Question

A random sample of 25 brand A cigarettes showed an average nicotine content of 5 milligrams, while a sample of 40 brand D cigarettes showed average nicotine of 4.8 milligrams. If the standard deviation of nicotine is 1.6 milligrams, would you say that brand D has a lesser nicotine content? Use a 0.01 level of significance. Assume the distribution of nicotine content to be normal

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1\le\mu_2$

$H_0:\mu_1>\mu_2$

This corresponds to a right-tailed test, and a z-test for two means, with known population standard deviations will be used.

Based on the information provided, the significance level is $\alpha = 0.01,$ and the critical value for a right-tailed test is $z_c = 2.3263$

The rejection region for this right-tailed test is $R = \{z: z > 2.3263\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{X}_1-\bar{X}_2}{\sqrt{\sigma_1^2/n_1+\sigma_2^2/n_2}}$$

$$=\dfrac{5-4.8}{\sqrt{1.6^2/25+1.6^2/40}}\approx0.4903$$

Since it is observed that $z = 0.49 \le z_c = 2.326,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=P(Z>0.4903)=0.311961,$ and since $p=0.311961>0.01,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu_1$is greater than $\mu_2,$ at the $\alpha = 0.01$ significance level.

Therefore, there is not enough evidence to claim that brand D has a lesser nicotine content, at the $\alpha = 0.01$ significance level.