In: Statistics and Probability
An experimental bumper was designed to reduce damage in
low-speed collisions. This bumper was installed on an experimental
group of vans in a large fleet, but not on a control group. At the
end of a trial period, there were 12 repair incidents (a "repair
incident" is an accident that resulted in a repair invoice) for the
experimental group and 9 repair incidents for the control group.
The dollar cost per repair incident is shown below.
Old bumper: | 1,800, | 2,347, | 638, | 1,017, | 1,024, | 2,315, | 1,522, | 1,848, | 1,714 | |||
New bumper: | 1,237, | 990, | 615, | 1,389, | 1,988, | 1,442, | 450, | 385, | 1,834, | 1,458, | 449, | 1,546 |
(a) Perform a two-tailed Mann-Whitney test at
α = .05 and find the z-statistic and
p-value. (Use the formulas in the textbook to
determine the z-statistic and use Excel to calculate the
p-value. If you use MegaStat to check your answers, you
must use version 10.2 Release 2.1 or higher and do NOT correct for
ties or continuity (leave boxes unchecked). Round your
z-statistic to 3 decimal places and p-value to 4
decimal places.)
z-statistic | |
p-value | |
(b) Is there a difference in medians?
No
Yes
(c) Perform a two-tailed parametric t
test for two independent sample means and find the t
statistic and the p-value. (Round your t
answer to 2 decimal places and p-value to 4 decimal
places.)
t | |
p-value | |
(d) Do these values support a conclusion of no
difference in means?
Yes
No
a)
Result 1 - U-value
The U-value is 30. The critical value of U at p < .05 is 26. Therefore, the result is not significant at p < .05.
Result 2 - Z-ratio
The Z-Score is -1.670. The p-value is .09492. The result is not significant at p < .05.
b) No, Because P value is >0.05. Also, Calculated U value is greater than critical value.
C)
Parametric test - t test for the following data
1800,2347,638,1017,1024,2315,1522,1848,1714
1237,990,615,1389,1988,1442,450,385,1834,1458,449,1546,
Significance Level α=0.05 and Two-tailed
test
Step-1:Null Hypothesis
H0 : There is no significant differentiating
between samples
Alternative Hypothesis
H1 : There is significant differentiating between samples
Step-2:
Calculation of S1^2, S2^2
Step 3: Calculate t
t=|x1-x2|/√S21n1+S22n2
=|1580.5556-1148.5833|/√347940.52789+313650.992412
=|431.9723|/√38660.0586+26137.5827
=|431.9723|/√64797.6413
=|431.9723|/254.5538
=1.697
d) Yes
Degree of Freedom
=n1+n2-2=9+9-2=19
df=19,t0.05=2.093
As calculated
t=1.697<2.093
Thank You! Please let me know if
there's any query.