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 = 18.15 kgf/cm² for the modified mortar (m = 42) and y = 16.82 kgf/cm² for the unmodified mortar (n = 31). Let μ₁ and μ₂ 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 σ₁ = 1.6 and σ₂ = 1.3, test H₀: μ₁ − μ₂ = 0 versus H_a: μ₁ − μ₂ > 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 suggests that the difference in average tension bond strengths exceeds 0.Reject H₀. The data does not suggest 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.Fail to reject H₀. 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 β = 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 σ₁ and σ₂ were unknown but s₁ = 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 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.Since n = 31 is a large sample, it would be more appropriate to use the t procedure. The appropriate conclusion would follow.

Answer 1

(a)

Test is right tailed. The test statistics is

The p-value using excel function "=1-NORMSDIST(3.91)" is 0.0000

Conclusion:

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

(b)

First we need to find the critical value of difference for which we will reject the null hypothesis.

The critical value of z for which we will reject the null hypothesis is 2.33. So

Now we need to find the z-score such that and so

The test is right tailed so the type II error is

(c)

The critical value of z for which we will reject the null hypothesis for is 1.645. So

For , the z-score is -1.28 so

(d)

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