The effectiveness of a blood-pressure drug is being investigated. An experimenter finds that, on average, the mean reduction in systolic blood pressure is 54.7 for a sample of size 1081 and standard deviation 17.6 . Estimate how much on average the drug will lower a typical patient's systolic blood pressure (using a \( 90 \% \) confidence level). a. Enter your answer as an open-interval (i.e., parentheses) accurate to one decimal place. Confidence interval = b. Express the same answer as a tri-linear inequality one decimal place. c. Express the same answer using the sample mean and margin of error. Give your answers to one decimal place.

To estimate how much, on average, the drug will lower a typical patient's systolic blood pressure using a 90% confidence level, we first calculate the confidence interval. The formula for the confidence interval is: \[ \text{Confidence Interval} = \bar{x} \pm z \left( \frac{s}{\sqrt{n}} \right) \] where: - \(\bar{x} = 54.7\) (sample mean), - \(s = 17.6\) (standard deviation), - \(n = 1081\) (sample size), - \(z\) is the z-value corresponding to the desired confidence level (for 90%, \(z \approx 1.645\)). Now, let's calculate the margin of error: \[ \text{Margin of Error} = z \left( \frac{s}{\sqrt{n}} \right) \] Calculating the standard error: \[ SE = \frac{s}{\sqrt{n}} = \frac{17.6}{\sqrt{1081}} \approx \frac{17.6}{32.9} \approx 0.534 \] Now calculate the margin of error: \[ \text{Margin of Error} \approx 1.645 \times 0.534 \approx 0.878 \] Now we can find the confidence interval: \[ \text{Confidence Interval} = 54.7 \pm 0.878 \] This gives us: - Lower limit: \(54.7 - 0.878 \approx 53.8\) - Upper limit: \(54.7 + 0.878 \approx 55.6\) Thus, the confidence interval is: 1. (53.8, 55.6) For the tri-linear inequality: 2. 53.8 < μ < 55.6 Expressing it with the sample mean and margin of error: 3. Mean = 54.7, Margin of Error = 0.9 So the summary of the results: a. Confidence interval = (53.8, 55.6) b. Tri-linear inequality = 53.8 < μ < 55.6 c. Mean = 54.7, Margin of Error = 0.9