For the study in Question 8 above, use R Markdown to find a 95% confidence interval for the true difference in the proportion of children with sepsis/death in the first 60 days after birth in the two groups (Control minus Synbiotic). Report the lower limit of the confidence interval to three decimal places. Make sure your answer is a difference of proportions, not a difference of percentages.

Question

For the study in Question 8 above, use R Markdown to find a 95% confidence interval for the true difference in the proportion of children with sepsis/death in the first 60 days after birth in the two groups (Control minus Synbiotic).  Report the lower limit of the confidence interval to three decimal places.  Make sure your answer is a difference of proportions, not a difference of percentages.

BlackTom AI · Accepted Answer

Let's parse what the question is asking and what information is needed to proceed.
- The task is to use R Markdown to find a 95% confidence interval for the true difference in the proportion of children with sepsis/death in the first 60 days after birth between two groups: Control minus Synbiotic.
- It explicitly asks to report the lower limit of that confidence interval to three decimal places and to express the result as a difference of proportions (i.e., p_control - p_synbiotic), not as a difference of percentages.

To compute a 95% CI for a difference in proportions, you typically need, for each group:
- the sample size (n_control and n_synbiotic), and
- the number of events (sepsis/death) in each group (x_control and x_synbiotic), which yields the estimated proportions p_control = x_control / n_control and p_synbiotic = x_synbiotic / n_synbiotic.

With those, you can compute the standard error for the difference in proportions. The common approach is:
- SE = sqrt( p_control(1 - p_control)/n_control + p_synbiotic(1 - p_synbiotic)/n_synbiotic ),
- CI_lower = (p_control - p_synbiotic) - z_{0.975} * SE, where z_{0.975} ≈ 1.96 for a 95% CI.
- CI_upper = (p_control - p_synbiotic) + z_{0.975} * SE.
And you would report CI_lower to three decimal places. Note that you must present the result as a difference of proportions, not as a percentage.

Important caveats and common pitfalls:
- If the question asks for the lower limit, you should present only CI_lower (to three decimals) after performing the calculation.
- If the event counts are small or if one of the proportions is near 0 or 1, alternative methods (e.g., Wilson or Agresti-Coull intervals) may be preferred for better accuracy, but the standard formula above is the typical starting point for a simple CI for a difference in proportions.
- Ensure you do not confuse a difference in proportions with a difference in percentages; the former is a unitless probability difference, while the latter would be scaled by 100.

In the current prompt, the provided data required to compute the interval (n_control, n_synbiotic, x_control, x_synbiotic) is not included. The answer field lists a numeric value (0.025), but without the underlying counts or proportions, we cannot verify or reproduce the CI or its lower limit. Therefore, with the information given, you cannot confirm whether 0.025 is the correct lower limit, nor can you compute the actual lower bound.

If you supply the group counts (n_control, x_control, n_synbiotic, x_synbiotic), I can walk through the calculation step by step and produce the 95% CI and its lower limit to three decimals, including the exact numeric value for the lower bound.

MAST90007_2025_WIN 2025 Quiz 2

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MAST90007_2025_WIN 2025 Quiz 2

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Similar Questions

Exhibit 10-10 The results of a recent poll on the preference of shoppers regarding two products are shown below. Product Shoppers Surveyed Shoppers Favoring This Product A 800 560 B 900 612 ​ ​ Refer to Exhibit 10-10. The standard error of is _____.

Questions 8 and 9 continued: Report the upper limit of the confidence interval to three decimal places. Make sure your answer is a difference of proportions, not a difference of percentages.

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