Air travel is probably not as fun as it used to be. Airlines have increased fares, reduced seat size, and now charge fees for everything from luggage to early boarding. Airlines can be fined by the federal transportation agencies for various "infractions". One issue subject to fines is consistently late arrivals and departures over scheduled times. Airline industry researchers believe charging passengers for luggage have resulted in passengers bringing larger and more luggage on board to avoid luggage fees. More on board luggage could potentially increase the amount of time it took to board all passengers and longer boarding times can result in delayed flights. A study was conducted to determine whether average boarding time was increasing. In previous studies conducted before baggage fees were implemented, average boarding time was recorded at 20 minutes with a known standard deviation of 5 minutes. In a recent sample (after fees) of 64 flights, the mean boarding time was determined to be 23 minutes. ANSWER THE FOLLOWING QUESTIONS: A. Perform the appropriate test at alpha = .05 to determine if average boarding times have increased.
What will be the z-score? please explain based on the example that I have because I know the formula but can you show me step by step. Thanks
To determine if there has been an increase in average boarding times, we can perform a one-sample z-test. This test compares the sample mean (23 minutes) to the population mean (previously recorded at 20 minutes) and assesses whether the difference is statistically significant.
Let's calculate the z-score step by step:
Step 1: Define the null and alternative hypotheses.
The null hypothesis (H0) assumes no difference in average boarding times:
H0: µ = 20
The alternative hypothesis (Ha) assumes there is an increase in average boarding times:
Ha: µ > 20 (one-tailed test)
Step 2: Set the significance level (α).
In this case, α = 0.05, which corresponds to a confidence level of 95%.
Step 3: Calculate the z-score.
The formula to calculate the z-score is:
z = (x - µ) / (σ / √n)
Where:
x = sample mean (23 minutes in this case)
µ = population mean (20 minutes)
σ = population standard deviation (5 minutes)
n = sample size (64 flights in this case)
Substituting the values into the formula:
z = (23 - 20) / (5 / √64)
z = 3 / (5 / 8)
z = 3 * (8 / 5)
z = 4.8
Step 4: Determine the critical value.
Since the alternative hypothesis is one-tailed (µ > 20), we need to find the critical value associated with a 5% level of significance. Consulting a z-table or using a statistics calculator, the critical z-value for a one-tailed test at α = 0.05 is approximately 1.645.
Step 5: Compare the z-score with the critical value.
Since the calculated z-score of 4.8 is greater than the critical z-value of 1.645, we can conclude that the result is statistically significant.
Step 6: Make a decision.
Considering that the z-score falls in the rejection region, we reject the null hypothesis. This suggests that there is sufficient evidence to conclude that the average boarding times have increased.
In conclusion, based on the sample data, the z-score of 4.8 indicates a significant increase in average boarding times, supporting the belief that charging for luggage has resulted in longer boarding durations.
To determine if there has been an increase in average boarding times after the implementation of baggage fees, we need to perform a hypothesis test. Specifically, we want to test if the mean boarding time is significantly different from the previously recorded value of 20 minutes.
Step 1: State the null and alternative hypotheses:
The null hypothesis (H0) states that there is no difference in the mean boarding time: μ = 20 minutes.
The alternative hypothesis (Ha) states that there is an increase in the mean boarding time: μ > 20 minutes.
Step 2: Select the appropriate test statistic:
Since we know the population standard deviation (5 minutes), we can use a z-test.
Step 3: Set the significance level (α):
Given α = 0.05, the significance level is 5%.
Step 4: Collect and analyze data:
In the recent sample of 64 flights, the mean boarding time was determined to be 23 minutes.
Step 5: Calculate the z-score:
The z-score is calculated using the formula:
z = (x - μ) / (σ / sqrt(n))
where:
x = sample mean (23 minutes)
μ = population mean (20 minutes)
σ = population standard deviation (5 minutes)
n = sample size (64 flights)
Plugging the values into the formula, we get:
z = (23 - 20) / (5 / sqrt(64))
z = (3) / (5 / 8)
z = 3 / 0.625
z = 4.8
So, the calculated z-score is 4.8.
Step 6: Determine the critical value:
Since our alternative hypothesis is μ > 20, we are conducting a one-tailed test.
To find the critical value, we need to look up the z-table for a right-tailed test at a significance level of 0.05.
From the z-table, the critical value for a right-tailed test at α = 0.05 is approximately 1.645.
Step 7: Make a decision:
If the z-score is greater than the critical value (4.8 > 1.645), then we reject the null hypothesis in favor of the alternative hypothesis.
In this case, since the z-score is greater than the critical value, we can conclude that there is a significant increase in average boarding times.