Answer to Question #179391 in Statistics and Probability for david Joshua Garcia

Question

A random sample of eight cigarettes of a certain

brand has average nicotine content of 4.2

milligrams and standard deviation of 1.4

milligrams. Is this in line with the manufacturer claim

that the average nicotine content does not

exceed 3.5 milligrams? Use 0.01 level of

significance.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\leq3.5$

$H_1:\mu>3.5$

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

Based on the information provided, the significance level is $\alpha=0.01,$ and the critical value for a right-tailed test and degrees of freedom $df=n-1=7$ is $t_c=2.99793.$

The rejection region for this right-tailed test is $R=\{t:t>2.99793\}.$  

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{4.2-3.5}{1.4/\sqrt{8}}=1.41421$$

Since it is observed that $t=1.41421<2.99793=t_c,$ it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is greater than 3.5, at the 0.01 significance level.

Using the P-value approach:

The p-value for right-tailed test, $df =7, t=1.41421$ is $p=0.100102,$ and since $p=0.100102>0.01=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is greater than 3.5, at the 0.01 significance level.