2. Clients at a transitional housing facility for recently released female priso

Question

2. Clients at a transitional housing facility for recently released female prisoners must return to prison if they are caught using illegal drugs or alcohol. A social worker at the facility is interested in the proportion of female prisoners who return to prison from using illegal drugs or alcohol. She analyzes the records of 137 clients that entered a transitional housing facility since it opened five years ago. Of these 137 clients, 29 returned to prison due to substance abuse.

(a) Explain in words what the parameter p is.

(b) Give the numerical value of the statistic “p-hat” that estimates p.

(c) Give a 99% confidence interval for p.

(d) Is there evidence that the proportion of all females in transitional housing that return to prison from substance abuse is over 20%? State hypotheses, calculate the test statistic, approximate the P-value, and state your conclusion in terms of the problem.

Explanation / Answer

a. proportion of female prisoners who return to prison from using
illegal drugs or alcohol

b.
Proportion of prisoners(x)=29
Sample Size(n)=137
Sample proportion = x/n =0.212

c.
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval

Confidence Interval = [ 0.212 ±Z a/2 ( Sqrt ( 0.212*0.788) /137)]
= [ 0.212 - 2.576* Sqrt(0.001) , 0.212 + 2.58* Sqrt(0.001) ]
= [ 0.122,0.302]

d.
Set Up Hypothesis
Null, H0:P=0.2
Alternate, H1: P>0.2
Test Statistic
No. Of Success chances Observed (x)=29
Number of objects in a sample provided(n)=137
No. Of Success Rate ( P )= x/n = 0.2117
Success Probability ( Po )=0.2
Failure Probability ( Qo) = 0.8
we use Test Statistic (Z) for Single Proportion = P-Po/Sqrt(PoQo/n)
Zo=0.21168-0.2/(Sqrt(0.16)/137)
Zo =0.3417
| Zo | =0.3417
Critical Value
The Value of |Z | at LOS 0.05% is 1.64
We got |Zo| =0.342 & | Z | =1.64
Make Decision
Hence Value of |Zo | < | Z | and Here we Do not Reject Ho
P-Value: Right Tail - Ha : ( P > 0.34174 ) = 0.36627
Hence Value of P0.05 < 0.36627,Here We Do not Reject Ho

we don't have way to that proportion of all females in transitional housing
that return to prison from substance abuse is over 20%

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