Question

Suppose that the mean stopping time of a particular model of car moving at 40 mph is listed

as 1.8 seconds, with a standard deviation of 0.25 seconds. A random sample of 12 cars obtains a

standard deviation of 0.35 seconds. Test the claim that o = 0.25 seconds at the 0.05 level of significance

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\sigma^2=0.25^2=0.0625$

$H_a:\sigma^2\not=0.25^2=0.0625$

This corresponds to a two-tailed test test, for which a Chi-Square test for one population variance will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ $df=n-1=11$ degrees of freedom, and the the rejection region for this two-tailed test is $R = \{\chi^2: \chi^2 < 3.8157 \text{ or } \chi^2 > 21.92\}.$

The Chi-Squared statistic is computed as follows:

$$\chi^2 =\dfrac{(n-)s^2}{\sigma^2}=\dfrac{(12-1)(0.35)^2}{(0.25)^2}=21.56$$

Since it is observed that $\chi_L^2 = 3.8157 \le \chi^2 = 21.56 \le \chi_U^2 = 21.92,$

it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population variance $\sigma^2$ is different than $(0.25)^2,$ at the $0.05$ significance level.