Answer to Question #273715 in Statistics and Probability for Jason

Question

A tobacco company advertises that the average nicotine content of its cigarettes is at

most 14 milligrams. A consumer protection agency wants to determine whether the

average nicotine content is in fact greater than 14. A random sample of 300 cigarettes

of the company’s brand yield an average nicotine content of 14.6 and a standard

deviation of 3.8 milligrams. Determine the level of significance of the statistical test of

the agency’s claim that 𝜇 is greater than 14. If 𝛼 = 0.01, is there significant evidence

that the agency’s claim has been supported by the data?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\leq 14$

$H_1:\mu>14$

This corresponds to a right-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 $\alpha = 0.01,$ $df=n-1=299$ degrees of freedom, and the critical value for a right-tailed test is $t_c =2.338884.$

The rejection region for this right-tailed test is $R = \{t: t > 2.338884.\}$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{14.6-14}{3.8/\sqrt{300}}$$

$$\approx 2.734817$$

Since it is observed that $t = 2.734817>2.338884 = t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for right-tailed, $\alpha=0.01, df=299$ degrees of freedom, $t=2.734817$ is$p=0.003307,$ and since $p=0.003307<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is greater than 14, at the $\alpha = 0.01$ significance level.