# Normality Testing Question

Six Sigma – iSixSigma Forums Old Forums General Normality Testing Question

Viewing 3 posts - 1 through 3 (of 3 total)
• Author
Posts
• #28613 Dan Hickey
Participant

I have this question that I need to answer for my Black Belt study. Im confident that question related to normality testing but am confused as to how it can be assessed given the lack of data in question needed to answer it??? Any ideas???
Background.Two underwriters, Adam and Joan, process a total of 200 Mortgage Insurance deals over a 1 week period. The deals are essentially the same apart from size (large or small). A large deal is more complex and takes longer to process. A Six Sigma team has been assigned to analyse the data, before they can – YOU must validate that the data collected is fair and representative.
Results.
Adam. Large deals = 21; Small deals = 94
Joan. Large deals = 27; Small deals = 58
Questions.
1. What type of data are you dealing with here ?
2. What type of statistical test do you propose to use ?
3. State your Hypothesis test in terms of Ho and Ha.
4. Using Minitab, what results do you get (for the p-value and the ) ?
5. Show me how you would calculate, by hand, this test result (not the p-value, the test result).
6. What does that tell you about the data, ie: do you reject Ha and accept Ho; or reject Ho and accept Ha ?

0
#71530 Erik L
Participant

Dan,
Based on the fact that you have a discrete counting scenario and the overall structure of the scenario, here’s my recommendation.  Form a 2×2 contingency table.  You can perform the requisite analysis via Chi-Square.
Regards,
Erik

0
#71535

Hi,
I think i will go with the Erick. For this kind of data it is better to do the contingency table approach. I am not sure whether i understand your question correct or not. I am trying to explain what i understood.
the resutls are
Large deals                small deals         total
Joan                           27(c)                          58(d)                     85
Total                           48                              152                       200
Null Hypothesis : P1 = P2
Aleternat           : P1 P2
predicted values are calculated as
a = 48*115/200 = 27.6          b = 115*152/200 = 87.4
c = 48*85/200 =  20.4           d = 152*85/200   = 64.6
Chi square Ch(cal) = Sum (observed – predicted)^2 /  perdicted
= 4.90
Chi square Ch(critical for df = 1, alpha = 0.05) = 3.85
Ch(cal) > Ch(critical) We reject the null hypothesis and conclude that the proportions of deals are not same for both adam and joan.
hope this helps you
regards
sridhar

0
Viewing 3 posts - 1 through 3 (of 3 total)

The forum ‘General’ is closed to new topics and replies.