Acme Airlines Flies Airplanes That Seat 12 Passengers From E

Acme Airlines Flies Airplanes That Seat 12 Passengers From Experience

Acme Airlines flies airplanes that seat 12 passengers. From experience, they have learned that, on average, 80% of the passengers holding reservations for a particular flight actually show up for the flight. If they book 13 passengers for a flight, what is the probability that 12 or fewer passengers holding reservations will actually show up for the flight? (Write a, b, or c as your answer.) A. less than 90% b. between 90 and 95%, inclusive C. more than 95% How do I calculate critical t? and df? Data: 3-a) Two-tailed test, N = 10, p = .10 3-b) Two-tailed test, N = 47, p = .05 3-c) One-tailed test, N = 80, p = .01

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The problem involving Acme Airlines' seating capacity and passenger show-up rate can be modeled using the binomial distribution, which describes the number of successes (passengers showing up) out of a fixed number of independent trials (reservations). Given the data that 80% of reservations typically show up and 13 passengers are booked, we are tasked with determining the probability that 12 or fewer actual passengers appear. This entails calculating the cumulative probability P(X ≤ 12) where X follows a binomial distribution with n=13 and p=0.8.

Understanding the Binomial Distribution and Its Application

The binomial distribution is suitable here because:

• The number of trials, n=13, is fixed.
• Each passenger's show-up behavior is independent of others.
• The probability of success (a passenger showing up), p=0.8, remains constant across trials.

The mean (expected value) of the binomial distribution is μ = n × p = 13 × 0.8 = 10.4, and the variance is σ² = n × p × (1-p) = 13 × 0.8 × 0.2 = 2.08. The standard deviation is √2.08 ≈ 1.44.

Calculating the Probability P(X ≤ 12)

Since calculating binomial probabilities directly can be cumbersome, especially for cumulative sums, the normal approximation to the binomial distribution is often used when n is sufficiently large. A common rule of thumb is that this approximation is appropriate when np(1-p) > 5, which in this case is true (2.08 ≈ 2.08). Although marginal, we proceed with the normal approximation for simplicity.

To apply the normal approximation, we use a continuity correction. The probability P(X ≤ 12) is approximated by P(Y ≤ 12.5), where Y is a normally distributed variable with mean μ = 10.4 and standard deviation ≈ 1.44.

Calculating the z-score:

• z = (12.5 - 10.4) / 1.44 ≈ 2.1 / 1.44 ≈ 1.46

From z-tables or standard normal distribution calculators, P(Z ≤ 1.46) ≈ 0.9279, or about 92.8%. This indicates that there is approximately a 92.8% chance that 12 or fewer passengers will actually show up, which falls within the "between 90 and 95%, inclusive" range. Therefore, the answer to the multiple-choice question is (b).

Calculating Critical t-value and Degrees of Freedom

The second part of the problem concerns calculating the critical t-value and degrees of freedom (df) for various hypothesis tests:

• 2-tailed test, N=10, p=0.10
• 2-tailed test, N=47, p=0.05
• 1-tailed test, N=80, p=0.01

Degree of Freedom (df)

Degrees of freedom depend on the type of test:

• For a t-test comparing a sample mean to a population mean (one-sample t-test), df = N - 1.
• For independent samples, df = N1 + N2 - 2.

In the context of single-sample tests, degrees of freedom are straightforward, i.e., N - 1.

Critical t-value Calculation

The critical t-value corresponds to the threshold beyond which the null hypothesis is rejected at a specified significance level (α). To find this value:

1. Determine the significance level: for a two-tailed test at p=0.10, α=0.10, split into two tails with α/2=0.05 each.

2. Find the t-value that corresponds to an area of 0.95 to the left for the upper tail (if upper tail); for two-tailed, the total area in the tails combined is 0.10.

3. Use a t-distribution table or calculator to find the t-value for the given N.

For example:

- For N=10, df=9. For a two-tailed test with α=0.10, the critical t-value is approximately ±1.833.

- For N=47, df=46, and α=0.05 (two-tailed), the critical t-value is approximately ±2.013.

- For N=80, df=79, and α=0.01 (one-tailed test), the critical t-value for the upper tail is approximately 2.390 (positive for upper tail, negative for lower tail).

These critical t-values are obtained from standard t-distribution tables or statistical software.

Conclusion

In conclusion, using the binomial distribution and normal approximation, the probability that 12 or fewer passengers show up out of 13 booked with an 80% show-up rate is approximately 92.8%, which aligns with answer choice (b). Regarding the t-distribution calculations, degrees of freedom are given by N-1 for the scenarios described, and critical t-values are obtained from tables or software based on the specified degrees of freedom and significance levels.

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