise.

a. Exercise 10.47

b. Exercise 10.50

Short answer

a). The null hypothesis is not rejected since the $95$percent lower confidence bound, $-16.92$, is not larger than or equal to zero. This is in line with the right-tailed hypothesis test's stated purpose.

b). The null hypothesis is rejected because the lower confidence bound is $99\%$, and $54.5$ is larger than zero. This is in line with the right-tailed hypothesis test's stated purpose.

Step-by-step solution

Part (a) Step 1: Given Information

Population $1:$ Fortified orange juice, ${\overline{x}}_1=9.0,s_1=37.4$, and $n_1=14$.

Population $2:$ Unfortified orange juice, ${\overline{x}}_2=1.6,s_2=34.6$, and $n_2=12$.

Significance level is $5\%$.

Part (a) Step 2: Explanation

A statement is expressed as - for a right-tailed hypothesis test.

The null hypothesis $H_0:{\mu }_1={\mu }_2$will be rejected in favour of the alternate hypothesis $H_a:{\mu }_1>{\mu }_2$at the significance level $\alpha$if and only if the $(1-\alpha )$-level lower confidence bound for ${\mu }_1-{\mu }_2$is larger than or equal to $0$.

The test statistic for exercise $10.47$does not fall in the rejection zone of the left-tailed hypotheses test at a significance level of $5$percent. As a result, null hypotheses are not ruled out.

The $95$percent confidence interval for exercise $10.53$is $-16.92$to $31.72$.

Part (b) Step 1: Given Information

Population $1:$ Lunch before recess, ${\overline{x}}_1=223.1,s_1=122.9$, and $n_1=889$.

Population 2 : Lunch after recess, ${\overline{x}}_2=156.6,s_2=108.1$, and $n_2=1119$.

Significance level is $1\%$.

Part (b) Step 2: Explanation

A statement is expressed as - for a right-tailed hypothesis test.

The null hypothesis $H_0:{\mu }_1={\mu }_2$will be rejected in favour of the alternate hypothesis $H_a:{\mu }_1>{\mu }_2$at the significance level $\alpha$if and only if the $(1-\alpha )$-level lower confidence bound for ${\mu }_1-{\mu }_2$is larger than or equal to $0$.

The test statistic for exercise $10.50$does not fall in the rejection zone of the left-tailed hypotheses test at a significance level of $5$percent. As a result, null hypotheses are not ruled out.

The $99\%$confidence interval for exercise $10.53$is $54.5\mathrm{gm}$to $78.5\mathrm{gm}$.