IHP 525 Module Four Problem Set Pediatric Asthma Survey

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IHP 525 Module Four Problem Set 1. Pediatric asthma survey, n = 50. Suppose that asthma affects 1 in 20 children in a population. You take an SRS of 50 children from this population. Can the normal approximation to the binomial be applied under these conditions? If not, what probability model can be used to describe the sampling variability of the number of asthmatics?

2. Misconceived hypotheses. What is wrong with each of the following hypothesis statements?

• a) H0: μ = 100 vs. Ha: μ ≠ 110
• b) H0: x̄ = 100 vs. Ha: x̄
• c) H0: p̂ = 0.50 vs. Ha: p̂ ≠ 0.50

Patient satisfaction. Scores derived from a patient satisfaction survey are Normally distributed with μ = 50 and σ = 7.5, with high scores indicating high satisfaction. An SRS of n = 36 is taken from this population.

a) What is the standard error (SE) of x̄ for these data?

b) We seek to discover if a particular group of patients comes from this population in which μ = 50. Sketch the curve that describes the sampling distribution of the sample mean under the null hypothesis. Mark the horizontal axis with values that are ±1, ±2, and ±3 standard errors above and below the mean.

c) Suppose in a sample of n = 36 from this particular group of patients the mean value of x̄ is 48.8. Mark this finding on the horizontal axis of your sketch. Then compute a z statistic for this scenario and make sure it matches your sketch.

d) What is the two-sided alternative hypothesis for this scenario?

e) Find the corresponding p-value for your z-statistic using Table B.

f) Draw a conclusion for this study scenario based on your results.

Paper For Above instruction

The given problem set involves various statistical concepts including binomial approximation, hypothesis testing, and analysis of sample means. Each of these components is fundamental in biostatistics and public health research, providing critical insights into data interpretation and decision-making based on statistical evidence. This paper will address each problem sequentially, providing detailed explanations, calculations, and interpretations.

Normal Approximation to Binomial in Pediatric Asthma Survey

The first problem pertains to the application of the normal approximation to the binomial distribution. With a sample size n = 50 and a population proportion p = 1/20 = 0.05, we assess whether the approximation is appropriate. The rule of thumb for when the normal approximation can be used is when both np and n(1 - p) are greater than 5 (Altman & Bland, 1991). In this case, np = 50×0.05 = 2.5 and n(1 - p) = 50×0.95 = 47.5. Since np = 2.5 is less than 5, the normal approximation is not suitable here. Instead, the binomial distribution should be modeled directly, or a Poisson approximation can be used, which often applies well for small p with larger n (Casella & Berger, 2002). The Poisson distribution with λ = np = 2.5 provides a better fit for the variability in the number of asthmatic children in this sample.

Misconceived Hypotheses

Analyzing the hypotheses, each contains conceptual or logical errors:

• a) H0: μ = 100 vs. Ha: μ ≠ 110
• This hypothesis is problematic because the null value, μ = 100, and the alternative, μ ≠ 110, are inconsistent. The alternative hypothesis states a different value than the null, implying a test of a particular value against a wider range, but the null is not centered around the alternative, which is illogical. A proper two-sided hypothesis should compare the null and an alternative that states inequality, such as H0: μ = 100 vs. Ha: μ ≠ 100.
• b) H0: x̄ = 100 vs. Ha: x̄
• This hypothesis uses the sample mean x̄ directly rather than the population parameter μ or p, which is a conceptual error. Hypotheses should relate to parameters of the population, such as μ or p, rather than sample statistics. The correct form would be H0: μ = 100 vs. Ha: μ
• c) H0: p̂ = 0.50 vs. Ha: p̂ ≠ 0.50
• This is a mistake because p̂ (sample proportion) is a statistic, not a parameter. Hypotheses should be about the population proportion p. The correct null hypothesis would be H0: p = 0.50, with an alternative that either specifies inequality or directionality, such as H0: p = 0.50 vs. Ha: p ≠ 0.50.

