Question

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

$\left({\overline{x}}_1-{\tilde{x}}_2\right)+t_a·S_p\sqrt{\left(1/n_1\right)+\left(1/n_2\right)}$

For a left-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 upper confidence bound for ${\mu }_1-{\mu }_2$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 10.45

b. Exercise 10.46

Short answer

a). The null hypothesis is rejected at the $95\%$upper confidence bound $-4.960$. This is in line with the left-tailed hypothesis test's stated purpose.

b). The null hypothesis is not rejected since the upper confidence bound of $95\%$, $8.84$ is not less than or equal to zero. This is in line with the left-tailed hypothesis test's stated purpose.

Step-by-step solution

Part (a) Step 1: Given Information

Population $1:$ Fraud offences, ${\overline{x}}_1=10.12,s_1=4.90$, and $n_1=10$.

Population 2 : Firearms offences, ${\overline{x}}_2=18.78,s_1=4.64$, and $n_2=10$.

Significance level is $5\%$.

Part (a) Step 2: Explanation

A statement is expressed as - for a left-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 upper confidence bound for ${\mu }_1-{\mu }_2$is less than or equal to $0$.

The test statistic falls in the rejection zone of the left-tailed hypotheses test starting with exercise $10.45$and a significance level of $5\%$. As a result, null hypotheses are ruled out.

The $95\%$confidence interval for exercise $10.51$is $-12.36$to $-4.96$.

Part (b) Step 1: Given Information

Population $1$: Male, ${\overline{x}}_1=37.6,s_1=38.5$and $n_1=30$.

Population $2$ : Female, ${\overline{x}}_2=55.8,s_2=48.3$, and $n_2=30$.

Significance level is $5\%$.

Part (b) Step 2: Explanation

A statement is supplied as for a left-tailed hypothesis test. A statement is given as - for a left-tailed-hypothesis test.

If and only if the $(1-\alpha )$-level upper confidence bound for ${\mu }_1-{\mu }_2$is less than or equal to $0$, 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 $\alpha$.

The test statistic for exercise $10.46$does not fall inside the rejection region of the left-tailed hypotheses test at a significance level of $5\%$. Null hypotheses are therefore not rejected.

The $95$percent confidence interval from exercise $10.52$to $8.84$is $-45.24$to $-45.24$.