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 18.17 kgf/cm2 for the modified m (m = 42) and y = 16.87 kgf/cm2 for the unmodified mortar (n = 30). Let ₁ and 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 ₁1.6 and ₂ 1.3, test Ho: #₁ #₂0 versus H₂: #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.) P-value= State the conclusion in the problem context. Reject Ho. The data suggests that the difference in average tension bond strengths exceeds 0. 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 average tension bond strengths exceeds 0. (b) Compute the probability of a type II error for the test of part (a) when #₁ #2 1. (Round your answer to four decimal places.)
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 18.17 kgf/cm2 for the modified m (m = 42) and y = 16.87 kgf/cm2 for the unmodified mortar (n = 30). Let ₁ and 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 ₁1.6 and ₂ 1.3, test Ho: #₁ #₂0 versus H₂: #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.) P-value= State the conclusion in the problem context. Reject Ho. The data suggests that the difference in average tension bond strengths exceeds 0. 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 average tension bond strengths exceeds 0. (b) Compute the probability of a type II error for the test of part (a) when #₁ #2 1. (Round your answer to four decimal places.)
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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 × = 18.17 kgf/cm² for the modified mortar
42
be the true average tension bond strengths for the modified and unmodified mortars, respectively. Assume that the bond strength distributions are both normal.
= 0 versus H₂: M₁ M₂ > 0 at level 0.01.
(m = 42) and y = 16.87 kgf/cm² for the unmodified mortar (n
=
30). Let and
M1
(a) Assuming that 0₁
= 1.6 and 0₂ 1.3, test Ho: M1
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.)
42
Z =
P-value =
State the conclusion in the problem context.
Reject Ho. The data suggests that the difference in average tension bond strengths exceeds 0.
O 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.
(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 p = 0.10 when μ₁ −μ₂ = 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 01 and
102
were unknown but s₁ = 1.6 and = 1.3?
52
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.
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.
You may need to use the appropriate table in the Appendix of Tables to answer this question.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdf83fd6e-e07d-4ede-9702-55cd0afe0119%2Fbe930c38-23ef-4ee2-8208-099e664e800e%2Ft46a6bd_processed.png&w=3840&q=75)
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 × = 18.17 kgf/cm² for the modified mortar
42
be the true average tension bond strengths for the modified and unmodified mortars, respectively. Assume that the bond strength distributions are both normal.
= 0 versus H₂: M₁ M₂ > 0 at level 0.01.
(m = 42) and y = 16.87 kgf/cm² for the unmodified mortar (n
=
30). Let and
M1
(a) Assuming that 0₁
= 1.6 and 0₂ 1.3, test Ho: M1
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.)
42
Z =
P-value =
State the conclusion in the problem context.
Reject Ho. The data suggests that the difference in average tension bond strengths exceeds 0.
O 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.
(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 p = 0.10 when μ₁ −μ₂ = 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 01 and
102
were unknown but s₁ = 1.6 and = 1.3?
52
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.
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.
You may need to use the appropriate table in the Appendix of Tables to answer this question.
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