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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 is6months.

Short Answer

Expert verified

At5% 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.

Step by step solution

01

Step 1. Given information.

Consider the given question,

Population standard deviation is 6months.

Sample size, n is 100.

Sample mean isxis17.8months.

02

Step 2. Determine the test statistics.

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

We have to test the hypotheses,

H0:μ=16.7vsHa: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

03

Step 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<-za2orz>za2, i.e., z<-1.96orz>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.

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