Solve: 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
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We are asked to test if the mean hospital stay in 2002 is less than the mean hospital stay in 2001 using a one-sample **t-test**. The given null and alternative hypotheses are:

- **Null Hypothesis (H₀):** The mean hospital stay in 2002 is not less than 2001, i.e., \(\mu \geq 7.1\).
- **Alternative Hypothesis (H₁):** The mean hospital stay in 2002 is less than 2001, i.e., \(\mu < 7.1\).

This is a **one-tailed** test with a significance level of \(\alpha = 0.05\).

### Step 1: Gather the given data.
- Population mean in 2001 (\(\mu_0\)) = 7.1 days.
- Sample mean in 2002 (\(\bar{x}\)) = 6.85 days.
- Sample size (\(n\)) = 40 patients.
- Sample standard deviation (\(s\)) = 7.01 days.
- Significance level (\(\alpha\)) = 0.05.

### Step 2: Calculate the **test statistic** using the t-test formula.

The formula for the t-statistic is:
\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
\]
where:
- \(\bar{x}\) is the sample mean,
- \(\mu_0\) is the population mean under the null hypothesis,
- \(s\) is the sample standard deviation,
- \(n\) is the sample size.

### Step 3: Calculate the t-statistic.
Let's calculate this now.The calculated **t-statistic** is approximately \(-0.226\).

### Step 4: Determine the critical t-value.
Since we are conducting a one-tailed test at \(\alpha = 0.05\) with 39 degrees of freedom (\(n-1 = 40-1\)), we need to find the critical t-value from the **t-distribution table** or use a statistical calculator.

Let's find this critical value now.The critical t-value for a one-tailed test at \(\alpha = 0.05\) with 39 degrees of freedom is approximately **1.685**.

### Step 5: Compare the t-statistic to the critical value.
- The calculated t-statistic is \(-0.226\).
- The critical t-value is **1.685**.

Since the t-statistic (\(-0.226\)) is greater than the critical value in the negative direction (i.e., it does not fall into the rejection region), we **fail to reject the null hypothesis**.

### Conclusion:
There is not enough statistical evidence at the \(\alpha = 0.05\) significance level to conclude that the mean hospital stay in 2002 is less than that in 2001.

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Would you like to ask for more details or have any questions?

Here are 5 related questions to expand on:
1. What are Type I and Type II errors in hypothesis testing?
2. How would the results change if the sample size was increased?
3. What are the assumptions behind using a t-test?
4. Can we use a z-test instead of a t-test in this case?
5. How would the test change if we conducted a two-tailed test instead?

**Tip:** Always verify the assumptions of normality and independence when conducting a t-test, especially with small sample sizes.

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.

Perform a one-sample t-test to determine if the mean hospital stay in 2002 is less than that in 2001. Analyze the provided sample of 40 patients with a mean of 6.85 days and a standard deviation of 7.01 days, and use a significance level of 𝛼=0.05. This problem involves hypothesis testing and uses the t-test formula.

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