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