Question

A car company says that the mean gas mileage for its luxury sedan is at least 24 miles per gallon​ (mpg). You believe the claim is incorrect and find that a random sample of 5 cars has a mean gas mileage of 23 mpg and a standard deviation of 5 mpg. At alpha equals 0.10​, test the​ company's claim. Assume the population is normally distributed.

Answer 1

Answer:

$t=\frac{23-24}{\frac{5}{\sqrt{5}}}=-0.447$    

$p_v =P(t_{(4)}$  

If we compare the p value and the significance level given $\alpha=0.1$ 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 reject the claim that the true mean is at least 24 mpg (null hypothesis) at 1% of signficance.  

Step-by-step explanation:

Data given and notation  

$\bar X=23$ represent the sample mean

$s=5$ represent the sample standard deviation for the sample  

$n=5$ sample size  

$\mu_o =24$ represent the value that we want to test

$\alpha=0.1$ 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 mean is lower than 24 (rejecting the claim proposed), the system of hypothesis would be:

Null hypothesis:$\mu \geq 24$  

Alternative hypothesis:$\mu < 24$  

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:  

$t=\frac{23-24}{\frac{5}{\sqrt{5}}}=-0.447$    

P-value

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

$df=n-1=5-1=4$  

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

$p_v =P(t_{(4)}$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.1$ 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 reject the claim that the true mean is at least 24 mpg at 1% of signficance.