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Consider the accompanying data on breaking load (kg/25 mm width) for various fabrics in both an unabraded (U) condition and a
Compute the test statistic value. (Round your answer to three decimal places.) State the conclusion in the problem context. O
Consider the accompanying data on breaking load (kg/25 mm width) for various fabrics in both an unabraded (U) condition and an abraded (A) condition. Use the paired t test to test Ho: μ D 0 versus Ha-Po > 0 at significance level 0.01, (Use μο μυΑ.) Note: The data below is formatted such that you can copy and paste it into R Fabric uc36.2, 55.0, 89 43.2, 488, 25.6, 49.9) A- 28.5, 20.0,46.0 34.0, 36.0.52.5 26.5, 46.5) Calculate the mean difference and standard deviation. Compute the test statistic value. (Round your answer to three decimal p-value
Compute the test statistic value. (Round your answer to three decimal places.) State the conclusion in the problem context. O Fail to reject Ho- The data does not suggest a significant mean difference in breaking load for the two fabric load conditions. Fail to reject Ho- The data suggests a significant mean difference in breaking load for the two fabric load conditions. Reject Ho-The data suggests a significant mean difference in breaking load for the two fabric load conditions. 0 Reject H0. The data does not suggest a significant mean difference in breaking load for the two fabric load conditions. You may need to use the t-table to complete this problem
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Answer #1
Sample #1 Sample #2 difference , Di =sample1-sample2 (Di - Dbar)²
36.2 28.5 7.7 0.131
55 20 35 765.214
51.1 46 5.1 5.006
38.9 34 4.9 5.941
43.2 36 7.2 0.019
48.8 52.5 -3.7 121.826
25.6 26.5 -0.9 67.856
49.9 46.5 3.4 15.504
sample 1 sample 2 Di (Di - Dbar)²
sum = 348.7 290 58.7 981.499

mean of difference ,    D̅ =ΣDi / n =   7.338
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    11.841
-----------------------------

std error , SE = Sd / √n =    11.8412   / √   8   =   4.1865     
t-statistic = (D̅ - µd)/SE = (   7.3375   -   0   ) /    4.1865   =   1.753
                          
Degree of freedom, DF=   n - 1 =    7                  
  
p-value =        0.062 [excel function: =t.dist.rt(t-stat,df) ]               
Decision:     p-value>α =0.01, Do not reject null hypothesis                      

conclusion: option A)

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