Calculations (SPA Table)

Below are the calculations from the frame work which are represents the SPA Table. This is just a section of the calculations which excludes graphs, Pie charts and Plots

Hypotheses Testing Framework

 To prove the calculations form the Facility Study a, Vining and Kowalski (2008)

Hypotheses Testing Framework has been implemented.

 A Hypotheses Testing Framework is a frame work that provides a set of established calculations that can provide a proven calculation based form the findings.

The frame work will consist of the following:

  • Confidence Internal for Population Mean
  • Significance Level P value
  • Hypotheses Test Confidence Interval
  • Probability Test Hypotheses
  • Test Confidence Interval (Product Device)

 Within the frame work steps of processes will be performed Caulcutt (2007) with regards to the Confidence Internal while from an engineering point of view the confidence interval approach is often more meaningful and the hypothesis testing framework is more general and widely use in statistics.

 The hypothesis testing process within the frame work will be concluding to the following:

  1.  Formulate hypothesis
  2. Define the test statistics
  3. Define the critical area
  4. Calculate the test statistic based on your sample
  5. Reach conclusions and state them in English

The above information will give the reader an indication to how the calculations have been controlled and how the frame work has been designed.

 The following sections within the report will indicate each section and give a non engineering description to the true meaning so the non engineering person can understand each area and who it related to the frame work.

Confidence Internal for Population Mean

 The Confidence Internal for Population Mean is a very simple equation it will give a calculation to state that the sample information is correct and the findings are based to an actual confidence level which is normally set at 95%.

 Confidence Interval for the mean based on s:

 [X-t n-1; σ/2*s / n]

 And

 [X*t n-1; σ/2*s / n]

 Taking α = 0.05 (i.e. 95% confidence interval]

 T11-1; 0.05 t distribution table = 1.812

 Mean = 3.12

σ = 1.76

 3.12 – 1.812 * 1.76/ 11, = 2.158

 3.12 + 1.823 * 1.76 / 11 = 4.087                      [2.158, 4.087]  = 3.12

Conclusion

 This can not be rejected.

For the non engineering person the phrase of “this will not be rejected” is a standard terminology with the meaning that the calculation has been accepted and is correct.

 The figure of 3.12 is the mean, as long of this figure is placed between the other two figures of 2.158, 4.087, then this calculation are correct and cannot be rejected.

To follow are two plots indicating the Confidence Internal for Population Mean. 

Significance Level

P value  

A P value test is to establish which type of hypothesis test would be the best to carry out from the findings this is normally from a Null Test (Null meaning; a true statement) or an alternative test. The alternative test is based on what information is known at the time of the findings.

P =       11/30.14   = 0.3649

           U

 P value = 0.3649

α; = 0.025 

 Null test would be preferred for a truer outcome of the findings due to that the null hypothesis is rejected if p < α;

Conclusion 

The Null hypothesis was rejected and an alternative hypothesis test will be carries out.

(Null meaning; a true statement)

The reason why this has been rejected is that the p value of 0.3649 is greater than the

α; = 0.025.

Hypotheses Test

Confidence Interval

A Confidence Interval is what it means it’s a figure that is calculated to give the confidence that the information and conclusions is correct and you are confident of that finding; this is normally set to percentage figure, of 99% confident or 95% confident.

The findings are below.

1. State

U = 34.40

  1. Ha 1 = U = 34..40 ( one sided )

 2. Establish the test Static’s

 t =       x –   U        =          3.12 – 34.40   =    – 2.17

S /  n                     1.76 / 11

3. State the critical region

Appropriate significance level α; = 0.05. (95% Confidence)

Ha1   =    t       x   –   U     =        x   –   34.40        t11-1; 0.05 =         1.812

                          S /  n                 S   /      n

Ha2   =    t       x   –   U     =        x   –   34.40        t11-1; 0.025 =      2.228

                          S /  n                 S   /      n

4. Experiment and calculate the test statistic

XB = 33.40; sB = 1.76 (X = estimated 33.40)

 =       33.40 – 34.40        =     -1.88

          1.76 /     11

5. Conclusions and State them in English

Ha1: t (-1.88) > ‐tcr (-1.812), hence we have to reject Ha1, at 5%, this will bring it within the critical region by 0.072, so we must reject.

