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Mathematics & Economics
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Case Study
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Case Study: Cloud Seeding

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Case Study Cloud Seeding to Increase Rainfall ? The data were collected in southern Florida between 1968 and 1972 to test a hypothesis that massive injection of silver iodide into cumulus clouds can lead to increased rainfall. In this experiment, on each of 52 days that were deemed suitable for cloud seeding, a random mechanism was used to decide whether to seed the target cloud on that day or to leave it unseeded as a control. An airplane flew through the cloud in both cases, since the experimenters and the pilot were themselves unaware of whether on any particular day the seeding mechanism in the plane was loaded or not (that is, they were blind to the treatment). Precipitation was measured as the total rain volume falling from the cloud base following the airplane seeding ran, as measured by radar. Did cloud seeding have an effect on rainfall in this experiment? If so, how much? Number of cases: 26 Variable Names: 1. Unseeded_Clouds: Amount of rainfall from unseeded clouds (in acre-feet) 2. Seeded_Clouds: Amount of rainfall from seeded clouds with silver nitrate (in acre-feet) The Data: Unseeded_Clouds Seeded_Clouds 1202.6 2745.6 830.1 1697.8 372.4 1656.0 345.5 978.0 321.2 703.4 244.3 489.1 163.0 430.0 147.8 334.1 95.0 302.8 87.0 274.7 81.2 274.7 68.5 255.0 47.3 242.5 41.1 200.7 36.6 198.6 29.0 129.6 28.6 119.0 26.3 118.3 26.1 115.3 24.4 92.4 21.7 40.6 17.3 32.7 11.5 31.4 4.9 17.5 4.9 7.7 1.0 4.1



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Case Study
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Question
Did cloud seeding have an effect on rainfall in this experiment? If so, how much?
The number of cases is 23
Variable
Unseeded: refers to rainfall from unseeded clouds (in acre-feet)
Seeded: the amount of rainfall from seeded clouds that contain silver nitrates (in acre-feet)
Cloud seeding helps to increase precipitation in dry regions and to reduce heavy rainfall and fog around airports.
Answer
Unseeded

Seeded


185.587

498.361

Mean

289.994

672.663

order

23

23

n

The hypothesis tested is
H0: U1=U2
H1: U1≠U2
The degree of freedom
= (61 2/n1 + 62 2/n2 (2/ C61 2/n1 /n1 )2/n1)
= 1+ (62 2/n2 E2/n2 -1)
The degree of freedom
x=( 289.9942 /23 +672.6632 /23)22a( 289.9942 /23)2 +672.6632 /23)2/23-1
x = 29
the test statistic,
t= (X1 – X2 )/S12n1+S22n2
t = (185.587 – 498.361) 289.994223+672.663223
t = 2.0473
the P.valve is 0.0497[using TDIDT (t value, df)] since the P-value (0.0497) is less than the significant leve...
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