-1.8 - 6x = 6(0.1 + 3x)
2 answers:
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Answer:
<em>There is no significant difference in the amount of rain produced when seeding the clouds.</em>
Step-by-step explanation:
Assuming that the amount of rain delivered by thunderheads follows a distribution close to a normal one, we can formulate a hypothesis z-test:
<u>Null Hypothesis </u>
: Average of the amount of rain delivered by thunderheads without seeding the clouds = 300 acrefeet.
<u>Alternative Hypothesis </u>
: Average of the amount of rain delivered by thunderheads by seeding the clouds > 300 acrefeet.
This is a right-tailed test.
Our z-statistic is
We now compare this value with the z-critical for a 0.05 significance level. This is a value such that the area under the Normal curve to the left of
is less than or equal to 0.05
We can find this value with tables, calculators or spreadsheets.
<em>In Excel or OpenOffice Calc use the function </em>
<em>NORMSINV(0.95) </em>
an we obtain a value of
= 1.645
Since 1.2845 is not greater than 1.645 we cannot reject the null, so the conclusion that can be drawn when the significance level is 0.05 is that there is no significant difference in the amount of rain produced when seeding the clouds.