Here is what I have so far:

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= 147.9223

Dev Std = 137.9713

N = 1674

H0: avg QiT <= 150, H1: avg QiT > 150.

= .05

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

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

Z= -.61612

Z critical value= 1.645

I will reject H0 if z > 1.645, hence do not reject H0

Conclusion: We reject the null hypothesis that the QiT is less than 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 < Avg ST PT/Avg ST PE - Avg ST PT < 0

I will reject H0 if Z < Z-critical Value

Alpha= .05

Z critical value = 1.645

I will reject H0 if z < - 1.645, hence do not reject H0

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 the Z critical value, -1.645; 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.

I would like to know i my calculations are correct. Can some one please review the work to ensure there are no mistakes?