Question

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, 30.7, 19.8, 27.1, 22.3, 18.8, 31.8, 23.4, 21.2, 27.9, 31.9, 27.1, 25.0, 24.7, 26.9, 21.8, 29.2, 34.8, 26.7, and 31.6. Can you support a claim that mean rainfall from seeded clouds exceeds 25 acre-feet? Use a 0.05 level of significance.

Answer 1

The Solution:

Given:

18.0, 30.7, 19.8, 27.1, 22.3, 18.8, 31.8, 23.4, 21.2, 27.9, 31.9, 27.1, 25.0, 24.7, 26.9, 21.8, 29.2, 34.8, 26.7, and 31.6.

Required:

To test the claim that the mean rainfall from seeded clouds exceeds 25 ace-feet.

Step 1:

Find the mean.

Step 2:

Find the standard deviation.

Step 3:

Hypothesis:

`\begin{gathered} H_0:\mu=25 \\ \\ H_1:\mu>25 \\ \\ \alpha=0.05 \end{gathered}`
`\begin{gathered} Z=\frac{\bar{x}-\mu}{(\sigma)/(√(n))} \\ \\ Where \\ n=20 \\ \mu=25 \\ \bar{x}=26.035 \\ \sigma=4.664 \\ \end{gathered}`
`Z=(26.035-25)/((4.664)/(√(20)))=(1.035)/(1.04290)=0.9924`

From the Z score tables,

`p-value=0.1605`

Since the p-value is greater than 0.05, we fail to reject the null hypothesis.

Therefore, there is no fact to support the claim that the mean rainfall exceeds 25 acre-feet.