Question

According to Reader's Digest, 46% of primary care doctors think their patients receive unnecessary medical care.

a. Suppose a sample of 260 primary care doctors was taken. Show the sampling distribution of the proportion of the doctors who think their patients receive unnecessary medical care. Use z-table. (to 4 decimals)

b. What is the probability that the sample proportion will be within +/- 0.03 of the population proportion. Round your answer to four decimals.

c. What is the probability that the sample proportion will be within +/- 0.05 of the population proportion. Round your answer to four decimals.

d. What would be the effect of taking a larger sample on the probabilities in parts (b) and (c)? Why? The probabilities would "increase" This is because the increase in the sample size makes the standard error "smaller"

Answer 1

a)

since np and n(1-p) both are greater than or eqaul to 4, sampling distribution of the proportion of the doctors who think their patients receive unnecessary medical care is approximately normal

with

estimate proportion=     μp=

0.4600

and

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

0.0309

b)

probability that the sample proportion will be within +/- 0.03 of the population proportion:

probability =P(0.43<X<0.49)=P((0.43-0.46)/0.031)<Z<(0.49-0.46)/0.031)=P(-0.97<Z<0.97)=0.834-0.166=0.6680

c)

probability =P(0.41<X<0.51)=P((0.41-0.46)/0.031)<Z<(0.51-0.46)/0.031)=P(-1.62<Z<1.62)=0.9474-0.0526=0.8948

d)

The probabilities would "increase" This is because the increase in the sample size makes the standard error "smaller