ST3003 Assignment The Normal Distribution and Confidence Intervals

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ST3003 Assignment

The Normal Distribution and Confidence Intervals

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Student Name

Walden University

ST3003

Professor Name

Date

The Normal Distribution and Confidence Intervals 

Part 1 — Confidence Intervals in Research

In this section, use the following article from the Walden Library as your reference:

References

Prevalence and predictors of depression, anxiety and stress among frontline healthcare workers at COVID-19 isolation sites in Gaborone, Botswana. PLoS ONE, 17(8), e0273052. https://doi.org/10.1371/journal.pone.0273052

All articles are available in full text in the Walden Library, which you may search by title or DOI. Make sure to explain each of your answers and support them with evidence from the text and Learning Resources.

Write responses to address the following in paragraph form by inserting your answers directly beneath the questions. 

Table 3 of the journal article shows the number of health care workers reporting different levels of depression, anxiety, and stress, presented both as counts and as percentages of the total sample (447). Confidence intervals are given for each percentage value. Choose ONE level of either depression, anxiety, or stress (e.g., normal depression) to use to address the following questions:

• Demonstrate how the provided percentage was calculated using the sample data.

Normal Anxiety was selected to demonstrate how the percentage was derived from the sample data in Table 3.

Total number of health care workers in the sample: N=447

Number classified as “Normal” on the Anxiety subscale: n=321

Formula for Sample Proportion (Percentage)

Percentage= (n/Total) ×100

321/447 × 100 = 71.81%

Hence, the table’s entry of 71.8 % under Normal for the Anxiety subscale arises directly from

321/447 × 100 = 71.81%

• Provide 1–2 sentences describing how to interpret the confidence interval.

The 95% confidence interval for the Normal Anxiety category (67.4%–75.9%) indicates that, based on the sample data, the true proportion of health care workers with normal anxiety in the population is estimated to fall within this range of 95% of the time (Siamisang et al., 2022). This reflects the reliability of the sample estimate in representing the broader population

• Based on this interval, provide one value that could be considered reasonable as the population proportion and one that would not be considered reasonable as the true population proportion. Fully explain your reasoning.

For the Normal Anxiety category (95% CI: 67.4%–75.9%), a value of 72% would be considered reasonable as the true population proportion because it falls squarely within the interval, indicating that the sample data support this proportion as plausible. Conversely, a value of 80% would not be considered reasonable, since it lies above the upper bound of 75.9%, implying that such a high percentage is not supported by the observed sample and would lie outside the range of proportions deemed plausible 95% of the time.

• If a subsequent study was completed that increased the sample size to 1,000 participants, explain one change you would expect to see in the confidence intervals. Describe the reasons you believe this change would take place. 

For the Normal Anxiety category, increasing the sample size to 1,000 participants would result in a narrower confidence interval around the estimated proportion, reflecting a smaller margin of error. This occurs because the standard error of the proportion, calculated as √p(1-p)/n , decreases when n increases, thereby reducing the width of the interval and yielding a more precise estimate of the population proportion.

Table 1 of this journal article provides a breakdown of the characteristics of the 447 participants in this study. Each participant total is presented, with the percentage indicated in (). 

• Provide one reason you believe a confidence interval was not presented for this data.

A confidence interval was likely not presented for these demographic breakdowns because Table 1 reports observed counts and percentages for the entire sample of 447 participants rather than an estimate of an unknown population parameter. In other words, these figures describe the actual study cohort (no sampling variability), so there is no need to quantify uncertainty around a “true” proportion with a confidence interval.

Part 2 — Using the Normal Distribution 

Use the data set you created from the larger BODY DATA (ST3001) data set to complete the following table.

Replace the text below with your answers using the table above to respond based on your individual data. For each question, write a 1-sentence explanation about how Excel was used to assist you in your computations.

• Determine the percentage of smokers expected to have a BMI above 25 (overweight).

Approximately 74.63% of smokers are expected to have a BMI greater than 25.

