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:
[tex]\begin{gathered} H_0:\mu=25 \\ \\ H_1:\mu>25 \\ \\ \alpha=0.05 \end{gathered}[/tex][tex]\begin{gathered} Z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}} \\ \\ Where \\ n=20 \\ \mu=25 \\ \bar{x}=26.035 \\ \sigma=4.664 \\ \end{gathered}[/tex][tex]Z=\frac{26.035-25}{\frac{4.664}{\sqrt{20}}}=\frac{1.035}{1.04290}=0.9924[/tex]
From the Z score tables,
[tex]p-value=0.1605[/tex]
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.
