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, Technometric) 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:

(Please INCLUDE how you got the answers, don't just post your final answer for parts a, and b)

1. 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: Ch 6.3.2 Paired t-Test of H (Seeded -Unseeded) T-Test of Ho: (Seeded unseeded)- o against HA: Seeded Unseeded) 0 t-Statistics 3.64 with 25 df P-Value 0.0006 Ch 6.3.3 2-SampzTest of u(Seeded) -u (Unseeded) How pulseeded)- u(Unseeded) o against HA: u(seeded) ulunseeded) o z-Statistics 1.9953 unseeded 277.1 seeded 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 a 0.01, state your conclusion in context of the problem.

Explanation / Answer

a) 2 sample Z test.

The paired t-test calculates the difference within each before-and-after pair of measurements, determines the mean of these changes, and reports whether this mean of the differences is statistically significant.

The before and after pair suggests that the seeded and unseeded clouds are same for each set of observation.

Hence, we use a 2 sample Z test, but this requires the prior knowledge of the standard deviation of the population.

b) The z-test conducted is for right tailed one with = 0.01

The p-value quoted is 0.0230 > 0.01 and therefore we retain the null hypothesis.

There is no sufficient evidence to suggest that the difference between means of seeded and unseeded clouds are greater than 0.

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