90-786 Intermediate Statistics
Assignment 7
MBS
4.57
(30 points)
a.
E ( X )    128.6
    128.6  11.34
b.
z
c.
x   41  128.6

 7.72

11.34
For a poisson random variable, we have:
k e  
k!
k 0
So, to calculate the given probability we would have to evaluate the
following sum:
n
P ( x  n)  
128.6 k e 128.6
k!
k 0
Which, incidentally, gives P(x122)0.3
The poisson model requires that the probability of the event occuring in a
given unit of time is the same for all units. Thus, if the probability of a
bank failure differed from one year to the next between 1988 and 1994,
the poisson model would be inappropriate. Also, the number of bank
failures in each year must be independent of the number in other years.
122
P( x  122)  
d.
7.75
(20 points)
Because the sample size n is large relative to the population size N, it is necessary
to adjust the standard error of the estimator by a finite population correction
factor:
pˆ 
x 50

 0.694
n 72
ˆ pˆ 
pˆ (1  pˆ ) N  n
.694(1  .694) 251  72

 0.0458
n
N
72
251
Thus a 95% confidence interval is given by:
pˆ  2ˆ pˆ  0.694  2(0.046)  0.694  0.092
Page 1 of 4
90-786 Intermediate Statistics
Assignment 7
Minitab Macro (25 points)
Here is the macro file:
GMACRO
Poisson
#
#
#
#
k1=counter
k2=number of iterations
k3=sample size
k4=lambda
Let k5=1/k4
Do k1=1:k3
Let C3(k1)=0
Enddo
name C2 'Time'
name C3 'Event'
Do k1=1:k2
Random k3 C1;
Exponential k5.
Let C2 = PARS(C1)
Plot C3*C2;
Symbol;
Title "Poisson Process";
ScFrame;
ScAnnotation.
Enddo
ENDMACRO
It is executed with the following commands in the command line editor:
let k2=5
let k3=10
let k4=.83
%d:\90-786\poisson
Page 2 of 4
90-786 Intermediate Statistics
Assignment 7
Here are the plots:
Poisson Process
Poisson Process
0.5
Event
Event
0.5
0.0
-0.5
-0.5
0
10
20
2
3
4
5
6
7
8
Time
Time
Poisson Process
Poisson Process
9
10
11
0.5
Event
0.5
Event
0.0
0.0
-0.5
0.0
-0.5
2
3
4
5
6
7
8
9
10
Time
11
0
5
10
Time
Poisson Process
Event
0.5
0.0
-0.5
0
5
10
Time
Page 3 of 4
90-786 Intermediate Statistics
Assignment 7
Chattergee – Mortgage Rates (25 points)
Here are the descriptive statistics by type:
Descriptive Statistics
Variable
Rate
0=Fixed
0
1
N
14
6
Mean
7.357
4.917
Median
7.313
4.750
TrMean
7.354
4.917
StDev
0.404
0.645
Variable
Rate
0=Fixed
0
1
SE Mean
0.108
0.264
Minimum
6.750
4.250
Maximum
8.000
6.000
Q1
7.062
4.438
Q3
7.594
5.437
The confidence interval is constructed as:
 s 
x  t  / 2 

 n
Where t/2 is based on (n-1) degrees of freedom.
For fixed rate mortgages we have:
t  / 2  2.160
 0.404 
C.I.  7.357  2.160
  7.357  0.233
 14 
For variable rate mortgages:
t  / 2  2.571
 0.645 
C.I.  4.917  2.571
  4.917  0.677
 6 
Page 4 of 4