|
Current Filter |
New Filter |
Difference |
|
7.6 |
7.3 |
0.3 |
|
5.1 |
7.2 |
-2.1 |
|
10.4 |
6.8 |
3.6 |
|
6.9 |
10.6 |
-3.7 |
|
5.6 |
8.8 |
-3.2 |
|
7.9 |
8.7 |
-0.8 |
|
5.4 |
5.7 |
-0.3 |
|
5.7 |
8.7 |
-3 |
|
5.5 |
8.9 |
-3.4 |
|
5.3 |
7.1 |
-1.8 |
Sample mean of the difference using excel function AVERAGE(),
xÌ…d = -1.4400
Sample standard deviation of the difference using excel function
STDEV.S() sd = 2.2406
Sample size, n = 10
Null and Alternative hypothesis:Â Â
Ho : µd =   0
H1 : µd <   0
Test statistic:Â Â
t = (xÌ…d)/(sd/√n) = (-1.44)/(2.2406/√10) =  Â
-2.0323
df = n-1 = Â Â 9
p-value :Â Â
Left tailed p-value = T.DIST(-2.0323, 9, 1) = Â Â
0.0363
Decision:Â Â
p-value < 0.05, Reject the null hypothesis Â
Conclusion:Â Â
There is enough evidence to conclude that  the new
ï¬ltration system is superior at 0.05 significance level.
