Question

Directions: For each question, you need to show each step of the hypothesis test, state your null and alternate hypothesis, identify if you are conducting a two-tailed or a one-tailed hypothesis test, identify the Zcrit and Zobt, graph the normal curve, label the critical value and the test statistic, shade the rejection region, tell whether we reject or retain the null and make a conclusion statement. You also need to calculate Cohen’s d, the probability of committing a type I error and type II error, and the strength of the effect size.

2. The national recidivism rate in the United States at three-years post-release is 67 with a standard deviation of 8. The warden at a local prison introduced a new reentry program aimed at reducing the rate that offenders return to prison. To test whether this new program is effective at reducing the recidivism rate of offenders at this local prison he selects 16 offenders to include in the sample and finds that the sample mean is 64. Use a nondirectional one-sample z test to test to determine the new program has an effect on recidivism.

Answer 1

a.
Given that,
population mean(u)=67
standard deviation, sigma =8
sample mean, x =64
number (n)=16
null, Ho: μ=67
alternate, H1: μ<67
level of significance, alpha = 0.05
from standard normal table,left tailed z alpha/2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 64-67/(8/sqrt(16)
zo = -1.5
| zo | = 1.5
critical value
the value of |z alpha| at los 5% is 1.645
we got |zo| =1.5 & | z alpha | = 1.645
make decision
hence value of |zo | < | z alpha | and here we do not reject Ho
p-value : left tail - ha : ( p < -1.5 ) = 0.067
hence value of p0.05 < 0.067, here we do not reject Ho
ANSWERS
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null, Ho: μ=67
alternate, H1: μ<67
test statistic: -1.5
critical value: -1.645
decision: do not reject Ho
p-value: 0.067
we do not have enough evidence to support the claim that whether this new program is effective at reducing the recidivism rate of offenders at this local prison
type 1 error is not possible because fail to reject the null hypothesis.so that it possible only
type2 error only.

b.
Given that,
Standard deviation, σ =8
Sample Mean, X =64
Null, H0: μ=67
Alternate, H1: μ<67
Level of significance, α = 0.05
From Standard normal table, Z α/2 =1.6449
Since our test is left-tailed
Reject Ho, if Zo < -1.6449 OR if Zo > 1.6449
Reject Ho if (x-67)/8/√(n) < -1.6449 OR if (x-67)/8/√(n) > 1.6449
Reject Ho if x < 67-13.1592/√(n) OR if x > 67-13.1592/√(n)
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Suppose the size of the sample is n = 16 then the critical region
becomes,
Reject Ho if x < 67-13.1592/√(16) OR if x > 67+13.1592/√(16)
Reject Ho if x < 63.7102 OR if x > 70.2898
Implies, don't reject Ho if 63.7102≤ x ≤ 70.2898
Suppose the true mean is 64
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(63.7102 ≤ x ≤ 70.2898 | μ1 = 64)
= P(63.7102-64/8/√(16) ≤ x - μ / σ/√n ≤ 70.2898-64/8/√(16)
= P(-0.1449 ≤ Z ≤3.1449 )
= P( Z ≤3.1449) - P( Z ≤-0.1449)
= 0.9992 - 0.4424 [ Using Z Table ]
= 0.5568
For n =16 the probability of Type II error is 0.5568

c.
effective size = X-U/s.d
effective size =(67-64)/8
effective size =0.375
effective size is less than 0.50
so that it is medium level.