The slant shear test is widely accepted for evaluating the b

The slant shear test is widely accepted for evaluating the bond of resinous repair materials to concrete; it utilizes cylinder specimens made of two identical halves bonded at 30degree. An article reported that for 12 specimens prepared using wire-brushing, the sample mean shear strength (N/mm^2) and sample standard deviation were 19.10 and 1.52, respectively, whereas for 12 hand-chiseled specimens, the corresponding values were 22.13 and 4.03. Does the true average strength appear to be different for th two methods of surface preparation? State and test the relevant hypotheses using a significance level of 0.05. (Use Mu1 for wire-brushing and Mu2 for hand-chiseling.) Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.) t= P-value = State the conclusion in the problem context. Fail to reject H0. The data suggests that the true average strength is not different for the two methods of surface preparation. Fail to reject H0. The data suggests that the true average strength is not different for the two methods of surface preparation. Reject Ho. The data suggests that the true average strength is not different for the two methods of surface preparation. Reject H0. The data suggests that the true average strength is different for the two methods of surface preparation. What are you assuming about the shear strength distributions? It is assumed that variances of distributions for both preparation types are equal. Li It is assumed that means of distributions for both preparation types are equal. It is assumed that distributions for both preparation types are normal. ) It is assumed that distributions for both preparation types are the same.

Solution

Set Up Hypothesis
Null Hypothesis , There Is No-Significance between them Ho: u1 = u2
Alternate Hypothesis, There Is Significance between them - H1: u1 != u2
Test Statistic
X(Mean)=19.1
Standard Deviation(s.d1)=1.52 ; Number(n1)=12
Y(Mean)=22.13
Standard Deviation(s.d2)=4.03; Number(n2)=12
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =19.1-22.13/Sqrt((2.3104/12)+(16.2409/12))
to =-2.44
| to | =2.44
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 11 d.f is 2.201
We got |to| = 2.43695 & | t | = 2.201
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != -2.4369 ) = 0.033
Hence Value of P0.05 > 0.033,Here we Reject Ho

ANS.
Ho: u1 = u2,H1: u1 != u2
to =-2.44
( P != -2.4369 ) = 0.033
Rejecct Ho, average strength is diffrent for the two methods of surface proportion
Assumed that both means are equal