ST3003 Assessment Template

Date: February 4, 2026

The Normal Distribution and Confidence Intervals

Part 1 — Confidence Intervals in Research

For this section, refer to the following article from the Walden Library:

References

Siamisang, K., Kebadiretse, D., Tjirare, L. T., Muyela, C., Gare, K., & Masupe, T. (2022). 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. Be sure to provide an explanation for each of your answers and include supporting evidence from the text and/or 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 displays the number of health care workers who report various levels of depression, anxiety, and stress. These numbers are presented as the number of workers as well as the percent of the total (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:

1. Show how the given percentage was derived based on the sample data.

The selected level is normal anxiety. The confidence interval is listed as: 71.8 (67.4–75.9). The total sample is 447, with 321 workers reporting normal anxiety levels. The CI percentage was derived by first finding the percentage of healthcare workers reporting normal anxiety as demonstrated below:

$$\frac{321}{447}*100 = 71.8\%\ or\ 0.718$$

The confidence interval for the proportion was calculated using the formula:

$\widehat{p} \pm z\sqrt{\frac{\widehat{p}\ (1 - \widehat{p})}{n}}$, where p̂ = 0.718, n = 447, and 95% CI (z = 1.96)

$$= 0.718 \pm 1.96\sqrt{\frac{0.718\ (1 - 0.718)}{447}}$$

Confidence Interval = 0.718 ± 0.04171 = (0.674, 0.759) or (67.4%, 75.9%)

2. Write 1–2 sentences explaining how the confidence interval would be interpreted.

The confidence interval, 67.4% to 75.9%, means that we are 95% confident that the true population proportion of healthcare workers experiencing normal anxiety lies within the range.

3. 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.

Based on the interval, a value that would be considered a reasonable estimate of the true population proportion would be 72% because it falls within the confidence interval. A value that would not be considered a reasonable estimate would be 80% because it falls outside the interval.

4. 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. Explain why you feel this change would occur.

A larger sample size means that the confidence interval will become narrower. The reason behind this is that increasing the sample size reduces the standard error of the proportion, and leads to a more precise estimate of the population proportion, even at the same confidence level (Hazra, 2017).

Table 1 of this journal article provides a breakdown of characteristics of the 447 participants in this study. The number of participants is given, followed by the percentage in ().

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

For Table 1, the confidence interval was not presented because the table includes summaries of observed characteristics of the study sample. Descriptive statistics are absolute for the dataset. There is no uncertainty, hence confidence intervals are irrelevant.

Part 2 — Using the Normal Distribution

The following should be based on the data set you created from the larger BODY DATA (ST3001) data set complete the following table:

	Smokers	Nonsmokers
Mean	28.10164	30.48525
Standard deviation	5.559931	8.476297
Sample size (n)	61	61

Replace the questions below with your response to the following:

Use the table above to answer the following questions based on your individual data. For each question, write a 1-sentence explanation about how Excel was used to assist you in your computations.

1. Find the percent of smokers expected to have a BMI of greater than 25 (overweight).

The percent of smokers with BMI greater than 25 is 71.15% or 0.7115

Since this is a normal distribution, the excel formula used was: NORM.DIST(x, mean, standard_deviation, cumulative). When calculating the probability that BMI>25, the formula was changed to: 1- NORM.DIST(x, mean, standard_deviation, cumulative=TRUE) to obtain the upper tail probability that represents overweight smokers.

2. Find the percent of nonsmokers expected to have a BMI of less than 18.5 (underweight).

The percent of nonsmokers with BMI of less than 18.5 is 7.87% or 0.0787

Using excel, this percentage was calculated using the NORM.DIST function with 18.5 as the x value, the nonsmokers mean and standard deviation, and setting cumulative argument to TRUE to return the lower-tail probability that represents the percentage of underweight individuals.

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

The percent of smokers with BMI of between 18.5 and 24.9 is 24.03% or 0.2403

The percent of nonsmokers with BMI of between 18.5 and 24.9 is 17.63% or 0.1763

Using excel, this percentage was calculated by subtracting the cumulative probability at x = 18.5 from the cumulative probability at x = 24.9 using the NORMDIST function, and the smokers’ mean and standard deviation. The same process was repeated with non-smokers mean and standard deviation to obtain the percentage of non-smokers expected to be within the range.

4. A researcher is interested in which BMI represents the 90^th percentile (90% are at this BMI level or lower). What BMI score represents the 90^th percentile cut-off rate?

90^th percentile cut-off BMI score for smokers: 35.23

90^th percentile cut-off BMI score for nonsmokers: 41.35

Using excel, the BMI score that represents the 90^th percentile cutoff rate was calculated using the NORM.INV function with a probability of 0.90. Separate calculations were conducted for the smokers and nonsmokers groups using their respective mean and standard deviations.

Part 3 — Creating Confidence Intervals

Use the table above when addressing the following.

Complete the following table:

Group	90% confidence interval	95% confidence interval	99% confidence interval
Smokers	26.93, 29.27	26.71, 29.50	26.27, 29.94
Nonsmokers	28.70, 32.27	28.36, 32.61	27.69, 33.28

Replace the questions below with your response to the following:

1. As the level of confidence increases, what happens to the width of the confidence interval? Does it increase or decrease? Explain one reason why this would happen.

As the level of confidence increases, the width of the confidence interval also increases (Hazra, 2017). For example, for smokers, the confidence interval width increased from 2.34 at 90% to 2.79 at 95%, then 3.67 at 99%. The reason behind this is that a higher confidence interval requires capturing a larger proportion of the sampling distribution, which leads to a higher critical value (z-score). The result is a larger margin of error and consequently, a wider interval.

2. 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 does that indicate about the differences in BMI between the two groups?

Using the 95% CI, the interval for smokers (26.71 to 29.50), overlaps with the confidence interval for nonsmokers (28.36 to 32.61). Overlapping of confidence intervals means that the range of values for two different parameter estimates, such as group means, share common values. This suggests that the difference between the two groups may not be statistically significant (Hazra, 2017).

Based on this information, the difference in mean BMI for smoker and nonsmoker groups may not be statistically significant at the 95% confidence interval. This means that while nonsmokers may have a higher average BMI, the difference between the two groups could be a result of sampling variability. However, a formal hypothesis test is required to prove whether there is a statistically significant mean difference.

3. 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.

The selection of the confidence interval level depends on whether the researcher wants to achieve certainty or precision. A 99% confidence interval may be selected in situations where the researcher wants to be more confident that the true value in the population is represented in the interval (Hazra, 2017). For example, I would select a 99% confidence interval when conducting high-stake research such as clinical drug trials. A higher interval reduces the risk of drawing incorrect conclusions regarding the safety or effectiveness of a drug.

A 90% confidence interval would be selected in situations where the goal is to obtain a narrower estimate while accepting a higher risk of error (Hazra, 2017). An example is preliminary study such as a pilot study where the focus is to quickly identify promising effects before conducting the real study.

References

Hazra, A. (2017). Using the confidence interval confidently. Journal of Thoracic Disease, 9(10), 4124–4129. https://doi.org/10.21037/jtd.2017.09.14