ASK YOUR TEACHER 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 = 18.11 kgf/cm2 for the modified mortar (m = 42) and y = 16.83 kgf/cm2 for the unmodified mortar (n = 30). 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 Ha: 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.)

Test statistics = 3.74

P-value = 0.0001

Step-by-step explanation:

We are given that an experiment to compare the tension bond strength of polymer latex modified mortar to that of unmodified mortar resulted in x = 18.11 kg f/ for the modified mortar (m = 42) and y = 16.83 kg f/ for the unmodified mortar (n = 30).

Assume that the bond strength distributions are both normal and assuming that σ1 = 1.6 and σ2 = 1.3.

Let = true average tension bond strengths for the modified mortars

      = true average tension bond strengths for the unmodified mortars

So, Null Hypothesis, :  or      {means that true average tension bond strengths for the modified and unmodified mortars are same}

Alternate Hypothesis, :  or      {means that the true average tension bond strengths for the modified mortars is greater than that for unmodified mortars}

The test statistics that will be used here is Two-sample z test statistics as we know about population standard deviations;

               T.S.  =    ~ N(0,1)

where, x = sample mean tension bond strengths for the modified mortars = 18.11 kg f/

         y = sample mean tension bond strengths for the unmodified mortars = 16.83 kg f/

          = population standard deviation for modified mortars = 1.6

          = population standard deviation for unmodified mortars = 1.3

         m = sample of modified mortars = 42

         n = sample of unmodified mortars = 30

So, test statistics  =  

                               =  3.74

Now, P-value is given by the following formula;

         P-value = P(Z > 3.74) = 1 - P(Z 3.74)  

                                            = 1 - 0.9999 = 0.0001

Here, the above probability is calculated by looking at the value of x = 3.74 in the z table which gives an area of 0.9999.

the answer to the expression is 83