Question 1185360: A journal article reported that the mean hospital stay following a particular surgical procedure
in 2001 was 7.1 days. A researcher feels that the mean hospital stay in 2002 should be less due
to initiatives aimed at reducing health care costs. A random sample of 40 patients undergoing
the same surgical procedure in 2002 had a mean length of stay of 6.85 days with a standard
deviation of 7.01 days. Run the appropriate statistical test at 𝛼 = 0.05.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to conduct a hypothesis test to see if the mean hospital stay in 2002 was less than in 2001:
**1. State the Hypotheses:**
* **Null Hypothesis (H₀):** The mean hospital stay in 2002 is the same as or greater than the mean hospital stay in 2001. μ ≥ 7.1 days
* **Alternative Hypothesis (H₁):** The mean hospital stay in 2002 is less than the mean hospital stay in 2001. μ < 7.1 days (This is a one-tailed test).
**2. Determine the Significance Level:**
α = 0.05
**3. Choose the appropriate test statistic:**
Since the sample size is greater than 30 and the population standard deviation is unknown, we will use a one-sample t-test.
**4. Calculate the test statistic:**
The t-statistic is calculated as:
t = (sample mean - population mean) / (sample standard deviation / √sample size)
t = (6.85 - 7.1) / (7.01 / √40)
t = -0.25 / (7.01 / 6.324)
t = -0.25 / 1.11
t ≈ -0.225
**5. Determine the degrees of freedom:**
Degrees of freedom (df) = sample size - 1 = 40 - 1 = 39
**6. Find the critical value (or p-value):**
* **Using a t-table:** For a one-tailed test with α = 0.05 and df = 39, the critical t-value is approximately -1.685.
* **Using a calculator or statistical software:** A calculator or statistical software can give a more precise p-value.
**7. Calculate the p-value:**
Using statistical software or a t-table with interpolation, with t ≈ -0.225 and df = 39, the p-value is approximately 0.412.
**8. Make a decision:**
* **Using the critical value:** Our calculated t-statistic (-0.225) is greater than the critical t-value (-1.685). Therefore, we fail to reject the null hypothesis.
* **Using the p-value:** The p-value (0.412) is greater than the significance level (0.05). Therefore, we fail to reject the null hypothesis.
**9. Conclusion:**
There is not sufficient evidence at the α = 0.05 level to conclude that the mean hospital stay in 2002 is less than 7.1 days.
|
|
|
