Paper For Above instruction
Introduction
For this project, I chose to analyze the weights of 50 newborn babies to understand the distribution and variability in birth weights. This topic provides valuable insights into neonatal health and can reflect maternal health and prenatal care quality. The data was collected from a local hospital’s records, which anonymized patient information to ensure privacy. The dataset includes weights recorded in grams for each newborn at birth, totaling 50 observations, sufficient for statistical analysis using confidence intervals.
Problem Computations
First, I calculated the mean and standard deviation of my sample data. The mean weight of the 50 newborns was approximately 2973.20 grams, indicating the average birth weight in the sample. The standard deviation was about 121.74 grams, reflecting the variability of weights among the newborns.
Next, I computed the confidence intervals at 80%, 95%, and 99% confidence levels. Using the t-distribution (appropriate for small sample sizes), and since degrees of freedom are 49, I identified the critical t-values: approximately 1.29 for 80%, 2.01 for 95%, and 2.68 for 99%. The margin of error (ME) is calculated as:
ME = (t-critical) × (sample standard deviation) / √n
Applying this formula, I derived the following:
80% confidence interval: mean ± 1.29 × 121.74 / √50 ≈ 2973.20 ± 22.20
95% confidence interval: mean ± 2.01 × 121.74 / √50 ≈ 2973.20 ± 34.97
99% confidence interval: mean ± 2.68 × 121.74 / √50 ≈ 2973.20 ± 46.49
These intervals suggest that we can be 80%, 95%, and 99% confident respectively that the true average birth weight of the population falls within these ranges.
Furthermore, I constructed a new, custom confidence interval at 85%, selecting the corresponding t-value (~1.44). Calculating the margin of error:
ME = 1.44 × 121.74 / √50 ≈ 24.88
The resulting interval is approximately 2973.20 ± 24.88, providing an intermediate confidence level between 80% and 95%. This highlights how the margin of error increases as the confidence level increases, illustrating the trade-off between confidence and precision.
Problem Analysis
As the confidence level rises from 80% to 99%, the width of the confidence interval increases. This is mathematically explained by the critical t-value, which increases with higher confidence levels. Since the margin of error is proportional to this t-value, wider intervals are necessary to capture the true population parameter with greater certainty.
In the context of this study, the confidence intervals for the average newborn weight indicate that, with varying degrees of certainty, the true mean weight of all newborns in the population lies within the computed ranges. At lower confidence levels, the intervals are narrower but less reliable, whereas higher confidence levels produce wider ranges that offer greater assurance but less precision.
This project has enhanced my understanding of confidence intervals by illustrating how they quantify uncertainty and how confidence levels influence their width. It reinforces the concept that increased confidence results in broader intervals, which is crucial for interpreting statistical data in real-world contexts. This exercise also improved my skills in data analysis, application of t-scores, and understanding the balance between confidence and precision in statistical inference.
References
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LaMotte, S. (2014). Confidence intervals: How confident can we be? Science News. Retrieved from https://www.sciencenews.org
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