(2) Forty-two percent of primary care doctors think their patients receive unnec

Question

(2) Forty-two percent of primary care doctors think their patients receive unnecessary medical care.

a). Suppose a sample of (100) primary care doctors was taken. What is the probability the proportion of primary care doctors who think their patients receive unnecessary medical care is between 45% and 50%?

b). Suppose a sample of (100) primary care doctors was taken. What is the probability the proportion of primary care doctors who think their patients receive unnecessary medical care is between 38% and 48%?

c). Within what symmetrical limits of the population proportion will 88% of the sample proportions fall?

Explanation / Answer

here p = 0.42
n = 100
SE = sqrt(p*(1-p)/n) = sqrt(0.42*(1-0.42)/100) = 0.0494

(a)
P(0.45<p<0.50) = P((0.45-0.42)/0.0494 < z < (0.5-0.42)/0.0494) = P(0.6073 < z < 1.6194) = 0.2191

(b)
P(0.38<p<0.48) = P((0.38-0.42)/0.0494 < z < (0.48-0.42)/0.0494) = P(-0.8097 < z < 1.2146) = 0.6787

(c)
For 88% CI, z-value = 1.55
lower limit = 0.42 - 1.55*0.0494 = 0.3434
upper limit = 0.42 + 1.55*0.0494 = 0.4966

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