Hello, I need help understanding the feedback given to me by the instructor, I am not sure of what he wants me change within the rejection criteria. Can someone can clarify what needs to be changed on my assignment? I have the written assignment and screen shots of the feedback, thanks in advance!

Part 1

In this study, we have performed two tests. The first is if the average QiT is lower than the market or if more resources should be allocated.

The second test was to assess whether the new protocol is better than the traditional one. For both tests, we carry out the four necessary steps. These are:

1.Determine the null hypothesis and the alternative hypothesis

2.Determine the significance level

3.Determine the critic Z-value

4.Conclusion

For the first case, the steps are:

Avg Qit[FPL1] = 147.9223

Dev Std = 137.9713

N = 1674

H0: avg QiT 150.

?= .05

Z= ((Avg QitT - AvG H0 ) / Stn Dev / sqr(N))

Z= (148 - 150) / 138/(1674^0.5) = -

Z= -.61612

Conclusion: We reject the null hypothesis that the QiT is 150 seconds, while the QiT (Average of 148 seconds) is not significantly lower than the average time of 150 seconds; it is still under the average time.

The second analysis pretends to determine if the new protocol (PE) had improved the average ST from the traditional protocol (ST).

Again, we do the same for steps than before but with the mean difference formula:

Avg ST PE= 149.28

Dev ST PE = 185.78

Var ST PE = 35,556.46

N = 853

Avg ST PT = 212.16

Dev PT = 190.46

Var ST PT = 36,320.90

N = 821

H0: Avg ST PE =Avg ST PT/Avg ST PE - Avg ST PT = 0

H1: Avg ST PE

I will reject H0 if Z

Z = ( (Avg ST PE - Avg ST PT) - 0 )/ (((Var PE / n1) + (Var PT / n2))^0.5)

Z= (149.28 - 185.78) / ((34515 / 853 + 36276 / 821)^0.5)

Z = -6.83

Z is lower than .05[FPL3]; we can reject H0. So, we can see that the new protocol has improved the service time. The average ST with service protocol PE is lower than with the PT protocol.

Summary

For the analysis of the ST, the z score and p-values were also used to determine whether to reject Ho or not. If the Z score's absolute value is larger than the critical value or the p-value is less than .05, the recommendation is to reject Ho, which is QiT

The data presented from the company's call centers had a z score of -6.4107 and a p-value of very close to zero. Using these statistics shows that the average ST time for the company is less than the previous month's average, which shows a positive PE effect. This statistical analysis indicates that no additional changes should be made to the existing customer service call center and serve its purpose if the company is content with its new protocol.

For the first case, the steps are: Frederick April 23, 2021 Avg Qit = 147.9223 The random variable is "Tme in Queue" of QiT (throughout this document). Dev Std = 137.9713 $ Reply Resolve N = 1674 HO: avg QiT 150. 0=.05 Z= ((Avg QitT - AVG HO ) / Stn Dev / squ(N)) Z= (148 - 150) / 138/(1674 0.5) = - Z= -.61612F Frederick I will reject HO if Z