Number 38 eylu giutely shoppers to fürther investigate their behavior. a. Show t

Question

Number 38 eylu giutely shoppers to fürther investigate their behavior. a. Show the sampling distribution of p, the proportion of groceries thrown out by your sample respondents What is the probability that your survey will provide a sample proportion wi of the population proportion? What is the probability that your survey will provide a sample proportion within ±.015 of the population proportion? th in ±.0 c. Forty-two percent of primary care doctors think their patients receive unnecessary medical a. Suppose a sample of 300 primary care doctors were taken. Show the sampling b. What is the probability that the sample proportion will be within ±.03 of the c. What is the probability that the sample proportion will be within ±05 of the popula- d. What would be the etfect of taking a larger sample on the probabilities in parts (b) and 38. care (Reader's Digest, December 2011/January 2012). distribution of the proportion of the doctors who think their patients receive unnecessary medical care. population proportion? tion proportion? (c)? Why? In 2008 the Better Business Bureau settled 75% of complaints they received (t March 2, 2009). Suppose you have been hired by the Bet SA Today, Suppose you have been hired by the Better Business Bureau to investi gate the complaints they received this year involving new car dealers. You plan to selec a sample of new car dealer complaints to estimate the proportion of complaints the Bette Business Bureau is able to settle. Assume the population proportion of complaints settle for new car dealers is .75, the same as the overall proportion of complaints settled

Explanation / Answer

38)a) here as n=300 and p=0.42 ; therefore np>10 and n(1-p)>10

for above condition true; sampling distribution of proportion will be approximately normal

with sample mean proportion p=0.42

and std error of proportion =(p*(1-p)/n)1/2 =0.0285

b) P(-0.03+p<X<0.03+p)=P(-0.03/0.0285<Z<0.03/0.0285)=P(-1.0528<Z<1.0528)=0.8538-0.1462 =0.7076

c)P(-0.05+p<X<0.05+p)=P(-0.05/0.0285<Z<0.05/0.0285)=P(-1.7547<Z<1.7547)=0.9603-0.0396 =0.9207

d)as we increase sample size ; the std error will reduce therefore probability in part (b) and (c) will increase as margin of error distance will increase from mean in terms of std deviation

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