Question

- vuestion3763015 30 TEA M YouTube on MS R Alication Status d eo 2264430 tamity Start o 30. 2/8 points Previous Answers BOUNDERSTAT12 7.4.012. I My Notes | Ask Your Teacher v Inorganic phosphorous is a naturally occurring in plants and anals with concentrations increasing progressively up the food chain ruit vegetables corps). Geochemical surveys take soil samples to determine phosphorous content in pom parts per million A hhor cont a ancient burlaste, food storage site, even a garbage dump Independend o gave the shophous measurements in ppm) Authe n ophorous mound-shaped and w ere for these two regions 1560 Region IX , 15 1550 120 25 30 23:40 1000 9101130 1450 1260 1010 Region , 14 10790 12:30 170 60 60 890 40 1180 1150 1050 1020 (a) Usator with me and Рpr ppm 60 to ha n d you wo ) ppm (0) Let te the population mean for, and let y be the population me confidence interval for H, M. (Round your answers to ano decimal place) At the eve of condence region more rest than the other from a chemic pretive Because the interval contano et positive and negative numbers, we can not say that and regions more interesting than the other

Answer 1

Solution: We need to construct the 95% confidence interval for the difference between the population means Hi-Ho, for the case that the population standard deviations are not known. The following information has been provided about each of the samples: Sample Mean 1 (X) = 1387.3 Sample Standard Deviation 1 (81) = 498.3 15 1039.3 Sample Size 1 (N) = Sample Mean 2 (X2) = Sample Standard Deviation 2 (82) = Sample Size 2 (N2) = 346.7 Based on the information provided, we assume that the population variances are equal, so then the number of degrees of freedom are df = n1 + n2 - 2 = 15+ 14- 2 = 27. The critical value for a = 0.05 and df = 27 degrees of freedom is te = ta/2:1-1 = 2.052. The corresponding confidence interval is computed as shown below: Since the population variances are assumed to be equal, we need to compute the pooled standard deviation, as follows: p = /(ni - 1)si + (n2 - 1)s n + n2- 2 (15 - 1) x 498.32 + (14 - 1) × 346.72 15 + 14 - 2 = 432 = 1 Since we assume that the population variances are equal, the standard error is computed as follows: 11 se = Sp V n. 12 = 1 432 x V 15 1 14 = 160.536 Now, we finally compute the confidence interval: CI = (X1 - X- tex se, X-X+te x se) = (1387.3 - 1039.3 - 2.052 x 160.536, 1387.3 - 1039.3 +2.052 x 160.536) = (18.607, 677.393) Therefore, based on the data provided the 95% confidence interval for the difference between the population means Hi-u2 is 18.607CM - Mg 677.393, which indicates that we are 95% confident that the true difference between population means is contained by the interval (18.607, 677.393).

Because the interval contains only positive numbers, we can say that region I is more interesting than region II.