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?

Z = (score-mean)/SEm

SEm = SD/√n

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.

can you please show me how to do it based on the example I have?

Thanks

To determine the z-score, we need to calculate the standard error of the mean and then use it to calculate the z-value.

First, let's define our hypotheses:

Null hypothesis (H0): The average boarding time has not increased (μ = 20 minutes).
Alternative hypothesis (Ha): The average boarding time has increased (μ > 20 minutes).

Next, we calculate the standard error of the mean, which measures the average amount of deviation from the sample mean to the population mean.

Standard error of the mean (SE) can be calculated using the formula:
SE = standard deviation / √(sample size)

In the previous study, the standard deviation was given as 5 minutes, and the recent sample consisted of 64 flights.

SE = 5 / √(64)
SE = 5 / 8
SE = 0.625 minutes

Now, we can calculate the z-value using the formula:
z = (sample mean - population mean) / (SE)

For our calculation:
Sample mean (x̄) = 23 minutes
Population mean (μ) = 20 minutes
Standard error (SE) = 0.625 minutes

z = (23 - 20) / 0.625
z = 3 / 0.625
z = 4.8

The calculated z-score is 4.8.

The z-score measures the number of standard deviations a particular value or observation is from the population mean. In this case, the z-score of 4.8 indicates that the sample mean of 23 minutes is about 4.8 standard deviations above the population mean of 20 minutes.