Chapter 14, Problem 14.1.6RE

To determine

The sample mean of unemployment rates and the sample mean of average weekly benefits of by using systematic sampling

To compare: The results with US figures.

Answer to Problem 14.1.6RE

The sample mean of unemployment rates is 5.78, and the sample mean of the average weekly benefits is 320.9.

There is no large difference between the sample mean and population mean of unemployment rates and also there is large difference between the sample mean and population mean of average weekly benefits

Explanation of Solution

Calculation:

Answers will vary. One of the possible answers is given below:

First arrange all the given states according to serial numbers.

State number	States	Unemployment rates (%)	Average weekly benefit ($)
1	AL	6.5	207
2	AK	6.5	250
3	AZ	8.0	221
4	AR	7.5	289
5	CA	8.9	301
6	CO	6.8	356
7	CT	7.8	345
8	DE	6.7	245
9	FL	7.2	231
10	GA	8.2	267
11	HI	4.8	424
12	ID	6.2	264
13	IL	9.2	324
14	IN	7.5	243
15	IA	4.6	337
16	KS	5.4	341
17	KY	8.3	292
18	LA	6.2	207
19	ME	6.7	285
20	MD	6.6	329
21	MA	7.1	424
22	MI	8.8	293
23	MN	5.1	367
24	MS	8.6	194
25	MO	6.5	242
26	MT	5.6	290
27	NE	3.9	276
28	NV	9.8	308
29	NH	5.3	287
30	NJ	8.2	398
31	NM	6.9	303
32	NY	7.7	308
33	NC	8.0	290
34	ND	2.9	396
35	OH	7.4	318
36	OK	5.4	293
37	OR	7.7	316
38	PA	7.4	360
39	RI	9.5	351
40	SC	7.6	248
41	SD	3.8	276
42	TN	8.2	235
43	TX	6.3	341
44	UT	4.4	345
45	VT	4.4	313
46	VA	5.5	295
47	WA	7.0	387
48	WV	6.5	275
49	WI	6.7	276
50	WY	4.6	359

Procedure for selecting $10$ states using systematic sampling.

• There are 50 states out of which 10 to be selected.
• The rule for systematic sample is selecting every fifth observation after selecting the first observation. Because it is obtained by dividing 50 by 10.
• Select any number from ‘1’ to ‘ $5$’ at random. In this case ‘1’ was Selected so, next number is 6 and so on. That is , $1$2 3 4 5 67 8 9 10 11 … 50
• The selected states are as follows

AL

CO

HI

KS

MA

MT

NM

OK

SD

VA

The corresponding unemployment rates and average weekly benefits of the 10 states are,

S NO	States	Unemployment rates (%)	Average weekly benefit ($)
1	AL	6.5	207
6	CO	6.8	356
11	HI	4.8	424
16	KS	5.4	341
21	MA	7.1	424
26	MT	5.6	290
31	NM	6.9	303
36	OK	5.4	293
41	SD	3.8	276
46	VA	5.5	295
Total		57.8	3,209

Thesample mean of unemployment rates is,

$\begin{array}{c}\text{Sample}\text{mean of unemployment rates}= \frac{\left(\begin{array}{l}\text{sum of observations of }\\ \text{unemployment rate in sample }\end{array}\right)}{\text{Number of observations in sample}}\\ =\frac{\left(\begin{array}{l}6.5+6.8+4.8+5.4+7.1+\\ 5.6+6.9+5.4+3.8+5.5\end{array}\right)}{10}\\ =5.78\end{array}$

The sample mean of the average weekly benefits is,

$\begin{array}{c}\text{Sample}\text{mean of the average weekly benefits} =\frac{\left(\begin{array}{l}\text{sum of observations of average}\\ \text{ weekly benefits in sample}\end{array}\right)}{\text{Number of observations in sample}}\\ =\frac{\left(\begin{array}{l}207+356+424+341+424+\\ 290+303+293+276+295\end{array}\right)}{10}\\ =320.9\end{array}$

The population mean of unemployment rate is,

$\begin{array}{c}\text{Population}\text{mean}\text{of}\text{unemployment}\text{rates}=\frac{\left(\begin{array}{l}6.5+6.5+8.0+7.5+8.9+6.8\\ +7.8+\mathrm{........}+6.7+4.6\end{array}\right)}{50}\\ =\frac{336.4}{50}\\ =6.728\end{array}$

The population mean of weekly average benefits is,

$\begin{array}{c}\text{Mean}\text{of}\text{average}\text{weekly}\text{benefits}=\frac{\left(\begin{array}{l}207+250+221+289+301+356+\\ 345+\mathrm{...}+275+276+359\end{array}\right)}{50}\\ =\frac{15,122}{50}\\ =302.44\end{array}$

Comparison of means:

The mean of unemployment rates is 5.78, population mean of unemployment rate is 6.728, the mean of the average weekly benefits is 320.9, andpopulation mean of weekly average benefits is 302.44.

It is clear that there is no large difference between the sample mean and population mean of unemployment rates and also there islarge difference between the sample mean and population mean of average weekly benefits.