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WITHOUT THE USE OF TECHNOLOGY. SHOW ALL STPES PLEASE.

Question 3.Use Data from Textbook Question 2.37, test the hypothesis that mean deflection temperatures are same for the two different formulations under the assumption that variances are different for thee two underlying populations. DO NOT USE TECHNOLOGY 2.37. In semiconductor manufacturing wet chemical etching is often used to remove silicon from the backs of wafers prior to metalization. The etch rate is an important characteristic of this process. Two different etching solutions are being evaluated. Eight randomly selected wafers have been etched in each solution, and the observed etch rates (in mils/min) are as follows. Solution 1 Solution2 9.9 10.6 4 10.3 9.3 9.8 10.2 10.0 10.7 10.5 10.6 10.2 10.4 10.3 10.0 10.3

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Answer #1

Unknown But Unequal Variances There ae situations where the analyst is not able to assume that q--σ2. Recall from Section 9.8 that, if the populations are normal, the statistic T 1-)-do has an approximate t-distribution with approximate degrees of freedom As a result, the test procedure is to not reject Ho when with v given as above. Again, as in the case of the pooled t-test, one-sided alter natives suggest one-sided critical regions.

s1 = 0.4504

s2 =3.1627

an The following null and alternative hypotheses need to be tested This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used (2) Rejection Region Based on the information provided, the significance level is 0.05, and the degrees of freedom are df 7.284. In fact, the degrees of freedom are computed as follows, assuming that the population variances are unequal: Hence, it is found that the critical value for this two-tailed test is tc-2.346, for 7.284. 0.05 and df The rejection region for this two-tailed test is R t: t2.346) (3 Since it is assumed that the population variances are unequal, the t-statistic is computed as follows 112 9.95-9.2 0.664 8-1)0.4504+(8-1)3.16271 8+8 Since it is observed th0.664e 2.346, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.5271, and since p = 0.5271 > 0.05, it is concluded that the null hypothesis is not rejected It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ| is different than μ , at the 0.05 significance level.

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