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 [12]. Construct a 90% confidence interval for the population proportion of children with special educational needs whose mothers have had more than 12 years of schooling. In 1980, 22% of all third-graders enrolled in the New York City public school system had mothers who had 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? Conduct the test at the 0.05 level of significance. What do you conclude?

Solution

a)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.088888889          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.042423174          
              
Now, for the critical z,              
alpha/2 =   0.05          
Thus, z(alpha/2) =    1.644853627          
Thus,              
Margin of error = z(alpha/2)*sp =    0.069779911          
lower bound = p^ - z(alpha/2) * sp =   0.019108978          
upper bound = p^ + z(alpha/2) * sp =    0.1586688          
              
Thus, the confidence interval is              
              
(   0.019108978   ,   0.1586688   ) [ANSWER]

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b)

Ho: p = 0.22
Ha: p =/= 0.22 [ANSWERS]

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c)

Formulating the null and alternatuve hypotheses,          
          
Ho:   p   =   0.22
Ha:   p   =/=   0.22
As we see, the hypothesized po =   0.22      
Getting the point estimate of p, p^,          
          
p^ = x / n =    0.088888889      
          
Getting the standard error of p^, sp,          
          
sp = sqrt[po (1 - po)/n] =    0.061752193      
          
Getting the z statistic,          
          
z = (p^ - po)/sp =    -2.123181459      
          
As this is a    2   tailed test, then, getting the p value,  
          
p =    0.033738649      

significance level =    0.05      

As P < 0.05, we REJECT THE NULL HYPOTHESIS.      

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d)

Thus, there is sigificant evidence that the proportion of mothers with more than 12 years of schooling in this special education program is different from 22%.