Question 1185360
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.