A newspaper reported on the results of an opinion poll in which adults were aske

Question

A newspaper reported on the results of an opinion poll in which adults were asked what one thing they are most likely to do when they are home sick with a cold or the flu. In the survey, 63% said that they are most likely to sleep and 18% said that they would watch television. Although the sample size was not reported, typically opinion polls include approximately 1,000 randomly selected respondents. a. Assuming a sample size of 1,000 for this poll, construct a 90% confidence interval for the true percentage of all adults who would choose to sleep when they are at home sick. The 90% confidence interval is (Round to the nearest hundredth as needed.) b. if the true percentage of adults who would choose to sleep when they are at home sick is 73%, would you be surprised? No Yes

Explanation / Answer

90% confidence interval

p = 0.63

n = 1000

confidence interval

(p - z*sqrt(pq/n) , p+  z*sqrt(pq/n))

now z0.05 = 1.645 ,q =1-0.63 = 0.37

sqrt(pq/n) = sqrt(0.63*0.37/1000) = 0.01526761

z*0.01526761 = 1.645*0.01526761 = 0.02511

hence required interval = (0.63 -0.02511 , 0.63+0.02511)

=(0.60489,0.65511)

b) Yes ,

since 0.7 is not in 90 % confidence interval, I would be surprised.

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