Using Excel, I calculated the z-score with = (Data Value – Mean) / Standard Deviation (25-28.5)/5.28, then found the probability with =1-NORM.S. DIST (z, TRUE), which gave 74.63%.

• Determine the percentage of nonsmokers expected to have a BMI below 18.5 (underweight).

The percentage of nonsmokers expected to have a BMI less than 18.5 is 10.27%.

The z-score was calculated in Excel using = (18.5 – 29.2)/8.45, then the probability was found with = NORM.S.DIST (z, TRUE), giving 10.27%.

• A normal BMI range is between 18.5 and 24.9. What percentage of smokers is expected to fall within this range? What percent of nonsmokers are expected to be within this range?

Percent of smokers with a BMI between 18.5 and 24.9: 21.71%

Percent of nonsmokers with BMI between 18.5 and 24.9: 20.29%

In Excel, z-scores were calculated for 18.5 and 24.9 using the formula = (X – Mean)/SD, then the probabilities were found using = NORM.S.DIST(z, TRUE) for each bound; the percentage in the normal BMI range is the difference between the two cumulative probabilities. This was done separately for smokers and nonsmokers using their respective means and standard deviations.

• A researcher is interested in which BMI represents the 90th percentile (90% are at this BMI level or lower). Which BMI score corresponds to the 90th percentile cut-off? 

90th percentile BMI for Smokers: 35.29

90th percentile BMI for Nonsmokers: 40.01

Using Excel, the z-score for the 90th percentile was found with =NORM.S.INV(0.90) which gives approximately 1.28, then the BMI cutoff was calculated by applying the formula = Mean + (z * SD) for smokers and nonsmokers separately.

Part 3 — Creating Confidence Intervals

Use the table above when addressing the following.

Complete the following table:

Enter your response to the following prompt in place of the questions below:

• How does raising the confidence level impact the width of the confidence interval? Does it increase or decrease? Explain one reason why this would happen.

As the confidence level rises from 90% to 95% to 99%, the interval width increases. For example, among smokers, the 90% interval spans 3.06 units (30.06 − 27.00), the 95% interval spans 3.68 units (30.37 − 26.69), and the 99% interval spans 4.96 units (31.01 − 26.05). A similar pattern holds for nonsmokers: the width grows from 4.44 to 5.34 to 7.14 as confidence increases. This happens because a higher confidence level requires a larger critical t‐value, which in turn multiplies the standard error by a bigger factor, expanding the margin of error so that we can be more certain the true mean lies within the interval (Shreffler & Huecker, 2025).

• Using the 95% confidence interval, compare the BMI of smokers vs. nonsmokers. Write a 2–3 sentence paragraph explaining whether these intervals overlap or not. What do these results suggest about the BMI differences between the two groups? 

Both 95% intervals overlap substantially, smokers’ interval (26.69, 30.37) and nonsmokers’ interval (26.52, 31.86) share the entire range from about 26.69 to 30.37. Because these intervals overlap, there is no clear evidence at the 5% level that mean BMI differs between smokers and nonsmokers. In other words, any observed difference in sample means could be due to sampling variability rather than a true population difference.

• In 1–2 sentences, explain the following: What is one situation in which you as a researcher might choose a 99% confidence interval? A 90% interval? Fully justify your choices. 

A 99% confidence interval would be most appropriate in medical research where decisions could affect patient safety, such as estimating the average BMI before prescribing weight-sensitive medication because it provides greater certainty. In contrast, a 90% confidence interval might be preferred in early-stage exploratory studies or marketing research, where quicker decisions are needed and a small margin of uncertainty is acceptable.

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References

Shreffler, J., & Huecker, M. R. (2025). In StatPearls. StatPearls Publishing. http://www.ncbi.nlm.nih.gov/books/NBK557421/

Prevalence and predictors of depression, anxiety and stress among frontline healthcare workers at COVID-19 isolation sites in Gaborone, Botswana. The Public Library of Science ONE, 17(8). https://doi.org/10.1371/journal.pone.0273052