Did I Get ThisResearchers from Dartmouth Medical School conducted a study in 2003 to look at the connection between watching actors smoking in movies and smoking initiation among adolescents. In the study, 6,522 U.S. adolescents ages 10-14 who had never tried smoking were randomly selected. Of those who subsequently tried smoking for the first time, 38% did so because of exposure to smoking in the movies.Estimate the proportion of all U.S. adolescents ages 10-14 who started smoking because of seeing actors smoke in movies by constructing a 95% confidence interval.Reset this ActivityDid I Get ThisHarris Interactive® conducted an online study from January 3 to 10, 2006 among a nationwide sample of 1,040 U.S. adults (ages 18 years and over), of whom 565 currently use organic foods in preparing meals. Figures for age by gender, race/ethnicity, education, region, and household income were weighted to reflect the total U.S. adult population. Propensity score weighting was used to adjust for respondents' propensity to be online. Construct a 99% confidence interval for the proportion of all U.S. adults who use organic foods in preparing meals.Reset this ActivityDid I Get ThisBased on the answer to the previous question, what is the margin of error of the confidence interval for the number of U.S. adults who use organic foods in preparing meals?0.010.040.080.543
Question
Did I Get ThisResearchers from Dartmouth Medical School conducted a study in 2003 to look at the connection between watching actors smoking in movies and smoking initiation among adolescents. In the study, 6,522 U.S. adolescents ages 10-14 who had never tried smoking were randomly selected. Of those who subsequently tried smoking for the first time, 38% did so because of exposure to smoking in the movies.Estimate the proportion of all U.S. adolescents ages 10-14 who started smoking because of seeing actors smoke in movies by constructing a 95% confidence interval.Reset this ActivityDid I Get ThisHarris Interactive® conducted an online study from January 3 to 10, 2006 among a nationwide sample of 1,040 U.S. adults (ages 18 years and over), of whom 565 currently use organic foods in preparing meals. Figures for age by gender, race/ethnicity, education, region, and household income were weighted to reflect the total U.S. adult population. Propensity score weighting was used to adjust for respondents' propensity to be online. Construct a 99% confidence interval for the proportion of all U.S. adults who use organic foods in preparing meals.Reset this ActivityDid I Get ThisBased on the answer to the previous question, what is the margin of error of the confidence interval for the number of U.S. adults who use organic foods in preparing meals?0.010.040.080.543
Solution
The text you provided contains three separate questions. Here's how to answer each one:
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To estimate the proportion of all U.S. adolescents ages 10-14 who started smoking because of seeing actors smoke in movies, you would use the data provided in the study. According to the study, 38% of the adolescents who tried smoking did so because of exposure to smoking in movies. To construct a 95% confidence interval, you would use the formula for a confidence interval: p̂ ± Z*√(p̂(1-p̂)/n), where p̂ is the sample proportion (0.38 in this case), Z is the Z-score for the desired confidence level (1.96 for 95% confidence), and n is the sample size (6,522 in this case).
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To construct a 99% confidence interval for the proportion of all U.S. adults who use organic foods in preparing meals, you would use the data provided in the study. According to the study, 565 out of 1,040 adults use organic foods in preparing meals. This gives a sample proportion of 565/1040 = 0.543. The Z-score for a 99% confidence interval is 2.576. Using the same formula as above, you can calculate the confidence interval.
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The margin of error of the confidence interval is the amount added and subtracted from the sample proportion to create the confidence interval. It is calculated as Z*√(p̂(1-p̂)/n). Using the data from the second study, the margin of error would be 2.576*√(0.543*(1-0.543)/1040).
Please note that these calculations assume that the samples are simple random samples, and that the conditions for constructing a confidence interval are met.
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