Answer to Question #98502 in Statistics and Probability for Juliet Beglaryan

Question

6. 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

Let $X=$ the gas mileage in miles per gallon

Assume that $\mu_0=50\ \text{miles per gallon}.$

$H_0:\mu=\mu_0$

$H_1:\mu\not=\mu_0$

Given that $\sigma=5.5\ \text{miles per gallon}, n=30, \bar{X}=47\ \text{miles per gallon}$

From the Central Limit Theorem, we know this sampling distribution is normal with a mean of 50 and standard error of

$\sigma/\sqrt{n}=5.5/\sqrt{30}$

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Let $\alpha=0.05$

The critical value for a two-tailed test is $z_c=1.96$

The rejection region for this two-tailed test is $R=\{{z:|z|>1.96}\}$

The z-statistic is computed as follows:

$$z={\bar{X}-\mu_0 \over \sigma/\sqrt{n}}$$

$$z={47-50 \over5.5/\sqrt{30}}\approx-2.9876$$

Since it is observed that $|z|=|-2.9876|=2.9876>1.96=z_c,$ it is then concluded that the null hypothesis is rejected.

The p-value

$$p=2\cdot P(Z<-2.9876)\approx2(0.0014)=0.0028$$

Since $p=0.0028<0.05,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is different than $50,$ at the $0.05$ significance level.

Confidence interval

$$\bar{X}\pm z_c{\sigma \over \sqrt{n}}$$

$$47\pm 1.96{5.5 \over \sqrt{30}}$$

$$47\pm 1.968$$

$$(45.032,48.968)$$