In a study of the relationship between pet ownership and physical activity in ol

Question

In a study of the relationship between pet ownership and physical activity in old adults, 580 subject reported that they owned pet, while 1916 reported that they did not. Give a 98% confidence interval for the proportion of older adults in this population who are pet owners. (Round your answers to tree decimal places.) One of your employees has suggested at your company develop a new product. You decide to take a random sample of y customers and k whether or not n the new product. The response is on a 1 to 5 scale the 1 indicating "definitely would not purchase: 2, "probably would n purchase: 3, "not sure 4, "probably would purchase" and 5, definitely would purchase "For an initial analysis, you will record the responses 2, and 3 as "No" and 4 and 5 as "Yes." What sample size would you use You wanted the 90% margin of error to be 0.15 or less? (Round your answer up to the next whole number.) An automobile manufacturer would like to know what portion of its customers are dissatisfied with the service recited from local dealer. The customer relations department will survey a random sample of customers and compute a 95% confidence interval for the proportion that are dissatisfied. From past studies, they believe that this proportion will be about 0.21. Find the sample size needed if the margin of error of the confidence interval is to be no more than 0.02. (Round your answer up to the next whole number.)

Explanation / Answer

27) estimated proportion p=580/(580+1916)=0.2324

std error =(p(1-p)/n)1/2 =0.0085

for 98% CI, z=2.3263

hence lower limit =p-z*std error =0.2127

upper limit =p+z*std error =0.2520

28)

here as estimated proportion not known p=0.5

margin of error E=0.15

and for 90% CI, z=1.645

hence sample size n=p(1-p)*(z/E)2 =~31

29)here margin of error E=0.02

for 95% CI, z=1.96

and p=0.21

hence sample size n=p(1-p)*(z/E)2 =~1594

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