Analysis of a random sample consisting of m = 20 specimens of cold rolled steel

Question

Analysis of a random sample consisting of m = 20 specimens of cold rolled steel to determine yield strengths resulted in a sample average strength of x - 29.6 ksi. A second random sample of n = 25 two sided galvanized steel specimens gave a sample average strength of y = 34.4 ksi. Assuming that the two yield strength distributions are normal with sigma_1 = 4.0 and sigma_2 = 5.0 +, does the data indicate that the corresponding true average yield strengths mu_1 and mu_2 are different? Let's carry out a test at significance level alpha = 0.01. The parameter of interact is mu_1 = mu_12, the difference between the true average strengths for the two typos or stool. The null hypothesis is mu_1 - mu_2 The alternative hypothesis is H_a: = mu_1 0; If H_a is true, then mu_1 and mu_2 are different. With Delta_0 = 0, the test statistic value is z = x - y/Squareroot sigma_1^2/m + sigma_2^2/n Substituting m =, x = 29.6, sigma_2 = 16.0, n = 25, y =, and sigma_2^2 = 25.0 into the formula for z yields the following. z = 29.6-34.4/Squareroot 16.0/20 + 25.0/25 = -4.9/1.34 - (rounded to two decimal places) That is, the observed value of x - y is more than 3 standard deviation below what would be expected were H_0 true. The notequalto equality in implies that test is appropriate. The P-value is 2(1 = Phi(3.58)] alpha 2(0) = 0 (software gives 0.00035). Since P-value = 0 lessthanorequalto 0.01 = a, H_0 is therefore at level 0.01 in favor of the conclusion that mu_1 notequalto mu_2. In fact, with a P-value this small, the null hypothesis would be rejected at any sensible significance level. The sample data strongly suggest that the true average yield strength for cold-rolled steel differs from that for galvanized steel. You may need to use the appropriate table in the Appendix of Tables to answer this question.

Explanation / Answer

2-Null hypothesis is H0: µ1-µ2=0

3-Alternative hypothesis is Ha: µ1-µ20

5-Substituting m=20,ybar=34.4 we get z=-3.58

6-The inequality in Ha implies that two tailed test is appropriate.

7-Since p-value<=0.01, H0 is therefore rejected

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