Question

6. 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.

   1. Set up a hypothesis test to decide whether the mean length of imprisonment for motor- vehicle-theft offenders in Sydney differs from the national mean in Australia.

   2. Explain what each of the following would mean

      1. Type I error

      2. Type II error

      3. Correct decision

   3. Now suppose that the results of carrying out the hypothesis test lead to reject the null hypothesis. Classify that conclusion by error type or as a correct decision if in fact the mean length of imprisonment for motor-vehicle-theft offenders in Sydney

      1. equals the national mean of 16.7 months.

      2. differs from the national mean of 16.7 months.

   4. One hundred randomly selected motor-vehicle-theft offenders in Sydney, Australia, had a mean length of imprisonment of 17.8 months. The population standard deviation of the length of imprisonment in Sydney is σ=6.0 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?

Answer 1

Given that,
population mean(u)=16.7
standard deviation, σ =6
sample mean, x =17.8
number (n)=100
null, Ho: μ=16.7
alternate, H1: μ!=16.7
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 17.8-16.7/(6/sqrt(100)
zo = 1.833
| zo | = 1.833
critical value
the value of |z α| at los 5% is 1.96
we got |zo| =1.833 & | z α | = 1.96
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 1.833 ) = 0.067
hence value of p0.05 < 0.067, here we do not reject Ho
ANSWERS
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null, Ho: μ=16.7
alternate, H1: μ!=16.7
test statistic: 1.833
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.067
we do not have enough evidence to support the claim that f in fact the mean length of imprisonment for motor-vehicle-theft offenders in Sydney
equals the national mean of 16.7 months.

Type 1 error is not possible because it reject the null hypothesis
Type 2 error is possible because it fails to reject the null hypothesis.
Given that,
Standard deviation, σ =6
Sample Mean, X =17.8
Null, H0: μ=16.7
Alternate, H1: μ!=16.7
Level of significance, α = 0.05
From Standard normal table, Z α/2 =1.96
Since our test is two-tailed
Reject Ho, if Zo < -1.96 OR if Zo > 1.96
Reject Ho if (x-16.7)/6/√(n) < -1.96 OR if (x-16.7)/6/√(n) > 1.96
Reject Ho if x < 16.7-11.76/√(n) OR if x > 16.7-11.76/√(n)
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Suppose the size of the sample is n = 100 then the critical region
becomes,
Reject Ho if x < 16.7-11.76/√(100) OR if x > 16.7+11.76/√(100)
Reject Ho if x < 15.524 OR if x > 17.876
Implies, don't reject Ho if 15.524≤ x ≤ 17.876
Suppose the true mean is 17.8
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(15.524 ≤ x ≤ 17.876 | μ1 = 17.8)
= P(15.524-17.8/6/√(100) ≤ x - μ / σ/√n ≤ 17.876-17.8/6/√(100)
= P(-3.7933 ≤ Z ≤0.1267 )
= P( Z ≤0.1267) - P( Z ≤-3.7933)
= 0.5504 - 0.0001 [ Using Z Table ]
= 0.5503
For n =100 the probability of Type II error is 0.5503