The members of a population have been numbered 1-500. A sample of size 9 is to be taken from the population, using systematic random sampling.

(a) Apply Procedure 1.1 on page 17 to determine the sample (i.e., the numbers corresponding to the members of the population that are included in the sample).

(b) Suppose that, in Step 2 of Procedure 1.1, the random number chosen is 48 (i.e., k = 48). Determine the sample.

The given statement is:

The members of a population have been numbered 1-500. A sample of size 9 is to be taken from the population, using systematic random sampling.

Below is one of the possible solutions.

To obtain a sample, use a systematic random sampling procedure as given below:

1. The sample size is divided by the population size, and the result is rounded to the nearest whole number (m).
2. The sample size is 9 and the population size is 500.

$$\begin{gathered} m=\frac{Populationsize}{Samplesize} \\ =\frac{500}{9} \\ =55.56\approx 55 \end{gathered}$$

Using MINITAB, find a number k that is between 1 and m.

Procedure for MINITAB:

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

Step 2: Double your selected sample size in the Number of rows of data to generate a section, just in case there are repeats. Enter 1 as the sample size.

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

Step 4: Enter 1 as the minimum value.

Step 5: Type 56 for the maximum value.

Step 6: Click the OK button.

Number 5is the MINITAB output.

As a result, the number between 1 andmis $k=5$.

Thus,

$$\begin{gathered} k=5 \\ k+m=5+55 \\ =60 \\ k+2m=5+\left(2\right)\left(55\right) \\ =5+110 \\ =115 \\ k+3m=5+\left(3\right)\left(55\right) \\ =5+165 \\ =170 \\ k+4m=5+\left(4\right)\left(55\right) \\ =5+220 \\ =225 \\ k+5m=5+\left(5\right)\left(55\right) \\ =5+275 \\ =280 \\ k+6m=5+\left(6\right)\left(55\right) \\ =5+330 \\ =335 \\ k+7m=5+\left(7\right)\left(55\right) \\ =5+385 \\ =390 \\ k+8m=5+\left(7\right)\left(55\right) \\ =5+440 \\ =445 \end{gathered}$$

As a result, the size 9 sample has the numbers 5, 60, 115, 170, 225, 280, 335, 390, and 445.

$$\begin{gathered} k=48 \\ k+m=48+55 \\ =103 \\ k+2m=48+\left(2\right)\left(55\right) \\ =48+110 \\ =158 \\ k+3m=48+\left(3\right)\left(55\right) \\ =48+165 \\ =213 \\ k+4m=48+\left(4\right)\left(55\right) \\ =48+220 \\ =268 \\ k+5m=48+\left(5\right)\left(55\right) \\ =48+275 \\ =323 \\ k+6m=48+\left(6\right)\left(55\right) \\ =48+330 \\ =378 \\ k+7m=48+\left(7\right)\left(55\right) \\ =48+385 \\ =433 \\ k+8m=48+\left(7\right)\left(55\right) \\ =48+440 \\ =488 \end{gathered}$$

As a result, the numbers 48, 103, 158, 213, 268, 323, 378, 433, and 488 make up the size 9 sample.