Question

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Over a decade ago, high levels of lead in the blood of children put many at risk for lead poisoning. After a concerted effort to removed from the environment the proportion of children at risk has decreased. In recent studies, the approximate proportion of all children at risk of high lead levels in the United States is 0.08. (a) Suppose a random sample of size 120 ottined from the population. Which of the following describes the samping distribution for p The distribution of we be normal with a mean of 0.0 and a standard deviation of 0.0006 The distribution of will be non normal with a mean of 0.08 and a standard deviation of 0.0006 The distribution of p will be normal with a mean of 0.08 and a standard deviation of 0.0248. The distribution of will be non-normal with a mean of 0.08 and a standard deviation of 0.0248. (D) What is the probability the proportion of children at risk from a random sample of size 120 is greater than 0.0897(Hound to three decimal places.) (c) Suppose a random sample of size 830 is obtained from the population Which of the following describes the sampling distribution for p? The distribution of wil be normal with a mean of 0.00 and a standard deviation of 0.0001. The distribution of will be non-normal with a mean of 0.08 and a standard deviation of 0.0094 The distribution of p will be non normal with a mean of 0.08 and a standard deviation of 0.0001, The distribution of p will be normal with a mean of 0.05 and a standard deviation of 0.0094. (0) What is the probability the proportion of children are from a random sample of sire is greater than 0.0097 Hound to three decimal places.) (e) What is the effect of increasing the sample size from 120 to 3307 Increasing n decreases the mean of Increasing increases the standard deviation of . Increasing n increases the mean of p.

Answer 1

a)

sample size       =n=	120
std error of proportion=σp=√(p*(1-p)/n)=	0.0248

the distribution of p^ will be normal with a mean of 0.08 and a standard deviation of 0.0248

b)

probability =P(X>0.089)=P(Z>(0.089-0.08)/0.025)=P(Z>0.36)=1-P(Z<0.36)=1-0.6406=0.359

c)

std error of proportion=σp=√(p*(1-p)/n)=

0.0094

the distribution of p^ will be normal with a mean of 0.08 and a standard deviation of 0.0094

d)

probability =P(X>0.089)=P(Z>(0.089-0.08)/0.009)=P(Z>0.96)=1-P(Z<0.96)=1-0.8315=0.169

e)

increaseing n decreases the standard deviation of p^