Question

uppose that an automobile manufacturer advertises that its new hybrid car has a mean gas mileage of 50 miles per gallon. You take a simple random sample of n = 30 hybrid vehicles and test their gas mileage. You find that in this sample, the average is 47 miles per gallon with a standard deviation of 5.5 miles per gallon. Is there enough evidence to support the advertised claim using α = 0.05? What is the test statistics for the problem?

Answer 1

Answer:

$t=\frac{47-50}{\frac{5.5}{\sqrt{30}}}=-2.988$    

$p_v =2*P(t_{(29)}$  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the true mean is significantly different from 50 mpg at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

$\bar X=47$ represent the sample  mean  

$s=5.5$ represent the sample standard deviation

$n=30$ sample size  

$\mu_o =50$ 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 50 or no, the system of hypothesis would be:  

Null hypothesis:$\mu = 50$  

Alternative hypothesis:$\mu \neq 50$  

If we analyze the size for the sample is = 30 but 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:  

$t=\frac{47-50}{\frac{5.5}{\sqrt{30}}}=-2.988$    

P-value

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

$df=n-1=30-1=29$  

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

$p_v =2*P(t_{(29)}$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the true mean is significantly different from 50 mpg at 5% of signficance.