Answer:
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 <u>Two-sample z test statistics</u> 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, <u><em>test statistics</em></u> =
= <u>3.74</u>
<u>Now, P-value is given by the following formula;</u>
P-value = P(Z > 3.74) = 1 - P(Z 3.74)
= 1 - 0.9999 = <u>0.0001</u>
<em>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.</em>