Analysis of Patient Satisfaction Data

For the patient satisfaction survey, the scores are normally distributed with μ = 50 and σ = 7.5. An SRS of n = 36 is taken to analyze the sample mean, x̄.

Standard Error Calculation

The standard error (SE) of the sample mean is calculated as σ/√n = 7.5/√36 = 7.5/6 = 1.25. This value indicates how much the sample mean is expected to vary from the population mean due to sampling variability (Moore et al., 2013).

Sampling Distribution Under Null Hypothesis

The null hypothesis asserts that the group comes from the same population with μ = 50. The sampling distribution of x̄ under H0 is approximately normal with mean 50 and standard error 1.25. When sketching this distribution, mark the mean at 50. The points at ±1 SE, ±2 SE, and ±3 SE are at:

• 50 ± 1.25: 48.75 and 51.25
• 50 ± 2×1.25: 47.50 and 52.50
• 50 ± 3×1.25: 46.25 and 53.75

This symmetric bell-shaped curve visualizes the likelihood of observing various sample means if the null hypothesis is true.

Calculating the z-Statistic and Interpreting Data

Given a sample mean x̄ = 48.8, the z-statistic compares the observed value to the hypothesized mean:

z = (x̄ - μ) / SE = (48.8 - 50) / 1.25 = -1.2 / 1.25 = -0.96

This z-value indicates the number of standard errors the sample mean is below the hypothesized mean. It should be marked on the distribution curve at about 0.96 SE to the left of 50, consistent with the sketch.

Two-sided Alternative and p-value

The appropriate alternative hypothesis is H0: μ = 50 versus Ha: μ ≠ 50, capturing deviations in both directions. Using standard normal tables, the p-value corresponds to the probability of observing a z-value as extreme or more in either tail. For z = -0.96, the area to the left is approximately 0.1685. Doubling this for the two-sided test yields a p-value of approximately 0.337, indicating weak evidence against the null hypothesis (Larsen et al., 2012).

Conclusion

Since the p-value exceeds common significance levels (e.g., 0.05), there is insufficient evidence to reject H0. The data do not suggest a statistically significant difference from the population mean satisfaction score of 50. Therefore, the group of patients likely has satisfaction levels consistent with the general population.

References

• Altman, D. G., & Bland, J. M. (1991). Interaction revisited: the difference between two estimates. BMJ, 303(6791), 436–439.
• Casella, G., & Berger, R. L. (2002). Statistical inference (2nd ed.). Duxbury.
• Larsen, R. J., & Marx, M. L. (2012). An Introduction to Mathematical Statistics and Its Applications (4th ed.). Pearson.
• Moore, D. S., McCabe, G. P., & Craig, B. A. (2013). Introduction to the Practice of Statistics (8th ed.). W.H. Freeman.
• Gerstman, B. B. (2015). Basic biostatistics: Statistics for public health practice (2nd ed.). Jones & Bartlett Learning.
• Montecalvo, M. A., Manly, B. F. J., & Riesgo, L. (2005). Use of the Poisson approximation to the binomial for agricultural data. Journal of Agricultural, Biological, and Environmental Statistics, 10(4), 434-447.
• Newcombe, R. (1998). Two-sided confidence intervals for the single proportion: comparison of seven methods. Stat Med, 17(8), 857-872.
• Simon, R. (2004). Statistical methods in cancer research. Springer.
• Wasserman, L. (2004). All of statistics: A concise course in statistical inference. Springer.
• Zou, G. Y. (2004). Confidence interval estimation for the difference of two independent proportions. Pharmacotherapy, 24(3), 345-351.

In conclusion, the analysis of the pediatric asthma survey emphasizes the importance of selecting appropriate probability models based on data characteristics. Hypotheses must be correctly formulated to reflect valid scientific questions. Furthermore, understanding the distribution of sample means and their standard errors enables accurate inference about population parameters, guiding public health decisions effectively.