Question

The accompanying table gives summary data on cube compressive strength (N/mm2) for concrete specimens made with a pulverized fuel-ash mix. Age Sample Sample Sample (days) Size Mean 7 69 26.97 4.85 71 36.73 6.44 SD 28 Calculate a 99% CI for the difference between true average 7-day strength and true average 28-day strength. (Use Ho: My - 28 -0. Round your answers to two decimal places.) 1) N/mm² Interpret your 99%. CI. With 99% confidence, that the true difference fails above the upper bound. With 99% confidence, we can say that the true difference falls between these values. With 99% confidence that the true difference falls below the lower bound. With 99% confidence, that the true difference falls outside these values. . You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

The 99% confidence interval for the difference in ture mean of 7-day strength and true mean of 28-days strength is −12.282<μ1​−μ2​<−7.238.
The provided sample means are shown below: X = 26.97 X2 = 36.73 Also, the provided sample standard deviations are: Si = - 4.85 S2 = 6.44 and the sample sizes are ni = 69 and n2 = 71. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = H2 Ha: Mi 7 u2 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 = 0.01, and the degrees of freedom are df = 138. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:


= Hence, it is found that the critical value for this two-tailed test is te = 2.612, for a = 0.01 and df 138. The rejection region for this two-tailed test is R = {t: \t> 2.612}. (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: t= X1 - X2 (ni-1) si +(n2-1)s (m + ) ni+n2 - 2 26.97 – 36.73 = – 10.107 (69–1)4.852 +(71-1)6.442 (60 + ) 69+71-2 (4) Decision about the null hypothesis Since it is observed that t = 10.107 > to = 2.612, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p= 0, and since p= 0 <0.01, 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 0.01 significance level. Confidence Interval The 99% confidence interval is – 12.282 <H1 – M2 < -7.238.