Question

(17 Points) Cloud seeding has been studied for many decades as a weather modification procedure. The rainfall in acre-feet from 20 clouds that were selected at random and seeded with silver nitrate follows: 18.0 31.9 30.7 27.1 19.8 25.0 27.1 24.7 22.3 26.9 18.8 21.8 31.8 29.2 23.4 34.8 21.2 26.7 27.9 31.6 Can you support the claim that the mean rainfall from seeded clouds exceeds 25 acre-feet? Use a 1% level of significance, you may assume the distribution of rainfall follows a normal distribution. Show all work for this problem! Step 1: Step 2: Step 3: Step 4: Step 5: P-value: Stan6 42

Answer 1

n Here we have given that Xi : rainfall in acre-feet from seeded with nitrate n= sample size = 20 ΣΧι 18.0 + 30.7 + ... + 26.7 + 31.6 ī = samplemean = 26.04 20 (Xi – 7)2 s = samplestandarddeviation = = 4.78 n - 1 (A) Here we have gien that Claim: To check wheather the mean rainfall from seeded clouds exceeds 25 acre-feet we have given that The hypothesis is Ho : = 25 V/S H1:11 > 25 Now test stististics is T - H = tstat = Vins 26.04 – 25 4.78 (20) = 0.97 Now we can calulate the Pvalue a = level of significance=0.01 degrees of freedom = n-1=20-1=19 Pvalue = 0.1728 (using excel =TINV( tstat,d.f)) Decision: here Pvalue > 0.01a degrees of freedom = n-1=20-1=19 Pvalue = 0.1728 (using excel =TINV( tstat,d.f)) Decision: here Pvalue > 0.01a that is we fail reject Ho hence the mean rainfall from seeded clouds nor exceeds 25 acre-feet (B) Now we check the rainfall is normally distributed or not Probability Plot of rainfall Normal - 95% CI 99 95 Mean 26.04 StDev 4.75 N 20 AD 0.214 P-Value 0.827 90 80 70 Percent 40 30 10 5 1 10 15 20 35 40 45 25 30 rainfal here all the data points are within the reference line that we say that the rainfall is normally distributed. Scanned with CamScanner

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