ST3003 Assignment Normal Distribution

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. For each of the given percentages, demonstrate how that percentage was obtained from the sample data.

Normal Anxiety was chosen to illustrate the calculation of the percentage of the sample data in Table 3.

Number of health care workers surveyed: N=447.

The number of individuals who scored in the “Normal” range on the Anxiety subscale: n=321

The formula for calculating the Sample Proportion (Percentage) is:

Percentage= (n/Total) ×100

321/447 × 100 = 71.81%

This is because the table reading for the Anxiety subscale is 71.8 % under the Normal column.

321/447 × 100 = 71.81%

2. Summarize in 1 – 2 sentences the interpretation of the confidence interval.

The 95% confidence interval for the Normal Anxiety category (67.4%–75.9%) tells us from the sample data that the true proportion of health care workers with normal anxiety in the population is expected to be in this range 95% of the time (Siamisang et al., 2022). This is an indication of how reliable the sample measure is as an estimate of the population measure.

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. Give a complete justification.

A value of 72% would be reasonable to consider the value of the true population proportion for the Normal Anxiety category (95% CI: 67.4%–75.9%), because it is inside the interval, meaning that the sample data support this proportion as plausible. On the other hand, if a value of 80% were observed, it would not be considered reasonable, because it exceeds the upper bound of 75.9%, indicating that it is not supported by the observed sample and is an extreme value based on the sample, given that it is unlikely to be observed 95% of the time.

For a second study, if more people (1,000) were included in the study, describe one thing you expect to see that is different in the confidence intervals. Describe why you believe this change takes place.

4. If the sample size for the Normal Anxiety category is increased to 1000, then the estimated proportion would have a tighter confidence interval, which would indicate a smaller margin of error. This is because the size of the sample n used to calculate the standard error of the proportion decreases as n increases, which makes the interval narrower and the estimate of the population proportion more precise: the standard error of the proportion = √p(1-p)/n.

Table 1 of this journal article provides a breakdown of the 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.

A confidence interval may not have been included for these demographic breakdowns, as they are simply observed counts and percentages of 447 respondents, rather than an estimate of an unknown population parameter. That is, these are the actual study population, without sampling variability, so there would be no need to quantify the uncertainty around a “true” proportion with a confidence interval.

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:

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. What percentage of smokers would you expect to have a BMI greater than 25 (i.e. t, to be overweight)? Of the smokers, about 74.63% are likely to be overweight.

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

2. Calculate the percentage of nonsmokers who are below 18.5 weight for height (underweight).

The proportion of the nonsmokers who are also underweight (BMI below 18.5%) is 10.27%.

In Excel, the Z score was calculated by using = (18.5 – 29.2)/8.45, and the probability was calculated by using = NORM.S.DIST (z, TRUE), which equals 10.27%.

3. A normal BMI range is between 18.5 and 24.9. What percentage of smokers can be expected to fall into this range? What percent of the nonsmokers do you expect to fall into this range?

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

Percent of nonsmokers who are underweight (BMI < 18.5): 2.27%

4. The z-scores for 18.5 and 24.9 were calculated using the formula = (X – Mean)/SD in Excel and the probabilities were determined by = NORM.S.DIST (z, TRUE) for each bound; the percentage in the normal range of BMI was the difference between the two cumulative probabilities. The means and standard deviations for smokers and nonsmokers were used separately in this.

A researcher wants to know what the 90th percentile of BMI is (90% of the students have a BMI at or below this value). What number of mg/ld. is the 90th percentile cutoff?

90th percentile BMI for Smokers: 35.29

90th percentile BMI for Nonsmokers: 40.01

Using Excel, the z-score for the 90th percentile for the smokers and nonsmokers was determined by =NORM.S.INV (0.90), and the BMI cutoff was calculated separately for smokers and nonsmokers using the formula = Mean + (z * the z score was calculated for the 90th percentile at =NORM.S.INV (0.90) using Excel and the BMI cutoff was calculated separately for smokers and nonsmokers by applying the formula = Mean + (z * SD).

Part 3 — Creating Confidence Intervals

Use the table above when addressing the following.

Complete the following table:

Replace the questions below with your response to the following:

What DOES happen to the width of the confidence interval as the level of confidence INCREASES? Does it increase or decrease? Describe one factor that might cause this to occur.

The interval width increases as the level of confidence gets larger, from 90% to 95% to 99%. 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). The same applies to non-smokers, with the width increasing from 444 to 5.4 to 7.14 as confidence rises. This is because the larger the confidence level is, the larger the critical t‐value will be, thus the larger the standard error will be multiplied by, which in turn makes the margin of error larger, meaning we will be more certain that the true mean is within the interval (Shreffler & Huecker, 2025).

2. using the 95% confidence interval, compare the BMI of smokers vs. nonsmokers. Write a 2–3 sentence paragraph about whether these intervals overlap or not. What does this suggest about the differences in BMI between the two groups?

The intervals for smokers and nonsmokers overlap significantly, one covering the whole interval from about 26.69 to 30.37, the other covering from 26.52 to 31.86. These intervals overlap, and at the 5% significance level, there is no evidence of a difference between smokers and nonsmokers in terms of the mean BMI. That is, any difference in sample means that may be observed may be due to sampling variation, as well as a real difference in population means.

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? Be sure to explain your decisions fully.

In medical research, where decisions may impact patient safety, a 99% confidence interval would be most appropriate, such as determining the average BMI before to prescribing weight-dependent medication to a patient, because it would give more confidence. A 90 percent confidence interval, on the other hand, could be used in an initial exploration or in marketing research where speed may be important, and there may be a lesser overall level of uncertainty acceptable.