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/cm for the modified mortar (m = 42) and y = 16.86 kgf/cm² for the unmodified mortar (n = 30). Let , and z 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 e- 1.6 and ez- 1.3, test Ho: H1 - H2 -0 versus Ha: - 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.) 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 does not suggest that the difference in average tension bond strengths exceeds 0. Fail to reject Ho: The data suggests that the difference in average tension bond strengths exceeds 0. Reject Ho. The data suggests that the difference in average tension bond strengths exceeds 0.

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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 = 16.86 kgf/cm for the unmodified mortar (n = 30). Let µ1 and Hz 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 o1 = 1.6 and o2 = 1.3, test Ho: µ1 - 42 = 0 versus H3: µ1 – 42 > 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.
o Reject Ho: The data does not suggest that the difference in average tension bond strengths exceeds 0.
Fail to reject Ho: The data suggests that the difference in average tension bond strengths exceeds 0.
o Reject Ho. The data suggests that the difference in average tension bond strengths exceeds 0.
(b) Compute the probability of a type II error for the test of part (a) when µ1 - µ2 = 1. (Round your answer to four decimal places.)
(c) Suppose the investigator decided to use a level 0.05 test and wished ß = 0.10 when u1 - 2 = 1. If m = 42, what value of n is necessary? (Round your answer up to the nearest whole number.)
n =
(d) How would the analysis and conclusion of part (a) change if o, and oz were unknown but s, = 1.6 and s, = 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 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.
o 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.
Transcribed Image Text: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 = 16.86 kgf/cm for the unmodified mortar (n = 30). Let µ1 and Hz 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 o1 = 1.6 and o2 = 1.3, test Ho: µ1 - 42 = 0 versus H3: µ1 – 42 > 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. o Reject Ho: The data does not suggest that the difference in average tension bond strengths exceeds 0. Fail to reject Ho: The data suggests that the difference in average tension bond strengths exceeds 0. o Reject Ho. The data suggests that the difference in average tension bond strengths exceeds 0. (b) Compute the probability of a type II error for the test of part (a) when µ1 - µ2 = 1. (Round your answer to four decimal places.) (c) Suppose the investigator decided to use a level 0.05 test and wished ß = 0.10 when u1 - 2 = 1. If m = 42, what value of n is necessary? (Round your answer up to the nearest whole number.) n = (d) How would the analysis and conclusion of part (a) change if o, and oz were unknown but s, = 1.6 and s, = 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 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. o 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.
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