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 = 18.18 kgf/cm2 for the modified mortar (m = 42) and y = 16.88 kgf/cm2 for the unmodified mortar (n = 30). Let uz and uz 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 01 = 1.6 and 2 = 1.3, test Ho: 41 - U2 = 0 versus Ha: 41 - U2 > 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. Fail to reject Ho. The data does not suggest that the difference in average tension bond strengths exceeds from 0. • Reject Ho. The data suggests that the difference in average tension bond strengths exceeds o. Fail to reject Ho. The data suggests that the difference in average tension bond strengths exceeds o. Reject Ho. The data does not suggest that the difference in average tension bond strengths exceeds o. (b) Compute the probability of a type II error for the test of part (a) when u - M2 = 1. (Round your answer to four decimal places.) - uy = 1. If m = 42, what value of n is necessary? (Round your answer up to the nearest whole number.) (c) Suppose the investigator decided to use a level 0.05 test and wished B = 0.10 when u n = (d) How would the analysis and conclusion of part (a) change if 04 and 2 were unknown but $1 = 1.6 and s2 = 1.3? Since n = 30 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 = 30 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. Since n = 30 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 = 30 is a large sample, it would be more appropriate to use the t procedure. The appropriate conclusion would follow.

Answer 1

a)

std error of difference σ1-2=				√(σ21/n1+σ22/n2)    =	0.3425
test stat z                                       =				(x1-x2-Δo)/σ1-2     =	3.80
p value : =					0.0001

b)

Hypothesized mean difference true mean difference first sample standard deviation second sample standard deviation first sample size second sample size standard error of mean =0 =(V10,/n1+0,?/n)) for 0.01 level and right tailed test critival value Za rejection region: X >= 40 +Za*ox or type II error =probability of not rejecting B = =nz 1.600 1.300 42.000 30.000 0.3425 2.33 0.7980 A)/ox)) P(Z <(0.798- = P(Z <-0.59) 0.2776

type II error =0.2776

c)

n=109