Question

U.S. House of Representatives. There are 435 representatives in the 113th session of the U.S. House of Representatives. On the website www.house.gov, you can find an alphabetized list of the 435 congresspersons. In 2013, the first representative listed is Robert Aderholt, a Republican from Alabama, and the last representative listed is Todd Young, a Republican from Indiana. Suppose that the alphabetized list is indexed 1 through 435.

(a) Use systematic random sampling to obtain a sample of 15 of the 435 representatives.

(b) Suppose that, in Step 2 of Procedure 1.1, the random number chosen is 12 (i.e., k = 12). Determine the sample.

Short answer

Part (a) 23, 52, 81, 110, 139, 168, 197, 226, 255, 284, 313, 342, 371, 400, and 429.

Part (b) 12, 41, 70, 99, 128, 157, 186, 215, 244, 273, 302, 331, 360, 389, and 418.

Step-by-step solution

Part (a) Step 1. Given information.

The given statement is:

There are 435 representatives in the 113th session of the U.S. House of Representatives. On the website www.house.gov, you can find an alphabetized list of the 435 congresspersons. In 2013, the first representative listed is Robert Aderholt, a Republican from Alabama, and the last representative listed is Todd Young, a Republican from Indiana. Suppose that the alphabetized list is indexed 1 through 435.

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 15 and the population size is 435.

$$\begin{gathered} m=\frac{\mathrm{Population}\mathrm{size}}{\mathrm{Sample}\mathrm{size}} \\ =\frac{435}{15} \\ =29 \end{gathered}$$

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 29 for the maximum value.

Step 6: Click the OK button.

Number 23is the MINITAB output.

As a result, the number between 1 andmis$k=23$.

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

$$\begin{gathered} k=23 \\ k+m=23+29 \\ =52 \\ k+2m=23+\left(2\right)\left(29\right) \\ =23+58 \\ =81 \\ k+3m=23+\left(2\right)\left(29\right) \\ =23+87 \\ =110 \\ k+4m=23+\left(4\right)\left(29\right) \\ =23+116 \\ =139 \\ k+5m=23+\left(5\right)\left(29\right) \\ =23+145 \\ =168 \\ k+6m \\ =23+\left(6\right)\left(29\right) \\ =23+174 \\ =197 \\ k+7m=23+\left(7\right)\left(29\right) \\ =23+203 \\ =226 \\ k+8m=23+\left(8\right)\left(29\right) \\ =23+232 \\ =255 \\ k+9m=23+\left(9\right)\left(29\right) \\ =23+261 \\ =284 \\ k+10m=23+\left(10\right)\left(29\right) \\ =23+290 \\ =313 \\ k+11m=23+\left(11\right)\left(29\right) \\ =23+319 \\ =342 \\ k+12m=23+\left(12\right)\left(29\right) \\ =23+348 \\ =371 \\ k+13m=23+\left(13\right)\left(29\right) \\ =23+377 \\ =400 \\ k+14m=23+\left(13\right)\left(29\right) \\ =23+406 \\ =429 \end{gathered}$$

As a result, the size 15 sample has the numbers 23, 52, 81, 110, 139, 168, 197, 226, 255, 284, 313, 342, 371, 400, and 429.

Part (b) Step 1. Use k=12 to determine the sample.

$$\begin{gathered} k=12 \\ k+m=12+29 \\ =41 \\ k+2m=12+\left(2\right)\left(29\right) \\ =12+58 \\ =70 \\ k+3m=12+\left(2\right)\left(29\right) \\ =12+87 \\ =99 \\ k+4m=12+\left(4\right)\left(29\right) \\ =12+116 \\ =128 \\ k+5m=12+\left(5\right)\left(29\right) \\ =12+145 \\ =157 \\ k+6m=12+\left(6\right)\left(29\right) \\ =12+174 \\ =186 \\ k+7m=12+\left(7\right)\left(29\right) \\ =12+203 \\ =215 \\ k+8m=12+\left(8\right)\left(29\right) \\ =12+232 \\ =244 \\ k+9m=12+\left(9\right)\left(29\right) \\ =12+261 \\ =273 \\ k+10m=12+\left(10\right)\left(29\right) \\ =12+290 \\ =302 \\ k+11m=12+\left(11\right)\left(29\right) \\ =12+319=331 \\ k+12m=12+\left(12\right)\left(29\right) \\ =12+348 \\ =360 \\ k+13m=12+\left(13\right)\left(29\right) \\ =12+377 \\ =389 \\ k+14m=12+\left(13\right)\left(29\right) \\ =12+406 \\ =418 \end{gathered}$$

As a result, the numbers 12, 41, 70, 99, 128, 157, 186, 215, 244, 273, 302, 331, 360, 389, and 418 make up the size 15 sample.