Based On The Following Exam Questions And Financial Calculat

Based On The Following Exam Questions And Financial Calculations

The following 13 questions (Q1 to Q13) are based on the following example: Patients recovering from an appendix operation normally spend an average of 6.3 days in the hospital. The distribution of recovery times is normal with a standard deviation of 2.0 days. The hospital is trying a new recovery program designed to lessen the time patients spend in the hospital. The first 16 appendix patients in this new program were released from the hospital in an average of 5.8 days. On the basis of these data, can the hospital conclude that the new program has a significant reduction of recovery time?

Test at the .05 level of significance.

Paper For Above instruction

Introduction

Hypothesis testing is an essential statistical method to determine if observed differences are statistically significant or due to random variation. In this scenario, the hospital aims to evaluate whether the new recovery program effectively reduces patients’ hospital stays compared to the traditional program. The data provided involves a sample mean of 5.8 days from 16 patients, with a known population standard deviation of 2.0 days and a population mean of 6.3 days from the traditional program.

Selection of the Appropriate Test

Given that the population standard deviation is known, and the sample size is relatively small (n=16), the appropriate statistical procedure is a z-test for the mean. The z-test is suitable for comparing a sample mean to a known population mean when the population standard deviation is known, which applies here.

One-tailed or Two-tailed?

As the research hypothesizes a reduction in recovery time, the test is directional, making it a one-tailed test. The alternative hypothesis tests whether the mean recovery time has decreased with the new program.

Null Hypothesis (Hypotheses in words)

The most appropriate null hypothesis states: "There is no difference in the average recovery time for appendix patients using the new program compared to the traditional recovery program."

(In words: H0: The mean recovery time in the new program equals that of the traditional program, which is 6.3 days.)

Null Hypotheses (In symbols)

H0: μ_new = 6.3

Alternative Hypothesis (in words): "The new program reduces recovery time."

H1: μ_new

Setting the Significance Level

The significance level (α) is 0.05, and for a one-tailed test, the critical z-value can be identified from z-tables.

Calculation of the Standard Error

Standard Error (SE) is calculated using the formula:

SE = σ / √n = 2.0 / √16 = 2.0 / 4 = 0.5

Calculating the Test Statistic

Z = (X̄ - μ0) / SE = (5.8 - 6.3) / 0.5 = -0.5 / 0.5 = -1.0

Decision Rule

The critical z-value at α = 0.05 for a left-tailed test is approximately -1.645. Since the calculated z-value (-1.0) is greater than -1.645, it does not fall into the rejection region.

Conclusion

Failing to reject the null hypothesis indicates that, based on the sample data, there isn't enough evidence at the 0.05 significance level to conclude that the new recovery program significantly reduces hospitalization time.

Implications

It is important to note that while the sample suggests a reduction, the evidence is not statistically significant. Further research with a larger sample size could provide more definitive results.

Additional Analyses

Calculating a 95% confidence interval for the mean recovery time provides insight into the possible range of the true mean. The formula for the confidence interval (CI) is:

CI = X̄ ± Z_(α/2) * SE

Where Z_(α/2) for 95% CI is approximately 1.96. Therefore:

CI = 5.8 ± 1.96 * 0.5 = 5.8 ± 0.98

Range: (4.82 days, 6.78 days)

This interval includes the population mean of 6.3 days, reinforcing the statistical conclusion of no significant reduction.

References

References

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  • Snedecor, G. W., & Cochran, W. G. (1989). Statistical Methods. Iowa State University Press.
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