d) Test to see if the means of the two samples differ significantly on the assumption that your answers to b)
and c) showed equal variances and Normal distributions. Use a test ratio, critical value or a confidence
interval (4) or all three (6). Your answers to all three should be almost identical.
From our computations of the variance x`  25.2718 and s12  177.724 . From the computer printout
x 2  17.73 s 22  9.112  82 .992 , n1  11 and n2  15 .   .10 .
H 0 : 1   2
H 1 : 1   2
d  x1  x 2  25.27 17.73  7.54 .
If we assume that the variances are equal
n  1s12  n2  1s 22 10177 .724   1482.992 

s p2  1

 122 .464 , so that
n1  n 2  2
24
 1
1 
11 
1 1
 15
  122 .464     122 .464 
s d2  sˆ 2p  

  122 .464 0.157575   19 .297 and
n
n
11
15
165
165




2 
 1
d  D0
7.54
1 
  1
  19.297  4.32928 . t 

 1.716 and
s d  s p2  
sd
4.32928
 n1 n 2 
df  n1  n 2  2  11  15  2  24 .
24
Make a diagram: Show an almost Normal curve with a center at zero and critical values at t .05
 1.711
24
and  t .05
 1.711 . Since the computed value of t is between these, do not reject the null hypothesis.

24
s d  7.54  1.7114.32928   7.54  7.41 . Since
Confidence Interval: D  d  t  2 s d  7.54  t 05

7.41 is smaller than 7.54, the interval does not include zero.
d  D0
d  D0
7.54
t

 1.716 Make a diagram: Show an almost
Test Ratio: t 
sd
4.32928
sd
24
24
 1.711 and  t .05
 1.711 . Since
Normal curve with a center at zero and critical values at t .05

the computed value of t is not between these, reject the null hypothesis.
Critical Value: d cv  D0  t  2 s d  7.41 . Make a diagram: Show an almost Normal curve with
a center at zero and critical values at 7.41 and -7.41. Since the computed value of
d  x1  x 2  7.54 is not between these critical values, reject the null hypothesis.
--------------------------------------------------------------------------------------------------------If we do not assume equal variances, use the following worksheet
sx21 
s12 1383 .84

 60 .1670
n1
23
s x22 
s 22 1980 .25

 165 .021
n2
12
s d2 
sd 
s12 s 22

n1 n 2
 225 .188
s12 s 22

 225.188  15.0063
n1 n 2
2
 s12 
 
 n1 
60 .1670 2
 

 164 .548
n1  1
22
2
 s12 
 
 n1 
165 .021 2
 

 2475 .63
n1  1
11


  s2 s2 2
  1  2 
  n1 n 2 
df  
2
2
  s2 
 s 22 
 
 1 
 n2 
  n1 
 


n2 1
 n1  1




2

s d2


225 .188 2

  19 .2069


2
2


2
2
164
.
548

2475
.
63
 sx


s x2
1


 n1 1 n 2  1


 
   
d  D0
 27 .6

 1.839 Make a diagram: Show
sd
15 .0063
19
19
an almost Normal curve with a center at zero and critical values at t .025
 2.093 . Since
 2.093 and  t .025
Round this down and use 19 degrees of freedom. t 
the computed value of t is between these, do not reject the null hypothesis.
2