Question

An insurance company was conducting performance analysis of their claims handling processes and process cycle time was one of their concerns. They collected a sample data of the process cycle time across a number of different claims handling processes over the past six months. However, the data followed a (non-normal) multimodal distribution instead of a normal distribution. Why? Explain what could be the reason(s) behind this? The company then focused on the CTP insurance claims handling processes and a sample data of the process cycle time of about 500 CTP insurance claims from the past six months. Such data followed a normal distribution characterised by mean=20.5 days and standard deviation=5.25 days. Answer the following questions based on the above information.

(a) Given the sample data, how often the cycle time of a CTP insurance claim process could fall within the range [15.25, 20.75] days? Why?

(b) If the expected mean cycle time of CTP insurance claims is 20.2 days, did the company meet this target in the past six months? Conduct an appropriate statistic test to draw conclusion

Answer 1

Solution

Let X = process cycle time (days) of CTP insurance claims. We are given: X ~ N(20.5, 5.25²)............................................ (1)

Back-up Theory

If a random variable X ~ N(µ, σ²), i.e., X has Normal Distribution with mean µ and variance σ², then,

Z = (X - µ)/σ ~ N(0, 1), i.e., Standard Normal Distribution and hence

P(X ≤ or ≥ t) = P[{(X - µ)/σ} ≤ or ≥ {(t - µ)/σ}] = P[Z ≤ or ≥ {(t - µ)/σ}] .……………........................................................…...…(2)

Empirical rule, also known as 68 – 95 – 99.7 percent rule:

P{(µ - σ) ≤ X ≤ (µ + σ)} = 0.68; ............................................................................……………………………………………….(3a)

P{(µ - 2σ) ≤ X ≤ (µ + 2σ)} = 0.95; ........................................................................……………………………………………….(3b)

P{(µ - 3σ) ≤ X ≤ (µ + 3σ)} = 0.997 .......................................................................……………………………………………….(3c)

i.e., Mean ± 1 Standard Deviation holds 68% of the observations; ......................……………………………………………….(3d)

Mean ± 2 Standard Deviations holds 95% of the observations ..........................…………......………………………………….(3e)

and Mean ± 3 Standard Deviations holds 99.7% of the observations ..........................………………………………………….(3f).

Now, to work out the solution,

Part (a)

Question says: [15.25, 20.75]. 20.75 is possibly a typo. It is probably meant to be 25.75.

However, solution given below covers both versions since the principle involved are different.

If it is [15.25, 25.75],

Vide (1), 15.25 = (µ - σ) and 25.75 = (µ + σ) and hence vide (3a) or (3d), the probability is 0.68. Thus, 68% of the time the cycle time of a CTP insurance claim process could fall within the range [15.25, 25.75] days Answer 1a

If it is as given, i.e., [15.25, 20.75],

The probability is:

P(15.25 ≤ X ≤ 20.75)

= P[{(15.25 – 20.5)/5.25} ≤ Z ≤ {(20.75 – 20.5)/5.25}] [vide (1) and (2)]

= P(- 1 ≤ Z ≤ 0.0476)

= P(Z ≤ 0.0476) - P(Z ≤ - 1)

= 0.5189 – 0.1586 [Using Excel Function: Statistical NORMSDIST]

= 0.3603

Thus, 36% of the time the cycle time of a CTP insurance claim process could fall within the range [15.25, 20.75] days Answer 1B

Part (b)

Let X = process cycle time (days) of CTP insurance claims.

Then given, X ~ N(µ, σ²)

Hypotheses:

Null H₀: µ = µ₀ = 20.2   Vs    Alternative H_A: µ ≠ 20.2

Test statistic:

t = (√n)(Xbar - µ₀)/s, = 1.277753

where

n = sample size;

Xbar = sample average;

s = sample standard deviation.

Summary of Excel Calculations is given below:

Given, n	500
µ0	20.2
Xbar	20.5
s	5.25
tcal	1.277753
Assuming α	0.05
tcrit	1.964729
p-value	0.201931

Distribution, Critical Value and p-value

Under H₀, t ~ t_{n – 1} [i.e., t-distribution with (n -1) degrees of freedom)

Critical value = upper (α/2)% point of t_{n – 1}.

p-value = P(t_{n – 1} > |tcal|)

Using Excel Functions: Statistical TINV and TDIST, tcrit and p-value are found to be as shown in the above table

Decision Criterion (Rejection Region)

Reject H₀, if | tcal | > tcrit or equivalently, p-value < α.

Decision:

Since | tcal | < tcrit, or equivalently, since p-value > α. H₀ is accepted.

Conclusion:

There is sufficient evidence to conclude that the company is able to meet the target. Answer 2

DONE