Healthcare administration leaders are asked to make evidence-based decisions on a daily basis. Sometimes, these decisions involve high levels of uncertainty, as you have examined previously. Other times, there are data upon which evidence-based analysis might be conducted.
This week, you will be asked to think of scenarios where building and interpreting confidence intervals (CIs) would be useful for healthcare administration leaders to conduct a two-sided hypothesis test using fictitious data.
For example, Ralph is a healthcare administration leader who is interested in evaluating whether the mean patient satisfaction scores for his hospital are significantly different from 87 at the .05 level. He gathers a sample of 100 observations and finds that the sample mean is 83 and the standard deviation is 5. Using a t-distribution, he generates a two-sided confidence interval (CI) of 83 +/- 1.984217 *5/sqrt(100). The 95% CI is then (82.007, 83.992). If repeated intervals were conducted identically, 95% should contain the population mean. The two-sided hypothesis test can be formulated and tested just with this interval. Ho: Mu = 87, Ha: Mu<>87. Alpha = .05. If he assumes normality and that population standard deviation is unknown, he selects the t-distribution. After constructing a 95% CI, he notes that 87 is not in the interval, so he can reject the null hypothesis that the mean satisfaction rates are 87. In fact, he has an evidence-based analysis to suggest that the mean satisfaction rates are not equal to (less than) 87.
Review the resources and consider how a CI might be used to support hypothesis testing in a healthcare scenario.
Describe of a healthcare scenario where a CI might be used, and then complete a fictitious two-sided hypothesis test using a CI and fictitious data.
PART II
Confidence intervals (CIs) and hypothesis tests assist healthcare administration leaders in executing decision making for a wide variety of conditions or experiences in a health services organization. Interpreting the significance of the information provided in hypothesis testing can ensure that healthcare administration leaders execute the best and appropriate measures possible to ensure effective healthcare delivery.
For this Assignment, review the resources for this week. Then, review your course text, and complete Problem 66 on page 455 and Case Study 9.4 on pages 459–460. Consider the type of analysis tools you may use to best fit the case study provided.
The Assignment: (3–5 pages)
- Complete Problem 66 on page 455 of your course text using SPSS.
- Complete Case Study 9.4, “Removing Vioxx From the Market,” on pages 459–460 of your course text.
- This case study requires only calculations and analysis of them, so you may complete this case study using any tool you choose. ATTACH ALL FILES
Albright, S. C., & Winston, W. L. (2015). Business analytics: Data analysis and decision making (5th ed.). Stamford, CT: Cengage Learning.
- Chapter 8, “Confidence Interval Estimation” (pp. 335–400)
- Chapter 9, “Hypothesis Testing” (pp. 401–459)
Fulton, L. V., Ivanitskaya, L. V., Bastian, N. D., Erofeev, D. A., & Mendez, F. A. (2013). Frequent deadlines: Evaluating the effect of learner control on healthcare executives’ performance in online learning. Learning and Instruction, 23, 24–32.
This week you will be asked to think of scenarios where building and interpreting confidence intervals (CIs)
**Part I: Scenario Description and Hypothesis Test**
**Scenario:**
Mary is a healthcare administration leader at a hospital who wants to determine whether there is a significant difference in the average length of stay for patients admitted with different types of insurance coverage. She gathers data from a random sample of patients with Medicaid and a random sample of patients with private insurance.
**Hypothesis Test:**
– Null Hypothesis (H0): The mean length of stay for patients with Medicaid coverage is equal to the mean length of stay for patients with private insurance coverage.
– Alternative Hypothesis (Ha): The mean length of stay for patients with Medicaid coverage is not equal to the mean length of stay for patients with private insurance coverage.
**Data:**
– Sample size for patients with Medicaid (n1) = 50
– Sample mean length of stay for patients with Medicaid (x̄1) = 5.3 days
– Sample standard deviation for patients with Medicaid (s1) = 1.2 days
– Sample size for patients with private insurance (n2) = 60
– Sample mean length of stay for patients with private insurance (x̄2) = 4.8 days
– Sample standard deviation for patients with private insurance (s2) = 1.0 days
**Calculation of Confidence Interval:**
– Using the formula for the two-sample t-test:
\[ \text{CI} = (\bar{x}_1 – \bar{x}_2) \pm t_{\alpha/2} \times \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]
– Degrees of freedom (df) = \( \frac{(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1} + \frac{(\frac{s_2^2}{n_2})^2}{n_2-1}} \)
– Using the t-distribution table or software, find the critical t-value for a two-sided test with the desired confidence level (e.g., 95% confidence level).
– Calculate the confidence interval.
**Interpretation:**
– If the confidence interval includes zero, we fail to reject the null hypothesis.
– If the confidence interval does not include zero, we reject the null hypothesis and conclude that there is a significant difference in the mean length of stay between patients with Medicaid and private insurance coverage.
**Part II: Analysis using SPSS and Case Study 9.4**
