91Ó°ÊÓ

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.

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

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.