Question

In: Operations Management

Employee Years Employed Salary 1 27 $65,487 2 20 $46,184 3 0 $32,782 4 12 $54,899...

Employee Years Employed Salary
1 27 $65,487
2 20 $46,184
3 0 $32,782
4 12 $54,899
5 7 $34,869
6 8 $35,487
7 5 $26,548
8 15 $32,920
9 5 $29,548
10 6 $34,231
11 0 $23,654
12 9 $39,331
13 6 $36,512
14 3 $35,467
15 25 $68,425
16 9 $35,468
17 4 $36,578
18 18 $39,828
19 6 $36,487
20 19 $37,548
21 3 $31,528
22 5 $34,632
23 14 $46,211
24 2 $29,876
25 6 $43,674
26 18 $38,985
27 25 $53,234
28 18 $51,698
29 22 $41,889
30 21 $38,791
31 22 $69,246
32 6 $48,695
33 9 $34,987
34 0 $28,985
35 6 $35,631
36 20 $54,679
37 9 $39,743
38 9 $41,255
39 9 $36,431
40 0 $26,578
41 15 $47,536
42 6 $36,571
43 12 $56,326
44 7 $31,425
45 6 $24,749
46 1 $26,452

The human resources manager of DataCom, Inc., wants to examine the relationship between annual salaries (Y) and the number of years employees have worked at DataCom (X). These data have been collected for a sample of employees and are given above.

1. Draw a normal probability plot of residuals by finding out the Z-score of the residuals. Use Excel. Plot Z-score at Y – axis and residuals at X – axis. Do the residuals seem normally distributed? Explain.

Solutions

Expert Solution

The initial graph for the data

Norml Probability Plot for X axis number of years employees have worked at DataCom.

Order the data according to the number of years employed. Then numbering is done in order from 1 - 46. Cumulative Probability is calculated.

i, number of the data . n = 46.

Cumulative Probability = (i-0.5)/n

Z score for the probability is calulated in excel using the formula NORMSINV(probability)

Years Employed Salary Number Cumilative Probability Z score
0 $     32,782 1 0.0109 -2.2949
0 $     23,654 2 0.0326 -1.8438
0 $     28,985 3 0.0543 -1.6041
0 $     26,578 4 0.0761 -1.4319
1 $     26,452 5 0.0978 -1.2940
2 $     29,876 6 0.1196 -1.1772
3 $     35,467 7 0.1413 -1.0745
3 $     31,528 8 0.1630 -0.9820
4 $     36,578 9 0.1848 -0.8973
5 $     26,548 10 0.2065 -0.8185
5 $     29,548 11 0.2283 -0.7446
5 $     34,632 12 0.2500 -0.6745
6 $     34,231 13 0.2717 -0.6076
6 $     36,512 14 0.2935 -0.5433
6 $     36,487 15 0.3152 -0.4811
6 $     43,674 16 0.3370 -0.4208
6 $     48,695 17 0.3587 -0.3619
6 $     35,631 18 0.3804 -0.3043
6 $     36,571 19 0.4022 -0.2477
6 $     24,749 20 0.4239 -0.1919
7 $     34,869 21 0.4457 -0.1367
7 $     31,425 22 0.4674 -0.0818
8 $     35,487 23 0.4891 -0.0272
9 $     39,331 24 0.5109 0.0272
9 $     35,468 25 0.5326 0.0818
9 $     34,987 26 0.5543 0.1367
9 $     39,743 27 0.5761 0.1919
9 $     41,255 28 0.5978 0.2477
9 $     36,431 29 0.6196 0.3043
12 $     54,899 30 0.6413 0.3619
12 $     56,326 31 0.6630 0.4208
14 $     46,211 32 0.6848 0.4811
15 $     32,920 33 0.7065 0.5433
15 $     47,536 34 0.7283 0.6076
18 $     39,828 35 0.7500 0.6745
18 $     38,985 36 0.7717 0.7446
18 $     51,698 37 0.7935 0.8185
19 $     37,548 38 0.8152 0.8973
20 $     46,184 39 0.8370 0.9820
20 $     54,679 40 0.8587 1.0745
21 $     38,791 41 0.8804 1.1772
22 $     41,889 42 0.9022 1.2940
22 $     69,246 43 0.9239 1.4319
25 $     68,425 44 0.9457 1.6041
25 $     53,234 45 0.9674 1.8438
27 $     65,487 46 0.9891 2.2949

In similar way, the normal probability plot fo rthe salary is drawn as shown below.

Employee Years Employed Salary Number Cumilative Probability Z score
11 0 $     23,654 1 0.0109 -2.2949
45 6 $     24,749 2 0.0326 -1.8438
46 1 $     26,452 3 0.0543 -1.6041
7 5 $     26,548 4 0.0761 -1.4319
40 0 $     26,578 5 0.0978 -1.2940
34 0 $     28,985 6 0.1196 -1.1772
9 5 $     29,548 7 0.1413 -1.0745
24 2 $     29,876 8 0.1630 -0.9820
44 7 $     31,425 9 0.1848 -0.8973
21 3 $     31,528 10 0.2065 -0.8185
3 0 $     32,782 11 0.2283 -0.7446
8 15 $     32,920 12 0.2500 -0.6745
10 6 $     34,231 13 0.2717 -0.6076
22 5 $     34,632 14 0.2935 -0.5433
5 7 $     34,869 15 0.3152 -0.4811
33 9 $     34,987 16 0.3370 -0.4208
14 3 $     35,467 17 0.3587 -0.3619
16 9 $     35,468 18 0.3804 -0.3043
6 8 $     35,487 19 0.4022 -0.2477
35 6 $     35,631 20 0.4239 -0.1919
39 9 $     36,431 21 0.4457 -0.1367
19 6 $     36,487 22 0.4674 -0.0818
13 6 $     36,512 23 0.4891 -0.0272
42 6 $     36,571 24 0.5109 0.0272
17 4 $     36,578 25 0.5326 0.0818
20 19 $     37,548 26 0.5543 0.1367
30 21 $     38,791 27 0.5761 0.1919
26 18 $     38,985 28 0.5978 0.2477
12 9 $     39,331 29 0.6196 0.3043
37 9 $     39,743 30 0.6413 0.3619
18 18 $     39,828 31 0.6630 0.4208
38 9 $     41,255 32 0.6848 0.4811
29 22 $     41,889 33 0.7065 0.5433
25 6 $     43,674 34 0.7283 0.6076
2 20 $     46,184 35 0.7500 0.6745
23 14 $     46,211 36 0.7717 0.7446
41 15 $     47,536 37 0.7935 0.8185
32 6 $     48,695 38 0.8152 0.8973
28 18 $     51,698 39 0.8370 0.9820
27 25 $     53,234 40 0.8587 1.0745
36 20 $     54,679 41 0.8804 1.1772
4 12 $     54,899 42 0.9022 1.2940
43 12 $     56,326 43 0.9239 1.4319
1 27 $     65,487 44 0.9457 1.6041
15 25 $     68,425 45 0.9674 1.8438
31 22 $     69,246 46 0.9891 2.2949

For the residuals seem normally distributed, it has to fall on the staright line. The plot Z-score at – axis is nomally distributed, where as the salary has more outliers and hence it is not normally distributed.

keosha answered 11 months ago
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