Inorganic phosphorous is a naturally occurring element in all plants and animals, with concentrations increasing progressively...

Inorganic phosphorous is a naturally occurring element in all plants and animals, with concentrations increasing progressively up the food chain (fruit < vegetables < cereals < nuts < corpse). 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. Independent random samples from two regions gave the following phosphorous measurements (in ppm). Assume the distribution of phosphorous is mound-shaped and symmetric for these two regions

Region I: x₁; n₁ = 15
855	1550	1230	875	1080	2330	1850	1860
2340	1080	910	1130	1450	1260	1010
Region II: x₂; n₂ = 14
540	810	790	1230	1770	960	1650	860
890	640	1180	1160	1050	1020

(a) Use a calculator with mean and standard deviation keys to verify that x₁, s₁, x₂, and s₂. (Round your answers to one decimal place.)

x₁	= ppm
s₁	= ppm
x₂	= ppm
s₂	= ppm

(b) Let μ₁ be the population mean for x₁ and let μ₂ be the population mean for x₂. Find a 90% confidence interval for μ₁ − μ₂. (Round your answers to one decimal place.)

lower limit	ppm
upper limit	ppm

Solutions

Expert Solution

	Region 1 ( X )	Σ ( X_i- X̅ )²	Region 2 ( Y )	Σ ( Y_i- Y̅ )²
	855	283378.7423	540	249286.2102
	1550	26460.4553	810	52571.9322
	1230	24753.7673	790	62143.3602
	875	262485.4103	1230	36371.9442
	1080	94453.7573	1770	533943.3882
	2330	888620.5073	960	6286.2222
	1850	214060.4753	1650	372971.9562
	1860	223413.8093	860	32143.3622
	2340	907573.8413	890	22286.2202
	1080	94453.7573	640	159429.0702
	910	227847.0793	1180	19800.5142
	1130	66220.4273	1160	14571.9422
	1450.0	3927.1153	1050	114.7962
	1260	16213.7693	1020	371.9382
	1010	142380.4193	14550	1562292.857
Total	20810	3476243.334

Mean X̅ = Σ Xi / n
X̅ = 20810 / 15 = 1387.3
Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 3476243.3335 / 15 -1 ) = 498.3

Mean Y̅ = ΣYi / n
Y̅ = 14550 / 14 = 1039.3
Sample Standard deviation S_Y = √ ( (Yi - Y̅ )² / n - 1 )
S_Y = √ ( 1562292.8568 / 14 -1) = 346.7

Part b)

Confidence interval :-

DF = 25

t(α/2, DF) = t(0.1 /2, 25 ) = 1.708

Lower Limit =
Lower Limit = 77.2
Upper Limit =
Upper Limit = 618.9
90% Confidence interval is ( 77.2 , 618.9 )

orchestra answered 1 year ago

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