A random sample of 134 patients, suffering from a particular disease are given a

Question

A random sample of 134 patients, suffering from a particular disease are given a new medicine. 101 of the patients report an improvement in their condition. a. construct a 90% (two-sided) confidence interval for the overall improvement rate of this medicine. Round your answers to 3 decimal places. 90% CI ( 0.693 , 0.815 ) ! 152-2939 b. Find the minimum sample size required if we want to estimate the improvement rate of this medicine to within 2% with 90% confidence if it is known that the improvement rate is between 70% and 90%. Enter an integer. You must round up. 173 . Tries 3/5 C. Find the minimum sample size required if we want to estimate the improvement rate of this medicine to within 2% with 90% confidence if we do not make any assumptions about the improvement rate. 1691 . Tries 1/5 d. Construct a 90% confidence lower bound for the overall improvement rate. Round your answer to 3 decimal places. Tries 0/5 e. construct a 95% confidence upper bound for the overall improvement rate. Round your answer to 3 decimal places. . Tries 0/5 f. The company that manufactures this medicine claims an 80% improvement rate. Based on your answer to part (e), do you find this claim believable? Explain your answer. (This part will be hand-graded.) Submission type: Submit entries below as answer to receive credit Save entries below (not submitted for credit yet)

Explanation / Answer

here p=101/134 =0.7537

std error =(p(1-p)/n)1/2 =0.0372

for 90% CI ; z =1.645

hence confidence interval =p -/+ z*std error =0.693 ; 0.815

b)here estimated p=0.8

margin of error E =0.02

therefore sample size n=p(1-p)*(Z/E)2 =~1083

c)for no assumption p=0.5

therefore sample size n=p(1-p)*(Z/E)2 =~1691

d)for 90% lower bound ; z=-1.28

therefore lower bound =p+z*std error =0.706

e) for 95% upper bound z =1.645

therefore upper bound =0.815

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