Answer to Question #250687 in Statistics and Probability for fiz

Question

A car is driven 20,000 km/year. To test the claim, a sample at 150 car owners are picked randomly and asked to state the distance that their car had traveled in a year. Do you agree to the claim of the following sample data: x = 22,500 km, σ = 3800 km. Use the significance level of 5%.

Show working on critical value approach and p-value approach.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=20000$

$H_1:\mu\not=20000$

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.05, df=n-1$

$=150-1=149$ ​degrees of freedom, and the critical value for a two-tailed test is$t_c = 1.976013.$

The rejection region for this two-tailed test is $R = \{t: |t| > 1.976013\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{22500-20000}{3800/\sqrt{150}}=8.0575$$

Since it is observed that $|t|= 8.0575>1.976013=t_c$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for two-tailed, $\alpha=0.05, df=149,$ $t=8.0575$ is $p\approx0,$ and since $p=0<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is different than 20000, at the $\alpha = 0.05$ significance level.