Question

Question 38 2 pts 15. 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 kings. 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. Regioni: 540 850 790 790 340 810 800 800 890 8401820 540 850 820 640 970 730 720 Region II: 520 820 750 740 380 800 820 810 870 820 815 560 820 840 Assume). Do we have evidence to conclude that region I is significantly more than region 2? Use a 5% level of significance.

Answer 1

Let X1: Phosphorus content for region 1

X2: Phosphorus content for region 2

X χ2 540 291600 850 722500 790 624100 790 624100 340 115600 810 656100 800 640000 800 640000 890 792100 840 705600 820 672400 540 291600 850 722500 820 672400 640 409600 970 940900 730 532900 720 518400 Σx = 13540 Σx2 = 10572400 ΣΧ Mean x = n 540 + 850 + 790 + 790 + 340 + 810 + 800 + 800 + 890 + 840 + 820 + 540 + 850 + 820 + 640 + 970 + 730 + 720 18 13540 18 = 752.2222 (Σx)2 (Σχ2 n Sample Standard deviation S = η -1 (13540)2 10572400. 18 17 10572400 - 10185088.8889 17 1387311.1111 17 = 22783.0065 = 150.9404

Σx: = 10365 Σx2 = 7950925 ΣΧ Mean i n 520 + 820 + 750 + 740 + 380 + 800 + 820 + 810 + 870 + 820 + 815 + 560 + 820 + 840 14 10365 14 = 740.3571 Σχ2. (Σx) 2 n Sample Standard deviation S = η - 1 (10365)2 (7950925. 14 13 (7950925 - 7673801.7857 13 1277123.2143 13 Ξ * V21317.1703 = 146.004

The provided sample means are shown below: X1 = 752.2222 X2 = 740.3571 Also, the provided sample standard deviations are: Si = 150.9404 S2 = 146.004 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: M1 = M2 Ha: U1 > U2 This corresponds to a right-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 = 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 right-tailed test is te = 1.697, for a = 0.05 and df = 30. The rejection region for this right-tailed test is R= {t:t > 1.697}. (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: t= X; - X2 (n1–1)s +(n2-1)s ni+n2-2 + ni n2 752.2222 – 740.3571 = = 0.22 (18-1)150.94042+(14-1)146.0042 18+14-2 ( + )

(4) Decision about the null hypothesis Since it is observed that t = 0.224 <tc = 1.697, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p= 0.4122, and since p = 0.4122 > 0.05, 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 the population mean Mi is greater than M2, at the 0.05 significance level. Confidence Interval The 95% confidence interval is –96.441 < Mi – M2 < 120.171.
Thus there is not sufficient evidence to conclude that phosphorus content for region 1 is more than region 2.

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