A random digit dialing sample of 1750 adults found that 1140 used some form of prescription medication. Of those who used prescription drugs, 940 said they are concerned about the current state of health care; 320 of the 610 nonusers agreed with this position as well. Let p₁ and p₂ be the proportion of all prescription drug users and nonusers, respectively, who are concerned regarding health care. Using technology, what is a 90% confidence interval for p₁-p₂?

Question

Quizplus · Accepted Answer

First, we need to calculate the sample proportion of prescription drug users who are concerned about health care: p̂₁ = 940/1140 ≈ 0.825 Next, we calculate the sample proportion of nonusers who are concerned about health care: p̂₂ = 320/610 ≈ 0.525 Now, we can calculate the standard error for the difference between two proportions: SE(p̂₁ - p̂₂) = sqrt((p̂₁(1-p̂₁))/n1 + (p̂₂(1-p̂₂))/n2) = sqrt((0.825(1-0.825))/1140 + (0.525(1-0.525))/610) ≈ 0.023 Using a 90% confidence level, we need to find the critical value from the standard normal distribution, which is approximately 1.645. Finally, we can construct the confidence interval: 0.825-0.525±1.645*0.023≈0.300±0.038 Therefore, the best choice is B with a margin of error of 0.038 in the confidence interval.

A Random Digit Dialing Sample of 1750 Adults Found That