Cloud seeding is used to increase precipitation by dispersing substances (silver

Question

Cloud seeding is used to increase precipitation by dispersing substances (silver iodide, dry ice, table salt) into the air that serve as cloud condensation. Simpson (1975, Technometrics) reported the results on 26 seeded and 26 unseeded clouds in order of the amount of rainfall (in acre-feet), largest amount first. Here are 2 possible tests to study the question of whether cloud seeding work: Paired t-Test of mu(Seeded - Unseeded) T-Test of H_0: mu(Seeded - Unseeded) = 0 against H_A: mu(Seeded - Unseeded) > 0 t-Statistics = 3.64 with 25 df P-Value = 0.0006 2-SampZTest of mu (Seeded) - mu (Unseeded) H_0: mu (Seeded) - mu (Unseeded) = 0 against H_A: mu (Seeded) - mu (Unseeded) > 0 z-Statistics = 1.9953 x bar_seeded - x bar_unseeded = 277.1 P-Value = 0.0230 a. Which of these 2 tests is appropriate for these data? Explain. b. Using the test you selected in part (a) and alpha = 0.01, state your conclusion in context of the problem.

Explanation / Answer

as we have 2 different samples of cloud ,

1 : 26 seeded clouds

2: 26 unseeded clouds , so both these samples are not the same , hence it is not a matched pair test. Had it been that we first selected 26 unseeded cluds , monitored their percipitation and then seeded the same 26 clouds (treatment), then it would have been a matched pair test. hence the 2nd test is appropriate

which tries to measure the statistical difference between the 2 independent groups

2) Now for the second test , we see that the p value is 0.023 and our alpha is 0.01

now as our p value is not less than 0.01 , hence we fail to reject the null hypothesis and conclude that there is no difference in the mean percipitation fof seeded and unseeded clouds

please note that

if calculated p value > alpha , then we fail to reject the null hypothesis

if calculated p value < alpha , then we reject the null hypothesis in favor of alternate hypothesis

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