Question

Please Help!!!

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 30°. An article reported that for 10 specimens prepared using wire- brushing, the sample mean shear strength (N/mm2) and sample standard deviation were 18.10 and 1.59, respectively, whereas for 10 hand-chiseled specimens, the corresponding values were 23.17 and 5.02. Does the true average strength appear to be different for the two methods of surface preparation? State and test the relevant hypotheses using a significance level of 0.05. (Use u1 for wire-brushing and up for hand-chiseling.) O Ho: M1 - M2 = 0 Ha: ui - U2 SO O Ho: M1 - M2 = 0 Ha: U1 - 4200 O Ho: M1 - U2 = 0 Ha: ui – u2 < 0 O Ho: Mi - U2 = 0 Ha: ui - U2 > 0 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. Reject Ho. 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 different for the two methods of surface preparation. Fail to reject Ho. The data suggests that the true average strength is not different for the two methods of surface preparation. Fail to reject Ho. The data suggests that the true average strength is not different for the two methods of surface preparation. What are you assuming about the shear strength distributions? It is assumed that means of distributions for both preparation types are equal. It is assumed that variances of distributions for both preparation types are equal. It is assumed that distributions for both preparation types are the same. It is assumed that distributions for both preparation types are normal.

Thank you!!!

Answer 1

Here we have: X 1 = 18.10 X2 = 23.17 S1 = 1.59 S2 = 5.02 n1 = 10 n2 = 10 a = 0.05 Hypothesis: Correct option "B" Ho; M1 – U2 = 0 Hai H1 - H270 Test statistics: t-distribution (o unknown)

(5 – #z) – (My – uz) = + S2 M ni II = -3.05 (18.10–23.17)-(0) 5.022 + 10 10 1.592 + sz n2 ni df = 2 Si ni $2 n2 + n - 1 n2 - 1

df 1.592 5.022 + -)2 10 10 1.592 -)2 2 10 + 10 10-1 10-1 = 11 5.022 P-value = 2p (t>/-3.05) = 0.011 Correct option "B"

Reject Ho. The data suggest that the true average strength is different for the two methods of surface preparation Correct option "D" It is assume that distribution for both preparation types are normal