Southern Florida is known as the "Lightning Capital" of the U.S. It averages 25.1 lightning strikes per year per square mile with a standard deviation of 5.25. A meteorologist believes the number of strikes has increased. Using a sample of the last 40 years with 26.5 strikes, does there seem to be a significant increase? Use alpha = .01.

Respuesta :

Answer:

[tex]z=\frac{26.5-25.1}{\frac{5.25}{\sqrt{40}}}=1.687[/tex]    

[tex]p_v =P(z>1.687)=0.0458[/tex]  

If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is higher than 25.1 (significant increase) at 1% of signficance.  

Step-by-step explanation:

Data given and notation  

[tex]\bar X=26.5[/tex] represent the sample mean

[tex]\sigma=5.25[/tex] represent the population standard deviation for the sample  

[tex]n=40[/tex] sample size  

[tex]\mu_o =25.1[/tex] represent the value that we want to test

[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.  

z 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 mean is higher than 25.1, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 25.1[/tex]  

Alternative hypothesis:[tex]\mu > 25.1[/tex]  

If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

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

z-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]z=\frac{26.5-25.1}{\frac{5.25}{\sqrt{40}}}=1.687[/tex]    

P-value

Since is a one right side test the p value would be:  

[tex]p_v =P(z>1.687)=0.0458[/tex]  

Conclusion  

If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is higher than 25.1 (significant increase) at 1% of signficance.