Answer to Question #323795 in Statistics and Probability for secret

Question

A random sample of 11 cigarettes of a certain brand has an average nicotine content of 6.6 milligrams and a standard deviation of 2.5 milligrams. Is this in line with the manufacturer's claim that the average nicotine content does not exceed 4.2 milligrams? Use 0.1 level of significance.

Answer 1

$\mu=4.2, \ n=11, \ \bar{x}=6.6, \ s=2.5,\ \alpha=0.1.$

The null and alternative hypotheses are

$H_0:\mu\le4.2,\\ H_1:\mu>4.2.$

Because $\sigma$ is unknown and he population is normally distributed, we use the t-test.

The test is a right-tailed test, the level of significance is $\alpha=0.1$ , and the degrees of freedom are

d.f. = 11 - 1 = 10. So, using t-table, the critical value is t₀ = 1.372. The rejection region is t > 1.372. The standardized test statistic is

$t=\cfrac{\bar{x}-\mu}{s/\sqrt{n}}=\cfrac{6.6-4.2}{2.5/\sqrt{11}}=3.184>1.372.$

Because t is in the rejection region, we reject the null hypothesis, the average nicotine content exceeds 4.2 milligrams.