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 themodified 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 distributionsare 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.

As per our guidelines, we are allowed to answer first three sub-parts only. Here part a) itself contains multiple sub-parts. Solving only first three . Please re-post remaining sub-parts. Thanks For Modified Mortar : Mean , x1 = 18.18 Standard deviation, σ1 = 1.6 sample size , n1 = 42 For Un-Modified Mortar : Mean , x2 = 16.86 Standard deviation, σ1 = 1.3 sample size , n1 = 30 Now, Null and Alternative Hypotheses : H0:μ1-μ2 = 0Ha:μ1-μ2 > 0 It is a right-tailed test Decision Rule: Significance level = α=0.01\alpha = 0.01 Critical value = zc=2.33z_c = 2.33 Rejection region = R={z:z>2.326}R = \{z: z > 2.326\} Reject H0 : if R={z:z>2.326}R = \{z: z > 2.326\} a)Test Statistics : z=Xˉ1−Xˉ2σ12/n1+σ22/n2=18.18−16.861.62/42+1.32/30=3.854 \begin{array}{ccl} z & = & \displaystyle \frac{\bar X_1 - \bar X_2}{\sqrt{ {\sigma_1^2/n_1} + {\sigma_2^2/n_2} }} \\\\ \\\\ & = & \displaystyle \frac{ 18.18 - 16.86}{\sqrt{ {1.6^2/42} + {1.3^2/30} }} \\\\ \\\\ & = & 3.854 \end{array} Decision : We noticed that z=3.854>zc=2.326z = 3.854 > z_c = 2.326 Hence null hypothesis is rejected b) P-value using Excel : =NORM.S.DIST(-3.854,TRUE) The p-value : p=0.0001p = 0.0001 c) Conclusion There is enough evidence to claim that the difference in average tension bond strengths exceeds 0 , at the α=0.01\alpha = 0.01 significance level