The slant shear test is widely accepted for evaluating the bond of resinous repa

Question

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 12 specimens prepared using wire-brushing, the sample mean shear strength (N/mm2) and sample standard deviation were 19.80 and 1.56, respectively, whereas for 12 hand-chiseled specimens, the corresponding values were 23.16 and 5.04. 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 ? 1 for wire-brushing and ?2 for hand- chiseling.) Hai 1- 2 # 0 Hai ??-?2 > 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.) P-value State the conclusion in the problem context. O Reject Ho- The data suggests that the true average strength is not O Reject Ho- The data suggests that the true average strength is different O Fail to reject Ho. The data suggests that the true average strength is not O Fail to reject Ho. The data suggests that the true average strength is not different for the two methods of surface preparation. for the two methods of surface preparation. different for the two methods of surface preparation. different for the two methods of surface preparation. Wha g about the shear strength distributions t are you assumin O It is assumed that variances of distributions for both preparation types are equal O It is assumed that means of distributions for both preparation types are equal O It is assumed that distributions for both preparation types are the same. O It is assumed that distributions for both preparation types are normal.

Explanation / Answer

The statistical software output for this problem is:

Two sample T summary hypothesis test:
?1 : Mean of Population 1
?2 : Mean of Population 2
?1 - ?2 : Difference between two means
H0 : ?1 - ?2 = 0
HA : ?1 - ?2 ? 0
(without pooled variances)

Hypothesis test results:

Hence,

Hypotheses: Option B is correct.

t = -2.21

P-value = 0.0456

Conclusion: Option B is correct.

Assumption: Option D is correct.

Difference Sample Diff. Std. Err. DF T-Stat P-value ?1 - ?2 -3.36 1.5230233 13.08854 -2.2061383 0.0458

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