A report on food science states that younger adults are more likely to see foods with genetically modified ingredients as being bad for their health than older adults. This statement is based on a representative sample of 175 adult Americans age 18 to 29 and a representative sample of 440 adult Americans age 50 to 64. Of those in the 18 to 29 age group, 48% said they believed these foods were bad for their health, while only 35% of those in the 50 to 64 age group believed this.
(a) Are the sample sizes large enough to use the large-sample confidence interval to estimate the difference in the
population proportions? Explain.
(b) Estimate the difference in the proportion of adult Americans age 18 to 29 who believe the foods made with
genetically modified ingredients are bad for their health and the corresponding proportion for adult Americans age
50 to 64. Use a 90% confidence interval. (Use p − p . Enter your answer using interval notation. Round your
numerical values to four decimal places.)
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(a) To determine if the sample sizes are large enough to use the large-sample confidence interval, we need to check if both sample sizes are at least 10. In this case, the sample size for the 18-29 age group is 175, which is larger than 10. The sample size for the 50-64 age group is 440, which is also larger than 10. Therefore, the sample sizes are large enough to use the large-sample confidence interval.
(b) To estimate the difference in the proportion of adult Americans age 18 to 29 who believe the foods made with genetically modified ingredients are bad for their health and the corresponding proportion for adult Americans age 50 to 64, we can use the formula for the confidence interval for the difference between two population proportions:
CI = (p1 - p2) ± Z * sqrt((p1 * (1 - p1) / n1) + (p2 * (1 - p2) / n2))
Where:
- p1 and p2 are the sample proportions
- n1 and n2 are the sample sizes
- Z is the Z-score for the desired confidence level
In this case, p1 = 0.48, p2 = 0.35, n1 = 175, n2 = 440, and we want a 90% confidence interval, so Z = 1.645 (corresponding to a 90% confidence level).
CI = (0.48 - 0.35) ± 1.645 * sqrt((0.48 * (1 - 0.48) / 175) + (0.35 * (1 - 0.35) / 440))
Calculating this expression yields:
CI = 0.13 ± 1.645 * sqrt((0.48 * (0.52) / 175) + (0.35 * (0.65) / 440))
CI = 0.13 ± 0.06351
Rounding the values to four decimal places, the 90% confidence interval is:
CI = (0.0665, 0.1935)
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