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Advanced Math/Quantitative analysis for managerial applications

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Question
Hello sir
i need answer for the following questions

1. A bank has a test designed to establish the credit rating of a loan applicant. Of the persons, who default (D), 90% fail the test (F). Of the persons, who will repay the bank (ND), 5% fail the test. Furthermore, it is given that 4% of the population is not worthy of credit; i.e., P(D) = .04. Given that someone failed the test, what is the probability that he actually will default (When given a loan)?

2.   Strength tests carried out on samples of two yarns spun to the same count gave the following results:
  Number in
sample   Sample
Mean   Sample
variance
Yarn A   4   50   42
Yarn B   9   42   56

The strengths are expressed in pounds. Does the difference in mean strengths indicate a real difference in the mean strengths of the yarn?

Answer
1. Pass     0.95  0.05
  No Pass  0.10  0.90
  
Let the chance of passing be P and the changce of not passing is 1-P.
This means that 0.05P + 0.90(1-P) = 0.04.
That works out to be 0.05P + 0.90 - 0.90P = 0.04.
Combining the P's and subtracting 0.90 from both sides gives -0.85P = -0.86.

This means that P = 85/86 of the people will pass, so 1/86 of the people will fail.


2. The strengths are expressed in pounds. Does the difference in mean strengths indicate a real difference in the mean strengths of the yarn?

The t test gives us the value of
t = (A1 - A2)/[S12*root(1/N1 + 1/N2)
A1 = average of set 1
A2 = average of set 2
S12 = sqrt[((n1-1)S1^2 + (n2-1)S2^2)/(n1+n2-2)]
n1 = number in sample 1
n2 = number in sample 2

n1 = 4; A1 = 4*50 = 200; n2 = 9; A2 = 9*42

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