The waiting time for patients at a local walk-in health clinic follows a normal distribution. One morning, the quality-assurance department measured 16 patients with a mean of 15 minutes and a standard deviation of 4 minutes. The quality-assurance department want to know what is the likelihood that the average waiting time was longer than 14.25 minutes with 99.9% confidence level.

To determine the likelihood that the average waiting time was longer than 14.25 minutes, we need to conduct a hypothesis test using the sample data provided. We are given a sample mean, sample size, and sample standard deviation, which allows us to use the \( t \)-distribution for our calculations.

### Step-by-Step Solution:

1. **State the null and alternative hypotheses:**
   - Null hypothesis (\( H_0 \)): \( \mu \leq 14.25 \) (The population mean waiting time is 14.25 minutes or less)
   - Alternative hypothesis (\( H_a \)): \( \mu > 14.25 \) (The population mean waiting time is greater than 14.25 minutes)

2. **Sample statistics:**
   - Sample mean (\( \bar{x} \)) = 15 minutes
   - Sample size (\( n \)) = 16
   - Sample standard deviation (\( s \)) = 4 minutes

3. **Calculate the standard error of the mean (SEM):**
   \[
   SEM = \frac{s}{\sqrt{n}} = \frac{4}{\sqrt{16}} = \frac{4}{4} = 1
   \]

4. **Calculate the \( t \)-score:**
   \[
   t = \frac{\bar{x} - \mu}{SEM} = \frac{15 - 14.25}{1} = 0.75
   \]

5. **Determine the degrees of freedom (df):**
   \[
   df = n - 1 = 16 - 1 = 15
   \]

6. **Find the critical \( t \)-value for a one-tailed test at the 99.9% confidence level:**
   - Using a \( t \)-distribution table or calculator, we find the critical \( t \)-value for \( df = 15 \) and a significance level (\( \alpha \)) of 0.001 for a one-tailed test. The critical \( t \)-value is approximately 3.733.

7. **Compare the calculated \( t \)-score to the critical \( t \)-value:**
   - The calculated \( t \)-score is 0.75, which is much less than the critical \( t \)-value of 3.733.

Since the calculated \( t \)-score (0.75) is less than the critical \( t \)-value (3.733), we fail to reject the null hypothesis. This means that there is insufficient evidence to conclude that the average waiting time was longer than 14.25 minutes with 99.9% confidence.

### Conclusion:
The likelihood that the average waiting time was longer than 14.25 minutes with 99.9% confidence level is very low, as the test result does not support rejecting the null hypothesis.

Would you like more details or have any questions about this process?

Here are some additional questions to explore the topic further:
1. What is the impact of increasing the sample size on the confidence interval?
2. How would the confidence level change if the standard deviation were higher?
3. What is the probability of a Type I error in this context?
4. How can the quality-assurance department improve the accuracy of their estimates?
5. What other statistical tests can be used to analyze waiting times?

**Tip:** Always check the assumptions of normality and sample size adequacy when performing hypothesis tests involving means.

Explore a statistical hypothesis test to determine the likelihood that the average patient waiting time at a health clinic was longer than 14.25 minutes. This problem involves key concepts like the t-distribution, sample mean, and a 99.9% confidence level. Suitable for undergraduate statistics students.

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