Question

In: Statistics and Probability

Listed below are amounts​ (in millions of​ dollars) collected from parking meters by a security service...

Listed below are amounts​ (in millions of​ dollars) collected from parking meters by a security service company and other companies during similar time periods. Do the limited data listed here show evidence of stealing by the security service​ company's employees?

Security Servvice Company 1.5 1.6 1.4 1.3 1.7 1.4 1.6 1.7 1.5 1.8

Other Companies 1.8 1.9 1.6 1.7 1.6 1.9 1.5 1.7 1.7 1.7 1.6

The coefficient of variation for the amount collected by the security service company is %

The coefficient of variation for the amount collected by the other companies is %

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.

Solutions

Expert Solution

Solution:

Coefficient of variation for the amount collected by the security service company is

Mean = ( 1.5 + 1.6 + 1.4 + 1.3 + 1.7 + 1.4 + 1.6 + 1.7 + 1.5 + 1.8)/10
= 15.5/10
Mean = 1.55

Standard Deviation σ = √(1/10 - 1) x (( 1.5 - 1.55)2 + (1.6 - 1.55)2 + (1.4 - 1.55)2 + (1.3 - 1.55)2 + (1.7 - 1.55)2 + (1.4 - 1.55)2 + (1.6 - 1.55)2 + (1.7 - 1.55)2 + (1.5 - 1.55)2 + (1.8 - 1.55)2)
= √(1/9) x ((-0.05)2 + (0.05)2 + (-0.15)2 + (-0.25)2 + (0.15)2 + (-0.15)2 + (0.05)2 + (0.15)2 + (-0.05)2 + (0.25)2)
= √(0.1111) x ((0.0025) + (0.0025) + (0.0225) + (0.0625) + (0.0225) + (0.0225) + (0.0025) + (0.0225) + (0.0025) + (0.0625))
= √(0.1111) x (0.225)
= √(0.0249975)
= 0.1581

Cofficient of Varaiance =σ/ μ
=0.1581/1.55
Coefficient of Variance = 0.102

Coefficient of variation for the amount collected by the other companies is

Mean = ( 1.8 + 1.9 + 1.6 + 1.7 + 1.6 + 1.9 + 1.5 + 1.7 + 1.7 + 1.7 + 1.6)/11
= 18.7/11
Mean = 1.7

Standard Deviation σ = √(1/11 - 1) x (( 1.8 - 1.7)2 + (1.9 - 1.7)2 + (1.6 - 1.7)2 + (1.7 - 1.7)2 + (1.6 - 1.7)2 + (1.9 - 1.7)2 + (1.5 - 1.7)2 + (1.7 - 1.7)2 + (1.7 - 1.7)2 + (1.7 - 1.7)2 + (1.6 - 1.7)2)
= √(1/10) x ((0.1)2 + (0.2)2 + (-0.1)2 + (0)2 + (-0.1)2 + (0.2)2 + (-0.2)2 + (0)2 + (0)2 + (0)2 + (-0.1)2)
= √(0.1) x ((0.01) + (0.04) + (0.01) + (0) + (0.01) + (0.04) + (0.04) + (0) + (0) + (0) + (0.01))
= √(0.1) x (0.16)
= √(0.016)
= 0.1265

Cofficient of Varaiance =σ/μ
=0.1265/1.7
Coefficient of Variance = 0.0744

Yes. There is a significant difference in the variation.

orchestra answered 1 year ago
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