Ha2: |t| (1.88) < ta (2.228), hence we cannot reject H0 (Ha2)

Ha2 does not fall within the critical region, so we can accept this at 0.025 (confidence interval)

The above information indicates that the first test was not accepted and the second test was, with a figure 1.182 was in the critical region of 1.88 and the second test was not with a figure of 2.228 both figures where based from the test statistic of 1.88.  

Probability

 A probability test is still part of the frame work it can give a good understanding if extra information is require and therefore calculates the probability.

 A percentage of 34.40 were recorded as secondary paper activities from a facility study of four departments, within the University of Bradford, assuming that a facility study was taken from 10 Departments within the University of Bradford.

 The probability test is as follows:

 A percentage of 34.40 were recorded as secondary paper activities from a facility study of four departments, within the University of Bradford, assuming that a facility study was taken from 10 Departments within the University of Bradford.

 What would the probability be if percentage was as follows?

 A, Stay at 34.40%

 B, Increase to 45%

 C, Increase to 48%

 D, Increase to 50%

 Using the normal approximation to binomial –      Z =        X – n . p

                                                                                                                  n . p ( 1 – p )

 A)       34.40 – 10 x 0.344     = 2.06 Z; 0.4803(1 – 0.4803) = 0.5197; 51.97%

       10 x 0.344 (1- 0.344)

B)       45.00 – 10 x 0.344     = 2.77 Z; 0.4972(1 – 0.4972) = 0.5028; 50.02%

       10 x 0.344 (1- 0.344)

C)       48.00 – 10 x 0.344     = 2.97 Z; 0.4985(1 – 0.4985) = 0.5013; 50.15%

       10 x 0.344 (1- 0.344)

D)       50.00 – 10 x 0.344     = 3.01 Z; 0.4987(1 – 0.4987) = 0.5013; 50.13%

       10 x 0.344 (1- 0.344)

For the normal approximation to binomial to work reasonably well. We require np>5 and n(1-p)>5-, both these conditions are satisfied here (100*0.344 = 34.40>5;100*(1-0.344) = 65.50 >> 5)

 The results show in the Probability Plot below indicating a sudden drop to 50.02% and then the results fluctuating around 50.00% this generally shows a downward trend and was expected and was not trying to shown that the percentage increase would be any different this confirms that if 10 facility studies where taken this could increase to possibility of approximately 50% although the actual Studies would result in a more accurate testing information. 

Confidence Interval (Product Device)

If an electronic product device was designed and manufactured, what confidence interval could we achieve to reach a total eradication of Secondary Paper activities, assuming that the secondary paper activities where based at 50.13% form the findings of the Non E flow activities and E flow activities, in the Secondary Paper activities?

Using the facility study information from Bradford Universities Departments, and designing to use, 11 different software program’s, and assuming to use the existing Facility Study information and the research information acquired form the Confidence Internal for Population Mean and the Probability finding.

 1. State

U = 50.13

Ha 1 = U = 50.13 ( one sided )

 2. Establish the test Static’s

t =       x –   U        =          7.81 – 51.13   =    – 10.54

S /  n                     0.84 / 11

3. State the critical region

Appropriate significance level α; = 0.01. (99% Confidence)

 Ha1   =    t       x   –   U     =        x   –   51.13        t11-1;0.01  =        2.764

                          S /  n                 S   /      n

4. Experiment and calculate the test statistic

XB = 49.50  ; sB = 0.84  ( X = estimated 49.50)

 =      49.50 – 50.13        =     -2.52

          0.84 /     11

5. Conclusions and State them in English

 Ha1: |t| (-2.52) < ta (2.764), hence we cannot reject H0 (Ha1)

Ha1 does not fall within the critical region, so we can accept this at 0.01 (confidence interval at 99% confidence)


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