Question

An automobile manufacturer who wishes to advertise that one of its models achieves 30 mpg (miles per gallon) decides to carry out a fuel efficiency test. Six nonprofessional drivers were selected, and each one drove a car from Phoenix to Los Angeles. The resulting fuel efficiencies (in miles per gallon) are given below:
27.3 29.4 31.2 28.4 30.2 29.6
Assuming that fuel efficiency is normally distributed under these circumstances, do the data contradict the claim that true average fuel efficiency is (at least) 30 mpg? Test the appropriate hypotheses at significance level 0.05. (Use a statistical computer package to calculate the P-value. Round your test statistic to two decimal places and your P-value to three decimal places.)

Answer 1

Answer:

$t=\frac{29.35-30}{\frac{1.365}{\sqrt{6}}}=-1.17$  

$p_v =P(t_{(5)}$  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.  

We can say that at 5% of significance the true mean is not significantly less than 30 so then the claim that true average fuel efficiency is (at least) 30 mpg makes sense

Step-by-step explanation:

Data given and notation  

The mean and sample deviation can be calculated from the following formulas:

$\bar X =\frac{\sum_{i=1}^n x_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)}{n-1}}$

$\bar X=29.35$ represent the sample mean  

$s=1.365$ represent the sample standard deviation  

$n=6$ sample size  

$\mu_o =30$ represent the value that we want to test  

$\alpha=0.05$ 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 lower than 30 or no, the system of hypothesis are :  

Null hypothesis:$\mu \geq 30$  

Alternative hypothesis:$\mu < 30$  

Since we don't know the population deviation, 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:  

$t=\frac{29.35-30}{\frac{1.365}{\sqrt{6}}}=-1.17$  

P-value  

We need to calculate the degrees of freedom first given by:  

$df=n-1=6-1=5$  

Since is a one-side left tailed test the p value would given by:  

$p_v =P(t_{(5)}$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.  

We can say that at 5% of significance the true mean is not significantly less than 30 so then the claim that true average fuel efficiency is (at least) 30 mpg makes sense