Question

The slant shear test is widely accepted for evaluating the bond of resinous repair materials to concrete; it utilizes cylinder specimens made of two identical halves bonded at 30°. An article reported that for 12 specimens prepared using wire-brushing the sample mean shear strength (N/mm') and sample standard deviation were 19.60 and 1.59, respectively, whereas for 12 hand-chiseled specimens, the corresponding values were 23.13 and 5.09. Does the true average strength appear to be different for the two methods of surface preparation State and test the relevant hypotheses using a significance level of 0.05. (Use M for wire-brushing and H2 for hand-chiseling.) Mon -- 0 Mini-SO - M: --- Calculate the test statistic and determine the value. (Round your test statistic to two decimal places and your value to three decimal places.) Pevalue 2:51 PM 7/30/2020

Answer 1

The provided sample means are shown below: X1 19.6 X2 = 23.13 Also, the provided sample standard deviations are: 81 = 1.59 S2 = 5.09 and the sample sizes are ni = 12 and n2 = 12. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: M1 = 42 Ha: Mi 7 42 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 a = 0.05, and the degrees of freedom are df = 13.127. 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 te = 2.158, for a = 0.05 and df 13.127. The rejection region for this two-tailed test is R = {t: 1t| > 2.158}. (3) Test Statistics Since it is assumed that the population variances are unequal, the t-statistic is computed as follows: X1 - 72 t= sa n2 ni 19.6 – 23.13 = -2.293 1.592 12 + 5.092 12 (4) Decision about the null hypothesis Since it is observed that \t\ = 2.293 > tc = 2.158, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p= 0.039, and since p = 0.039 0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean Mi is different than 42, at the 0.05 significance level.

please like ??