A study was conducted to investigate the relationship betwee

A study was conducted to investigate the relationship between maternal smoking during pregnancy and the presence of congenital malformations in the child. Among children who suffer from an abnormality other than Down\'s syndrome or an oral cleft, 32.8% have mothers who smoked during pregnancy [10]. This proportion is homogeneous for children with various types of defects. If you were to select repeated samples of size 25 from this population, what could you say about the distribution of sample proportions? List three properties. Among the samples of size 25, what fraction has a sample proportion of 0.45 or higher? What fraction has a sample proportion of 0.20 or lower? What value of;? cuts off the lower 10% of the distribution?

Solution

a)

1. It is bell shaped.

2. It has mean 0.328.

3. It has standard deviation of

s(p) = sqrt(p(1-p)/n) = sqrt(0.328*(1-0.328)/25) = 0.093896965.

************************

b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.45      
u = mean =    0.328      
          
s = standard deviation =    0.093896965      
          
Thus,          
          
z = (x - u) / s =    1.299296521      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.299296521   ) =    0.096921094 [ANSWER]

************************

c)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.2      
u = mean =    0.328      
          
s = standard deviation =    0.093896965      
          
Thus,          
          
z = (x - u) / s =    -1.36319635      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.36319635   ) =    0.086410319 [ANSWER]

**********************

d)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.1      
          
Then, using table or technology,          
          
z =    -1.281551566      
          
As x = u + z * s,          
          
where          
          
u = mean =    0.328      
z = the critical z score =    -1.281551566      
s = standard deviation =    0.093896965      
          
Then          
          
x = critical value =    0.207666198   [ANSWER]  

 A study was conducted to investigate the relationship between maternal smoking during pregnancy and the presence of congenital malformations in the child. Amon
 A study was conducted to investigate the relationship between maternal smoking during pregnancy and the presence of congenital malformations in the child. Amon

Get Help Now

Submit a Take Down Notice

Tutor
Tutor: Dr Jack
Most rated tutor on our site