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 = 18.18 kgf/cm² for the modified mortar (m = 42) and y = 16.89 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 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.

Fail to reject H₀. The data does not suggest that the difference in average tension bond strengths exceeds from 0.

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 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 = 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 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 = 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.

Answer 1

a)

z = 3.767

p-value = 0.0001

since p-value < alpha

we reject the null hypothesis and The data suggests that the difference in average tension bond strengths exceeds 0.

option B) is correct

b)

type ii error = fail to reject the null when the null hypothesis is not true

= P( Z < 2.325- 1 / sqrt(1.6^2/42 + 1.3^2/30))

= P(Z < -0.59496 )

= 0.276

d)

option C)

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