Q. 9.87

Question

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 of-fenders 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 6months.

Step-by-Step Solution

Verified
Answer

At 5% significance level, the data do not provide sufficient evidence to conclude that the mean length of imprisonment of motor-vehicle-theft offenders differs from the national mean.

1Step 1. Given information.

Consider the given question,

Population standard deviation is months.

Sample size, n is 100.

Sample mean is x is 17.8 months.

2Step 2. Determine the test statistics.

Assume μ to be the mean length of imprisonment for motor-vehicle-theft offenders.

We have to test the hypotheses,

H0:μ=16.7 vsHa:μ16.7

On performing the test at 5% level of significance, i.e., α=0.05.

Test statistics z=x-μ0σn

z=17.8-16.76100=1.83

3Step 3. Write the critical values.

As the test is two tailed test with σ is 0.05, the critical values are ±z0.025=±1.96.

Here, the rejection region is z<-za2 or z>za2, i.e., z<-1.96 or z>1.96


Here, -1.96<z<1.96.

Therefore, we do not reject H0 at 5% level of significance as the value of the test statistic, z does not falls in the rejection region.

Hence, at 5% significance level, the data do not provide sufficient evidence to conclude that the mean length of imprisonment of motor-vehicle-theft offenders differs from the national mean.