An experiment to compare the tension bond strength of polymer latex modified mor

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 bar = 18.14 kgf/cm^2 for the modified mortar (m = 42) and y bar = 16.86 kgf/cm^2 for the unmodified mortar (n = 32). Let mu_1 and mu_2 be the true average tension bond strengths for the modified and unmodified mortars, respectively. Assume that the bond strength distributions are both normal. Assuming that sigma_1 = 1.6 and sigma_2 = 1.3 test H_0: mu_1 - mu_2 = 0 versus H_a: mu_1 - mu_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.) z = _ P-value = ____ State the conclusion in the problem context. Reject H_0. The data suggests that the difference in average tension bond strengths exceeds 0. Reject H_0. The data does not suggest that the difference in average tension bond strengths exceeds 0. Fail to reject H_0. The data does not suggest that the difference in average tension bond strengths exceeds from 0. Fail to reject H_0. The data suggests that the difference in average tension bond strengths exceeds 0. Compute the probability of a type I error for the test of part (a) when mu_1 - mu_2 = 1. (Round your answer to four decimal places.) _ Suppose the investigator decided to use a level 0.05 test and wished beta = 0.10 when mu_1 - mu_2 = 1. If m = 42 what value of n is necessary? (Round your answer up to the nearest whole number.) n = _______________

Explanation / Answer

(a) xbar = 18.14 kgf/cm2 , Ybar = 16.86 kgf/cm2

alpha = 0.01

Standard error of sampling se0 =sqrt [12 /n1 + 22 /n2] = sqrt [1.62 /42 + 1.32 /32] = 0.3373

z = (xbar - Ybar)/se0 = (18.14 - 16.86)/ 0.3373 = 3.80

P - value as it is directional test => P - value = 0.000072

Zcritical = 2.33

(b) If 1 -2 =1

Type II error occurs when there we failed to reject the null hypothesis, even if it is true.

so the critical range below which null hypothesis is true is 0 + 2.33* 0.3373 = 0.786

so The region in which we will accept the null hypothesis is where Pr(1 -2 <= 0.786 ; 1; 0.3373)

Z value = (0.786 - 1)/ 0.3373 = -0.6344

so Probability of type II error = (-0.6344) = 0.263

(c) Here = 0.10 and alpha = 0.05

so we have to calculate standard error of sampling

as = 0.10 then Pr( 1 -2 <= x ; 1;se0 ) = 0.10

from Z - table relative Z - value = -1.28

(x -1)/se0 = -1.28

here x = 1.645 (critical z at 0.05 one directional) * se0

1.645 se0 -1 = -1.28 se0

2.925 se0 =1

se0 = 0.342

sqrt [ 1.62 /42 + 1.32 /n] = 0.342

1.62 /42 + 1.32 /n = 0.11

n =30.22 or say n = 30

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