A brand new airline has 172 seats on an airplane and books 179 passengers for its inaugural flight to account for flight check-in rate of 93.5% at the airport.

For the inaugural flight, 159 out of 179 booked passengers showed-up for a rate of 88.8%.

The airline would like to model the P-value for a hypothesis test claiming that their overall check-in rate is below 93.5% at the 0.01 level of significance using a standard normal distribution with their inaugural flight as evidence.

To model the P-value for the hypothesis test regarding the airline's check-in rate, we need to follow these steps:

### **1. State the Hypotheses:**
- **Null Hypothesis (\(H_0\))**: The overall check-in rate is 93.5%, or \( p = 0.935 \).
- **Alternative Hypothesis (\(H_1\))**: The overall check-in rate is less than 93.5%, or \( p < 0.935 \).

### **2. Calculate the Sample Proportion:**
From the inaugural flight, the number of passengers who showed up is 159 out of 179. The sample proportion (\( \hat{p} \)) is:

\[
\hat{p} = \frac{159}{179} \approx 0.888
\]

### **3. Determine the Standard Error:**
The standard error (SE) of the sample proportion under the null hypothesis is calculated using the formula:

\[
SE = \sqrt{\frac{p_0(1 - p_0)}{n}}
\]

where:
- \( p_0 = 0.935 \) (the assumed population proportion under the null hypothesis),
- \( n = 179 \) (the sample size).

\[
SE = \sqrt{\frac{0.935 \times (1 - 0.935)}{179}} \approx \sqrt{\frac{0.935 \times 0.065}{179}} \approx 0.0178
\]

### **4. Calculate the Z-Score:**
The Z-score is a measure of how many standard errors the sample proportion is away from the hypothesized population proportion. It is calculated as:

\[
Z = \frac{\hat{p} - p_0}{SE}
\]

\[
Z = \frac{0.888 - 0.935}{0.0178} \approx \frac{-0.047}{0.0178} \approx -2.64
\]

### **5. Find the P-Value:**
The P-value corresponds to the probability of observing a Z-score as extreme as the one calculated, under the null hypothesis. Since this is a left-tailed test (because we're testing if the proportion is less than 93.5%), we look at the area to the left of \( Z = -2.64 \) on the standard normal distribution.

Using standard normal distribution tables or a calculator:

\[
P(Z < -2.64) \approx 0.0041
\]

### **6. Conclusion:**
Compare the P-value to the significance level (\( \alpha = 0.01 \)):

- If the P-value \( \leq 0.01 \), we reject the null hypothesis.
- If the P-value \( > 0.01 \), we fail to reject the null hypothesis.

Here, \( P = 0.0041 \), which is less than the significance level of 0.01. Therefore, we **reject the null hypothesis** and conclude that there is sufficient evidence to support the claim that the airline's overall check-in rate is below 93.5%.

### Summary
The P-value is approximately 0.0041, and the hypothesis test concludes that the check-in rate is likely below 93.5%.

---

Do you have any questions or need further details on any part of the explanation? 

Here are some related questions you might find useful:

1. How would the P-value change if the sample size was larger?
2. What is the impact of using a two-tailed test in this context?
3. How would the conclusion change if the significance level was 0.05 instead of 0.01?
4. How is the standard error affected if the hypothesized proportion \( p_0 \) was different?
5. Can you explain how the Z-score relates to the standard normal distribution?
6. What if the sample proportion \( \hat{p} \) was exactly 0.935?
7. How does the significance level (\( \alpha \)) affect the hypothesis test?
8. What would be the type I error in this context?

**Tip:** Always consider the context of the data and the consequences of rejecting or failing to reject the null hypothesis, especially when dealing with real-world scenarios like airline operations.

Learn how to model the P-value for a hypothesis test on an airline's check-in rate using the inaugural flight data. Understand the steps involved in hypothesis testing, including calculating the sample proportion, standard error, Z-score, and P-value. Suitable for advanced college-level students and professionals.

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