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.

Question

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?

Answer 1

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√n

If only one SEm is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.

I hope this helps.

Answer 2

To determine the z-score, we need to calculate the standard error and the difference between the sample mean and the population mean. Then, we divide the difference by the standard error.

Given:
Population mean (before baggage fees): μ = 20 minutes
Sample mean (after baggage fees): x̄ = 23 minutes
Sample size: n = 64
Standard deviation (before baggage fees): σ = 5 minutes

First, we calculate the standard error using the formula:
Standard Error (SE) = σ / √n

SE = 5 / √64
SE = 5 / 8
SE = 0.625

Next, we find the difference between the sample mean and the population mean:
Difference = x̄ - μ
Difference = 23 - 20
Difference = 3

Now, we calculate the z-score using the formula:
z = Difference / SE

z = 3 / 0.625
z = 4.8

Therefore, the z-score is 4.8.

The z-score measures how many standard deviations an observation or sample mean is from the population mean. In this case, the z-score of 4.8 indicates that the sample mean of 23 minutes is 4.8 standard deviations above the population mean of 20 minutes.