Question

The amount of time it takes for a given insurance company to process each insurance claim is normally distributed. To better gauge the processing time, a sample of 16 insurance claims were collected. The sample mean of processing times was 5.4 days and the sample standard deviation was 2.2 days. Based on this sample

a) Find the probability that the standard deviation of the processing time of insurance claims is smaller than 3 days.

b) Find the probability that the average processing time of insurance claims is larger than 6 days.

c) Build a 90% confidence interval for the average processing times of insurance claims through the following steps:

(i) Specify the pivotal statistic and its sampling distribution

(ii) Construct the confidence interval

Answer #1

Data given is:

Sample size n = 16

Sample mean m = 5.4

Sample SD, S = 2.2

(a)

Calculating the statistic for the chi-squared test:

T = (n-1)*(S/S')^2

Here, S' = 3

So,

T = (16-1)*(2.2/3)^2 = 8.067

Degrees of freedom, df = 15

The corresponding probability is: 0.92

(b)

At X = 6, we have:

z = (X-m)/S = (6-5.4)/2.2 = 0.273

So,

P(z > 0.273) = **0.392**

(c)

The 90% CI for average processing time is:

m-(1.64*(S/(n^0.5)) < < m+(1.64*(S/(n^0.5))

5.4-(1.64*(2.2/(16^0.5)) < < 5.4+(1.64*(2.2/(16^0.5))

**4.498 <
< 6.302**

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