Question

14. Inorganic phosphorous is a naturally occurring element in all plants and animals, with concentrations increasing progressively up the food chain. Geochemical surveys take soil samples to determine phosphorous content in ppm, parts per million). A high phosphorous content may or may not indicate an ancient burial site, food storage site, or even a garbage dump. The Hill of Tara is a very important archaeological site in Ireland. It is by legend the seat of Ireland's ancient high kingst. Independent random samples from two regions in Tara gave the following phosphorous measurements (ppm). Assume the population distributions of phosphorous are mound-shaped and symmetric for these two regions. Region 1: 540 850 790 790 340 810 800 800 890 840 820 540 850 820 640 970 730 720 18 Region II: 520 820 750 740 380 800 820 810 870 820 815 560 820 840 Do we have evidence to conclude that region I is significantly different than region 2? (assume x = 752, S = 151, x2 = 740.4, s2 = 147.0) (Chapter 11) a. Write both null and alternative. Indicate the tail. b. Find test statistics (show formula and final answer with two decimals) c. Find P-value (round to 4 decimal places) d. Write the conclusion in the contest of the problem e. Find a 99% confidence interval for ui - uz (write formula and final answer): Mi - U2 = (-134.7, 157.92) f. Interpret your intervals. There is 99% chance that ui - uz is between -134.7 and 157.92 ce for the given sample data.

Answer 1

Solution

The provided sample means are shown below: X1 = 752 X2 = 740.4 Also, the provided sample standard deviations are: 81 = 151 82 = = 147 and the sample sizes are ni = 18 and n2 = 14. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = 12 Ha: li 12 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 = 30. 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 to = 2.75, for a = 0.01 and df = 30 The rejection region for this two-tailed test is R= {t: t > 2.75} (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: X1 - X t= (n1 -1)+(n2 - 1): (+ a) + V ni+n2 - 2 752 - 740.4 = 0.218 (18–1)1512 +(14–191472 (1s + ) 18+14-2 (4) Decision about the null hypothesis Since it is observed that t = 0.218 <te = 2.75, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p=0.8289, and since p= 0.8289 > 0.01, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that region 1 is significantly different than region 2​, at the 0.01 significance level.

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