An experiment to compare the tension bond strength of polymer latex modified mor

Question

An experiment to compare the tension bond strength of polymer latex modified mortar (Portland cement mortar to which polymer latex emulsions have been added during mixing) to that of unmodified mortar resulted in x = 18.11 kgf/cm2for the modified mortar (m = 42) and y = 16.89kgf/cm2 for the unmodified mortar (n = 31). Let 1and 2 be the true average tension bond strengths for the modified and unmodified mortars, respectively. Assume that the bond strength distributions are both normal.

(a) Assuming that 1 = 1.6 and 2 = 1.3, testH0: 1  2 = 0 versus Ha: 1  2 > 0 at level 0.01. 
Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) 

z = 
P-value = 

State the conclusion in the problem context.

Reject H0. The data does not suggest that the difference in average tension bond strengths exceeds 0.
Fail to reject H0. The data suggests that the difference in average tension bond strengths exceeds 0.    
Reject H0. The data suggests that the difference in average tension bond strengths exceeds 0.
Fail to reject H0. The data does not suggest that the difference in average tension bond strengths exceeds from 0.

(b) How would the analysis and conclusion of part (a) change if 1 and 2 were unknown buts1 = 1.6 and s2 = 1.3?

Since n = 31 is a large sample, it would no longer be appropriate to use the large sample test. Any other test can be used, and the conclusions would stay the same.Since n = 31 is a large sample, it would be more appropriate to use the t procedure. The appropriate conclusion would follow.    Since n = 31 is not a large sample, it still be appropriate to use the large sample test. The analysis and conclusions would stay the same.Since n = 31 is not a large sample, it would no longer be appropriate to use the large sample test. A small sample t procedure should be used, and the appropriate conclusion would follow.

Explanation / Answer

Part-a

Test statistic Z=(xbar1-xbar2)/sqrt(12/n1+22/n2)

=(18.11-16.89)/sqrt(1.6*1.6/42+1.3*1.3/31)

=3.59

p-value=P(Z>3.59)=0.0002 using excel function =1-NORMSDIST(3.59)

Reject H0. The data suggests that the difference in average tension bond strengths exceeds 0.

Part-b

Since n = 31 is a large sample, it would be more appropriate to use the t procedure. The appropriate conclusion would follow.

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