Statistics ,  Example of Two sample procedure, by hand.

10 Polyester cloth samples are buried in a landfill, in marked locations .   A random 5 are dug up after 2 weeks, the other 5 after 16 weeks.
To compare the mean breaking strength (Does the cloth decay?)
2 wks|  |16 wks      Breaking    2 wks: 118, 126, 126, 120, 129
     | 9|8           strength:  16 wks: 124, 98, 110, 140, 110 pounds
     |10|                       
    8|11|00                     "Normal" seems reasonable for both samples.  Spreads are different.
 9660|12|4
     |13|                         (Doesn't look like evidence for decay is very strong.  Do the math...)
     |14|0
 
To do it by hand:  (Find SE  OR  Sample variance, 
not both.)
Group
Popu-
lation
Sample 
size
Sample
mean
Sample 
standard dev.
Standard Error 
of the mean = SE 
Sample 
variance
 (SE)2 
2 wks 
 1 
 n1 = 5 
xbar1= 123.8
 s1 = 4.6043
s1/sqrt(n1) =2.0591
s12 =21.1996
s12/n1 =4.2399
16 wks 
2 
n2 = 5 
xbar2= 116.4
 s2 = 16.0873
s2/sqrt(n2) =7.1944
s22 =258.8012
s22/n2 =51.7602
.
    
 diff = 7.4
      
sum = 56.0001
SE diff  = sqrt(56.0001) = 7.4833
                                                  For hand work, d.f. = smaller of (5-1) and (5-1) = 4
   H0: µ1 - µ2 = 0, same as µ1 = µ2 , "no difference"
   Ha: µ1 - µ2 > 0, same as µ1 > µ2 , shorter burial gives stronger cloth.
   diff is in the correct direction, but how strong is the evidence?
t = diff/SE diff = 7.4 / 7.4833 = .989
                Find t(4) in the t-table.   .989 is between .941 and 1.190,  P-value is between .15 and .20

CI:  (xbar1 - xbar2) +t*.SEdiff   , 
                          diff +t*.SEdiff        For 90% CI, t*(4) = 2.132
                    7.4 + 2.132 .7.4833
            7.4 + 15.96:     -8.56   to 23.36

Sievers home  Math251-Fall05/twosampexample.htm 
11/18/05
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