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In: Statistics and Probability

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 ( Xi- X̅ )2 Region 2 ( Y ) Σ ( Yi- 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 SX = √ ( (Xi - X̅ )2 / n - 1 )
SX = √ ( 3476243.3335 / 15 -1 ) = 498.3

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