The Cola Preference Survey
A beverage company conducts a taste test to determine if consumers prefer a new cola formula over the original. In a random sample of 150 consumers, 90 preferred the new formula. The company wants to estimate the proportion of all consumers who prefer the new formula with a 95% confidence interval. However, a statistician notes that the sample proportion is 0.6, and the margin of error might be too large for the company's needs. The company requires a margin of error of no more than 0.05. They also want to test if the proportion is greater than 0.5 at the 5% significance level.
1. Construct and interpret a 95% confidence interval for the population proportion. Check the conditions.
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Conditions: Random sample: met. Independence: The sample size (150) is less than 10% of all consumers (assuming more than 1500 consumers), so independence is reasonable. Normality: np̂ = 150(0.6) = 90 ≥ 10 and n(1-p̂) = 150(0.4) = 60 ≥ 10, so the sampling distribution of p̂ is approximately normal. The 95% confidence interval is p̂ ± z* sqrt(p̂(1-p̂)/n) = 0.6 ± 1.96 * sqrt(0.6*0.4/150) = 0.6 ± 1.96 * 0.04 = 0.6 ± 0.0784, which is (0.5216, 0.6784). We are 95% confident that the true proportion of all consumers who prefer the new formula is between 0.522 and 0.678.
2. The company wants a margin of error of 0.05. What sample size is needed for a 95% confidence interval? Use the conservative estimate for p̂.
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Using the conservative estimate p̂ = 0.5, the required sample size is n = (z*/ME)^2 * p̂(1-p̂) = (1.96/0.05)^2 * 0.25 = (39.2)^2 * 0.25 = 1536.64 * 0.25 = 384.16. Round up to 385. So a sample of 385 consumers is needed to achieve a margin of error of 0.05 at 95% confidence.