Question

Question 1

A call center for a major insurer recently undertook an extensive staff training program to reduce call times for staff establishing insurance contracts for new customers. The call centre manager examined the records for a random sample 10 recent new customers that purchased motor vehicle insurance for a single vehicle. The time for each of these calls, recorded in seconds (e.g. 600 seconds is 10 minutes), were:

601 523 450 525 552 654 623 579 505 560

Suppose that average call duration prior to the training was 600 seconds and that the known standard deviation of call times was 60 seconds. Test at 5% level if the training has increased workers’ productivity on average using both the:

(a)p-value approach, and

(b)critical value approach.

Question 2 In relation to this test, outline

(a)the Type I -error

(b)what assumption has been made about the distribution of call times?

Answer 1

Solution Q1

Note

The base work is common to both parts, which is first given below. The relevant portion alone are given under the sub-parts (a) and (b)

Let X = call times (in seconds) after the training program. Then, we assume X ~ N(µ, σ²), where σ is known to be 60.

Claim: The training program has improved the workers’ productivity, i.e., the average call time has reduced.

Hypotheses:

Null H₀: µ = µ₀ = 600   Vs Alternative H_A: µ < 600 [claim]

Test statistic:

Z = (√n)(Xbar - µ₀)/σ, where n = sample size; Xbar = sample average; σ = known population standard deviation.

Data

i	xi
1	601
2	523
3	450
4	525
5	552
6	654
7	623
8	579
9	505
10	560

Summary of Excel Calculations is given below:

n	10
Xbar	557.2
Given
µ0	600
σ	60
Zcal	-2.25576

Part (a) p-value approach

Distribution, and p-value

Under H₀, Z ~ N(0, 1)

p-value = P(Z < Zcal)

Using Excel Function, Statistical NORMSDIST, p-value is found to be: 0.012043

Decision:

Since p-value < α = 0.05 i.e., 5% (given), H₀ is rejected.

Conclusion:

There is sufficient evidence to support the claim that the training program has improved the workers’ productivity, i.e., the average call time has reduced. Answer 1

Part (b) Critical Value Approach

Distribution, Critical Value

Under H₀, Z ~ N(0, 1)

Critical value = lower α% point of N(0, 1).

Using Excel Function, Statistical NORMSINV, Zcrit is found to be: -1.64485 at α = 5% or 0.05.

Decision:

Since Zcal < Zcrit, H₀ is rejected.

Conclusion:

There is sufficient evidence to support the claim that the training program has improved the workers’ productivity, i.e., the average call time has reduced. Answer 2

Q2

Part (a)

Type I –error is the error of rejecting the null hypothesis when it is actually true. In this case it refers to concluding the training program has improved workers’ productivity when in reality it has not.

Given the level of significance as 5%, probability of Type I –error is 5% itself. Answer 3

Part (d)

Assumption

As pointed at the very start, we need to assume that the call times follow Normal Distribution since sample size is small. Answer 4

DONE

[Going beyond,

If the sample size were not small, Normality assumption would not have been necessary because in that case Central Limit Theorem would ensure Normality.]