Question

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. This proportion is homogeneous for children with various types of defects.
A) If you were to select repeated samples of size 25 from this population, what could you say about the distribution of the sample proportions? List 3 properties.



B) Among the samples of size 25, what fraction has a sample proportion of .45 or higher?



C) What fraction has a sample proportion of .20 or lower?



D) What value of p cuts off the lower 10% of the distribution?

Answer 1

Answer:

a) In the explanation.

b) P=0.09714

c) P=0.0866

d) p=0.208

Step-by-step explanation:

a) The sampling distribution will have this properties:

1 - The sampling distribution of a proportion is equal to the binomial distribution.

2 - The mean of the sampling distribution will be equal to the proportion of the population (0.328).

$\mu=\pi=0.328$

3 - The standard deviation will be equal to the square root of the product of the proportion and its complementary, divided by the sample size.

$\sigma_s=\sqrt{\pi(1-\pi)/n}=\sqrt{0.328*(1-0.328)/25}=0.094$

B) The fraction with sample proportion of 0.45 or higher is

$z=(p-\pi)/\sigma=(0.45-0.328)/0.094=1.298\\\\P(p\geq0.45)=P(z\geq 1.298)=0.09714$

C) The fraction with sample proportion of 0.20 or lower is

$z=(p-\pi)/\sigma=(0.2-0.328)/0.094=-1.362\\\\P(p\leq0.2)=P(z\geq -1.362)=0.0866$

D) The value of p that cuts off the lower 10% of the distribution corresponds to z=-1.28.

Then we have

$z=(p-\pi)/\sigma\\\\p=\pi+z\sigma=0.328+(-1.28)*0.094=0.328-0.120=0.208$