In: Statistics and Probability
Listed below are amounts (in millions of dollars) collected from parking meters by a security service company(SSC) and other companies during similar time periods. Do the limited data listed here show evidence of stealing by the security service company's employees? SCC: 1.4, 1.7, 1.5, 1.6, 1.6, 1.3, 1.5, 1.7, 1.4, 1.7 Other Companies: 1.9, 1.8, 1.5, 1.7, 1.6, 1.8, 1.7, 1.6, 1.8, 1.7 Find the coefficient of variation for each of the two samples, then compare the variation. (Round to one decimal place as needed.) Do the limited data listed here show evidence of stealing by the security service company's employees? Consider a difference of greater than 1% to be significant.
Case1: SCC:
Mean = = 15.4/10 =1.54
x | x - | (x - )2 |
1.4 | - 0.14 | 0.0196 |
1.7 | 0.16 | 0.0256 |
1.5 | - 0.04 | 0.0016 |
1.6 | 0.06 | 0.0036 |
1.6 | 0.06 | 0.0036 |
1.3 | - 0.24 | 0.0576 |
1.5 | - 0.04 | 0.0016 |
1.7 | 0.16 | 0.0256 |
1.4 | - 0.14 | 0.0196 |
1.7 | 0.16 | 0.0256 |
Total = | 0.184 |
Variance = s2 = 0.184/9 = 0.0204
So,
Standard Deviation = s =
So,
Coefficient of Variation (CV) is given by:
%
Case 2: Other companies:
Mean = = 17.1/10 =1.71
x | x - | (x - )2 |
1.9 | 0.19 | 0.0361 |
1.8 | 0.09 | 0.0081 |
1.5 | - 0.21 | 0.0441 |
1.7 | - 0.01 | 0.0001 |
1.6 | - 0.11 | 0.0121 |
1.8 | 0.09 | 0.0081 |
1.7 | - 0.01 | 0.0001 |
1.6 | - 0.11 | 0.0121 |
1.8 | 0.09 | 0.0081 |
1.7 | - 0.01 | 0.0001 |
Total = | 0.129 |
Variance = s2 = 0.129/9 = 0.0143
So,
Standard Deviation = s =
So,
Coefficient of Variation (CV) is given by:
%
Difference in CV = 9.29 % - 7.00 % = 2.29 %
So,
the difference is significant.
So,
the limited data listed here show evidence of stealing by the security service company's employees.