Q 1.60. - Elementary Statistics

Question

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.

Step-by-Step Solution

Answer

Part (a) 5, 60, 115, 170, 225, 280, 335, 390, and 445. 

Part (b) 48, 103, 158, 213, 268, 323, 378, 433, and 488.

Part (a) Step 1. Given information.

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.

Part (a) Step 2. Determine the sample 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. 

$$\begin{gathered} m=\frac{Population size}{Sample size} \\ =\frac{500}{9} \\ =55.56\approx 55 \end{gathered}$$

Part (a) Step 3. Determine the sample using systematic random sampling.

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 5 is the MINITAB output.

As a result, the number between 1 and m is $k=5$.

Part (a) Step 4. Select the numbers k, k+m, ..., k+8m to acquire the sample

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.

Part (b) Step 1. Use k = 48 to determine the sample.

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

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