Question

The Weather Underground reported that the mean amount of summer rainfall for the northeastern US is 8.5 inches. Mr. Weatherman, a meteorologist does not believe that. He randomly selected 9 cities in the northeast and found the mean rainfall amount to be 7.5 inches with a standard deviation of 1.5 inches. Test the hypothesis at 1% significance level.

Answer 1

Answer:

$t=\frac{7.5-8.5}{\frac{1.5}{\sqrt{9}}}=-2$    

$p_v =P(t_{(8)}$    

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ 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 actually its significant different from 8.5 in at 1% of signficance.  

Step-by-step explanation:

Data given and notation  

$\bar X=7.5$ represent the sample mean

$s=1.5$ represent the sample standard deviation

$n=9$ sample size  

$\mu_o =8.5$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

$p_v$ 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 true mean is different form 8.5 inches, the system of hypothesis would be:  

Null hypothesis:$\mu = 8.5$  

Alternative hypothesis:$\mu \neq 8.5$  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (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:  

$t=\frac{7.5-8.5}{\frac{1.5}{\sqrt{9}}}=-2$    

P-value

The first step is calculate the degrees of freedom, on this case:  

$df=n-1=9-1=8$  

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

$p_v =P(t_{(8)}$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ 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 actually its significant different from 8.5 in at 1% of signficance.