We will test the hypothesis that the new policies were effective in reducing the wait time using a one-sample t-test. Here's the procedure:

Step 1: Define the hypotheses

• Null Hypothesis ($H_0$): The mean wait time has not decreased, i.e., $\mu = 45$ seconds.
• Alternative Hypothesis ($H_a$): The mean wait time has decreased, i.e., $\mu < 45$ seconds.

Step 2: Set the level of significance

$\alpha = 0.05$

Step 3: Test statistic formula

The formula for the t-test statistic is:

$t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$

Where:

• $\bar{x} = 44.4$ (sample mean)
• $\mu_0 = 45$ (hypothesized mean)
• $s = 4.2$ (sample standard deviation)
• $n = 250$ (sample size)

Step 4: Compute the test statistic

Substitute the values:

$t = \frac{44.4 - 45}{4.2 / \sqrt{250}}$

$t = \frac{-0.6}{4.2 / 15.81}$

$t = \frac{-0.6}{0.2658} \approx -2.26$

Step 5: Determine the critical value or p-value

For a one-tailed t-test with $df = n-1 = 250-1 = 249$, use a t-distribution table or calculator:

• $t_{critical}$ at $\alpha = 0.05$: approximately $-1.65$ (negative because it's one-tailed).

Since $t = -2.26$ is less than $-1.65$, we reject the null hypothesis.

Alternatively, calculate the p-value using a t-distribution calculator. For $t = -2.26$ and $df = 249$, the p-value is approximately 0.012.

Step 6: Conclusion

1. Statistical Significance: Since $p = 0.012 < \alpha = 0.05$, there is sufficient evidence to reject the null hypothesis. The new policies were effective in reducing wait time.
2. Practical Significance: The reduction from 45 seconds to 44.4 seconds may not be practically significant, depending on the context. A reduction of 0.6 seconds might not noticeably impact customer experience.

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Summary

• Test statistic ($t$): $-2.26$
• p-value: 0.012
• Decision: Reject $H_0$ at $\alpha = 0.05$.
• Conclusion: There is evidence that the new policies reduced the mean wait time. Practical significance depends on the airline's goals.

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Do you want further details or have any questions?

Relative Questions:

1. What are the key differences between statistical and practical significance?
2. How would the results change if the sample size were smaller?
3. What other tests could be used for this scenario if assumptions of the t-test are not met?
4. How can effect size be calculated to assess practical significance?
5. What are the implications of reducing $\alpha$ to 0.01 in this context?

Tip: Always consider the context of the problem when interpreting practical significance alongside statistical significance!