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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 16.7months. One hundred randomly selected motor-vehicle-theft offenders in Sydney. Australia, had a mean length of imprisonment of localid="1653225892125" 17.8months. At the localid="1653225896253" 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 localid="1653225901113" 6.0months.

Short Answer

Expert verified

The information does not get enough information to infer that the mean duration of sentence of motor vehicle theft offenders differentiates from of the nation mean in .at the 5%level of significance.

Step by step solution

01

Introduction

Let μbe the average sentence duration for motor vehicle theft offenses.

For a reason,

σ=6months populations standard deviation

The theories must be tested.

H0:μ=16.7months Vs

Ha:μ16.7months

We'll run the test at a significance threshold of 5%, or α=0.05.

n=100is the sample size.

x¯=17.8months, arithmetic mean.

02

Explanation

Statistical test,z=x¯-μ0σn

=17.8-16.76100

=1.83

03

Explanation

The crucial values are ±zα/2=±z0.025because the test is a two-tailed test with α=0.05.

=±1.96

A regression line is z<-zα/2or z>zα/2in this case.

i.e., z<-1.96 or z>1.96

04

Explanation

05

Explanation

Perhaps z<-1.96or z1.96(Actually in fact -1.96<z<1.96)

As that of the amount of a statistical test,z, doesn't really fall as in rejection zone, we need not reject H0at the 5%significance level.

06

Explanation

The records may not provide credible evidence to infer that its average duration of sentence for motor vehicle theft offenses varies first from nation mean in .just at 5%level of significance.

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