Question

On-Time Airlines. From the FlightStats On-time Performance Report Summary, we found that, in July 2013, the top five North American airlines in terms of percentage of on-time arrivals were Hawaiian (H), Horizon (Z), Compass (C), Alaska (A), and Jazz (J).

(a) List the 10 possible samples ( without replacement ) of size 3 that can be obtained from the population of five airlines. Use the parenthetical abbreviations in your list.

(b) If a simple random sampling procedure is used to obtain a sample of three of these five airlines, what are the chances that it is the first sample on your list in part (a) ? the second sample? the tenth sample?

(c) Describe three methods for obtaining a simple random sample of three of these five airlines.

(d) Use one of the methods that you described in part (c) to obtain a simple random sample of three of these five airlines.

Short answer

Part (a) (H, Z, C), (H, Z, A), (H, Z, J), (H, C, A), (H, C, J), (H, A, J), (Z, C, A), (Z, C, J), (Z, A, J), (C, A, J)

Part (b) 0.1 for each.

Part (c) Random number generator, Simple random sample without replacement, and Simple random sample with replacement.

Part (d) Sample: 4 1 2

Step-by-step solution

Part (a) Step 1. Given information.

The given statement is:

In terms of percentage of on-time arrivals, the top five North American airlines were:

Hawaiian (H), Horizon (Z), Compass (C), Alaska (A), and Jazz (J).

Part (a) Step 2. List the 10 possible samples ( without replacement ) of size 3 t.

The 10 samples are:

(H, Z, C), (H, Z, A), (H, Z, J), (H, C, A), (H, C, J), (H, A, J), (Z, C, A), (Z, C, J), (Z, A, J), (C, A, J)

Part (b) Step 1. chances that the selected sample is the first sample from the list.

Total samples are ten.

Every sample has an equal chance to be selected.

Therefore, the probability that the first sample is selected is:

$$\begin{gathered} =\frac{1}{10} \\ =0.1 \end{gathered}$$

Every sample has an equal chance to be selected. Therefore the chance that the second sample is selected is 1.

Similarly, the probability that the second sample is selected is:

$$\begin{gathered} =\frac{1}{10} \\ =0.1 \end{gathered}$$

In the same way, the chance that the tenth sample is selected is 1.

Therefore, the probability that the tenth sample is selected is:

$$\begin{gathered} =\frac{1}{10} \\ =0.1 \end{gathered}$$

Part (c) Step 1. Three methods for obtaining a simple random sample.

Method 1:

Take five slips of paper and fold them in half. Each airline should be written on each slip, and three slips should be chosen at random to make a sample of size three.

Method 2:

Pick three airlines at random using a random number generator.

Method 3:

Take ten slips of paper and fold them in half. Part (a) specifies how to write each airline combination. To make a three-size sample, draw a slip at random.

Part (d) Step 1. Obtain a simple random sample of three of the given five airlines.

Using MINITAB, create a three-size random sample from five airlines.

PROCEDURE FOR MINITAB:

Step 1: Select Calc > Random Data > Integer from the menu bar.

Step 2: Twice your selected sample size in the Number of rows of data to create a section, just in case there are repeats. Enter 2 as the sample size.

Step 3: In the Store the column, type Sample in column C1.

Step 4: Enter 1 as the minimum value.

Step 5: Enter 5 as the maximum value.

Step 6: Select OK from the drop-down menu.

MINITAB output:

The resulted Sample is: 4 1 2