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, 65% said that they are most likely to sleep and 11% said that they would watch television. Although the sample size was not reported, typically opinion polls include approximately 1,000 randomly selected respondents. 16. a. Assuming a sample size of 1,000 for this poll, construct a 95% confidence interval for the true percentage of all adults who would choose to sleep when they are at home sick The 95% 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 72%, would you be surprised? O Yes No

Explanation / Answer

The statistical software output for this problem is:

One sample proportion summary confidence interval:
p : Proportion of successes
Method: Standard-Wald

95% confidence interval results:

Hence,

a) 95% confidence interval is:

(0.62, 0.68)

b) Yes since 72% is not in the confidence interval.

Proportion Count Total Sample Prop. Std. Err. L. Limit U. Limit p 650 1000 0.65 0.015083103 0.62043766 0.67956234

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