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, 60 ​% said that they are most likely to sleep and 20 ​% said that they would watch television. Although the sample size was not​ reported, typically opinion polls include approximately​ 1,000 randomly selected respondents.

Answer 1

SOLUTION:

From given data,

In the​ survey, 60 ​% said that they are most likely to sleep and 20 ​% said that they would watch television. Although the sample size was not​ reported, typically opinion polls include approximately​ 1,000 randomly selected respondents.

Sample size = n = 1000

proportion of adults who would like to sleep is , () = 60% = 60/100 = 0.60

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 .

A 90% confidence interval has significance level = 0.10 and critical value is,

= 1.645

90% confidence interval for population proportion (p) is,

0.60-1.645* < P < 0.60 + 1.645 *

0.60-0.02549 < P <  0.60+0.02549

0.5745 < P < 0.6255

In percentage : 57.45% < P < 62.55%

Therefore , a 90% confidence interval for the true percentage of all adults who would choose to sleep when they at home sick is, (57.45 , 62.55) %

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