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
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.

Question

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

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.

UpStudy AI · Accepted Answer

To estimate how much, on average, the blood-pressure drug will lower a typical patient's systolic blood pressure with a $$ 90\% $$ confidence level, we'll construct a confidence interval using the given sample statistics.

### Given:
- **Sample Mean ($$ \bar{x} $$)**: 54.7
- **Sample Size ($$ n $$)**: 1081
- **Sample Standard Deviation ($$ s $$)**: 17.6
- **Confidence Level**: 90%

### Steps to Calculate the Confidence Interval:

1. **Determine the Z-Score for 90% Confidence:**
   - For a $$ 90\% $$ confidence level, the critical value ($$ z^* $$) is approximately **1.645**.

2. **Calculate the Standard Error (SE):**
   $$
   SE = \frac{s}{\sqrt{n}} = \frac{17.6}{\sqrt{1081}} \approx \frac{17.6}{32.893} \approx 0.5345
   $$

3. **Compute the Margin of Error (ME):**
   $$
   ME = z^* \times SE \approx 1.645 \times 0.5345 \approx 0.879
   $$
   
   Rounded to one decimal place: **0.9**

4. **Construct the Confidence Interval:**
   $$
   \text{Confidence Interval} = \left( \bar{x} - ME, \bar{x} + ME \right) = (54.7 - 0.9, 54.7 + 0.9) = (53.8, 55.6)
   $$

### Answers:

**a. Confidence interval as an open interval:**
$$
\boxed{(53.8,\ 55.6)}
$$

**b. Confidence interval as a tri-linear inequality:**
$$
\boxed{53.8\ <\ \mu\ <\ 55.6}
$$

**c. Confidence interval using the sample mean and margin of error:**
$$
\boxed{\mu = 54.7 \pm 0.9}
$$
**a. Confidence interval:** $$ \boxed{(53.8,\ 55.6)} $$ **b. Tri-linear inequality:** $$ \boxed{53.8\ <\ \mu\ <\ 55.6} $$ **c. Sample mean and margin of error:** $$ \boxed{\mu = 54.7 \pm 0.9} $$

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