To calculate a 95% confidence interval (CI) for the true mean systolic blood pressure (SBP) in this hypothetical study, we can use the formula for the confidence interval of the mean when the sample size is large (n = 300):

$\text{CI} = \bar{X} \pm z_{\alpha/2} \times \frac{s}{\sqrt{n}}$

Where:

• $\bar{X}$ is the sample mean (110.5),
• $s$ is the sample standard deviation (15),
• $n$ is the sample size (300),
• $z_{\alpha/2}$ is the critical value for a 95% confidence level, which is 1.96 (since we are assuming the sample is large enough for the z-distribution to apply).

Step-by-step Calculation:

1. Standard error (SE): This is the standard deviation divided by the square root of the sample size. $SE = \frac{s}{\sqrt{n}} = \frac{15}{\sqrt{300}} \approx \frac{15}{17.32} \approx 0.866$

2. Critical value: For a 95% confidence level, $z_{\alpha/2} = 1.96$.

3. Margin of error: Multiply the standard error by the critical value. $\text{Margin of error} = 1.96 \times 0.866 \approx 1.697$

4. Confidence interval: Finally, calculate the confidence interval by subtracting and adding the margin of error to the sample mean. $\text{CI} = 110.5 \pm 1.697$

This gives the interval: $\text{CI} = (110.5 - 1.697, 110.5 + 1.697) = (108.803, 112.197)$

Final Answer:

The 95% confidence interval for the true mean systolic blood pressure is approximately (108.80, 112.20).

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Would you like more details or explanations on any part of the calculation? Here are some related questions:

1. How would the CI change if the sample size were smaller (e.g., n = 50)?
2. What is the difference between using a z-score and a t-score in confidence intervals?
3. How does the confidence level (e.g., 90%, 99%) affect the width of the CI?
4. What assumptions are made when using a z-distribution for confidence intervals?
5. How would the standard deviation affect the width of the confidence interval?

Tip: When working with confidence intervals, always check whether the sample size is large enough (typically n > 30) to use the z-distribution; otherwise, the t-distribution is a better choice.