Q. 3.87 Serving Time. According to the B... [FREE SOLUTION]

Chapter 8: Q. 3.87 (page 381)

Serving Time. According to the Bureau of Crime Statistics and Research of Australia, as reported on Lawlink, the mean length of imprisonment for motor-vehicle theft offenders in Australia is 16.7 months. One hundred randomly selected motor-vehicle-theft offenders in Sydney, Australia, had a mean length of imprisonment of 17.8 months. At the 5% significance level, do the data provide sufficient evidence to conclude that the mean length of imprisonment for motor-vehicle theft offenders in Sydney differs from the national mean in Australia? Assume that the population standard deviation of the lengths of imprisonment for motor-vehicle-theft offenders in Sydney is 6.0 months.

Short Answer

Expert verified

Ans: Since $z$ does not fall in the rejection region.

Thus, role="math" localid="1652215192012" $H_{0}$ is not rejected at a role="math" localid="1652215197979" $5 %$ level of significance of the test value statistic.

At the $1 %$ significance level, the data provided is sufficient evidence for concluding that the mean post-work heart rate for easting workers exceeds the normal resting heart rate of role="math" localid="1652215186430" $72 b p m$ .

Step by step solution

Step 1. Given information.

given,

Assume that the population standard deviation of the lengths of imprisonment for motor-vehicle-theft offenders in Sydney is 6.0 months.

Step 2. Let's assume the mean post-work heart rate for casting workers to be μ.

Given that,

Population standard deviation is,

$蟽 = 6.8$ years

Now, test the hypotheses,

$\begin{aligned} H_{0} : 渭 = 72 b.p.m \\ H_{a} : 渭 > 72 b.p.m \end{aligned}$

Perform the test at a $5 %$ level of significance i.e., $伪 = 0.05$

The sample size was $n = 29$

The sample mean is,

$x = 78.3 b p m$

Step 3. Now,

$\begin{aligned} Test statistic, z = \frac{\overset{炉}{x} 鈭� 渭_{0}}{\frac{蟽}{\sqrt{n}}} \\ = \frac{78.3 鈭� 72}{\frac{11.2}{\sqrt{29}}} \\ = 3.03 \end{aligned}$

Since the test is left the tailed test with $伪 = 0.05$ ,

and the critical value is,

localid="1651233032648" $z_{伪} = z_{0.05} = 1.645$

Here, the region is,

$z > z_{0.05}$

i.e., $z > 1.645$

Step 4. Then,

Step 5. Here,

$z = 3.03 > z_{0.05} = 1.645$

Since $z$ does not fall in the rejection region.

Thus, $H_{0}$ is not rejected at a $5 %$ level of significance of the test value statistic.

At the $1 %$ significance level, the data provided is sufficient evidence for concluding that the mean post-work heart rate for easting workers exceeds the normal resting heart rate of $72 b p m$ .