Question

In: Statistics and Probability

An experimental bumper was designed to reduce damage in low-speed collisions. This bumper was installed on...

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

Solutions

Expert Solution

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.


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