Question

Nationally, the per capita monthly fuel oil hill is $ 110. A random sample of 26 cities in the Southeast average $78, with a standard deviation of $ 4. Is the sample different with the national average? Summarize your conclusions in a few words.

Answer 1

Answer:

$t=\frac{78-110}{\frac{4}{\sqrt{26}}}=-40.792$  

$p_v =2*P(t_{25}$    

If we compare the p value and for any significance level assumed for example $\alpha=0.05$ we see that $p_v$ so we can reject the null hypothesis, and the actual true mean is significantly different from the national average at 5% of significance.

Step-by-step explanation:

Data given and notation    

$\bar X=78$ represent the sample mean  

$s=4$ represent the standard deviation for the sample    

$n=26$ sample size    

$\mu_o =110$ represent the value that we want to test    

$\alpha$ 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 determine if the mean is equal to 110 or no, the system of hypothesis would be:    

Null hypothesis:$\mu = 110$    

Alternative hypothesis:$\mu \neq 110$    

We don't know the population deviation, so for this case 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:    

$t=\frac{78-110}{\frac{4}{\sqrt{26}}}=-40.792$  

Calculate the P-value    

First we need to find the degrees of freedom given by:

$df=n-1=26-1=25$  

Since is a two tailed test the p value would be:    

$p_v =2*P(t_{25}$    

Conclusion    

If we compare the p value and for any significance level assumed for example $\alpha=0.05$ we see that $p_v$ so we can reject the null hypothesis, and the actual true mean is significantly different from the national average at 5% of significance.