Question

Suppose 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 ̄x = 47 miles per gallon with a standard deviation of s = 5.5 miles per gallon. Does this indicate that the advertiser’s statement is true or false? Let a=.0.05

Answer 1

We have $\mu_0=50$

For the sample, we have

$n=30, \bar{X}=47$ and

 $S=5.5$

Null Hypothesis, $H_0:\mu=\mu_0$

Alternate hypothesis, $H_1:\mu\not=\mu_0$

Test statistics, $t=(\bar{x}-\mu_0)/(S/\sqrt{n})$

$=(47-50)/(5.5/\sqrt{30})$

$=-3/1.004=-2.9875$

Since the population standard deviation is unknown and the sample size is also small, we will use t-test.

Degree of freedom (DOF) for this t-test will be equal to $n-1=30-1=29$ .

The t-critical values for a two-tailed test, for a significance level of $\alpha= 0.05$ and $DOF=29$ will be

$t_c<-2.045$ and $t_c>2.045$.

As we can see that our test statistics lies in the critical region, we will have to reject the null hypothesis.

So, we can say that at $\alpha =0.05$ the advertiser's statement is false.