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Meteorologists in Texas want to increase the amount of rain delivered by thunderheads by seeding the clouds. Without seeding, thunderheads produce, on average, 300 acrefeet. The meteorologists randomly selected 30 clouds which they seeded with silver iodide to test their theory that average acrefeet is more than 300. The sample mean is 370.4 with a sample standard deviation of 300.1. rainfall-Oct13 What conclusion can be drawn when the significance level is 0.05? The data does not provided statistical evidence that the average acrefeet from seeded clouds is more than 300. Two of the above are correct. We can not draw conclusions based on the p-value because the conditions are not met due to the extreme outliers.

Respuesta :

Answer:

[tex]t=\frac{370.4-300}{\frac{300.1}{\sqrt{30}}}=1.285[/tex]  

[tex]p_v =P(t_{(29)}>1.285)=0.104[/tex]  

We can say that at 5% of significance the average acrefeet from seeded clouds is NOT significantly higher than 300.

The data does not provided statistical evidence that the average acrefeet from seeded clouds is more than 300.

Step-by-step explanation:

Data given and notation  

[tex]\bar X=370.4[/tex] represent the sample mean  

[tex]s=300.1[/tex] represent the sample standard deviation  

[tex]n=30[/tex] sample size  

[tex]\mu_o =300[/tex] represent the value that we want to test  

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the average acrefeet from seeded clouds is more than 300 :  

Null hypothesis:[tex]\mu \leq 300[/tex]  

Alternative hypothesis:[tex]\mu > 300[/tex]  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

[tex]t=\frac{370.4-300}{\frac{300.1}{\sqrt{30}}}=1.285[/tex]  

P-value  

We need to calculate the degrees of freedom first given by:  

[tex]df=n-1=30-1=29[/tex]  

Since is a one-side right tailed test the p value would given by:  

[tex]p_v =P(t_{(29)}>1.285)=0.104[/tex]  

Conclusion  

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.  

We can say that at 5% of significance the average acrefeet from seeded clouds is NOT significantly higher than 300.

The data does not provided statistical evidence that the average acrefeet from seeded clouds is more than 300.

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