In New York City a study was conducted to evaluate whether a

In New York City, a study was conducted to evaluate whether any information that is available at the time of birth can be used to identify children with special educational needs. In a random sample of 45 third-graders enrolled in the special education program of the public school system, 4 have mothers who have had more than 12 years of schooling.

A) Construct a 90% CI for the population proportion of children with special educational needs whose mothers have had more than 12 years of schooling.

B) In 1980, 22% of all third-graders enrolled in the NYC public school system had mothers who had more than 12 years of schooling. Suppose you wish to know whether this proportion is the same for children in the special education program. What are the null and alternative hypotheses of the appropriate test?

C) Conduct the test at the 0.05 level of significance.

Solution

a)
Confidence Interval For Proportion
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
Mean(x)=4
Sample Size(n)=45
Sample proportion = x/n =0.0889
Confidence Interval = [ 0.0889 ±Z a/2 ( Sqrt ( 0.0889*0.9111) /45)]
= [ 0.0889 - 1.64* Sqrt(0.0018) , 0.0889 + 1.64* Sqrt(0.0018) ]
= [ 0.0193,0.1585]

b)
Set Up Hypothesis
Null, H0:P=0.22
Alternate, H1: P!=0.22
c)
Test Statistic
No. Of Success chances Observed (x)=4
Number of objects in a sample provided(n)=45
No. Of Success Rate ( P )= x/n = 0.0889
Success Probability ( Po )=0.22
Failure Probability ( Qo) = 0.78
we use Test Statistic (Z) for Single Proportion = P-Po/Sqrt(PoQo/n)
Zo=0.08889-0.22/(Sqrt(0.1716)/45)
Zo =-2.1232
| Zo | =2.1232
Critical Value
The Value of |Z | at LOS 0.05% is 1.96
We got |Zo| =2.123 & | Z | =1.96
Make Decision
Hence Value of | Zo | > | Z | and Here we Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != -2.12318 ) = 0.03374
Hence Value of P0.05 > 0.0337,Here we Reject Ho