To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nicotine found in all commercial brands of cigarettes. A new cigarette has recently been marketed. The FDA tests on this cigarette gave a mean nicotine content of 26.8 milligrams and standard deviation of 2.7 milligrams for a sample of nequals9 cigarettes. The FDA claims that the mean nicotine content exceeds 29.9 milligrams for this brand of cigarette, and their stated reliability is 95%. Do you agree?

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and is not enough evidence to conclude that the claim is true.

Step-by-step explanation:

Data given

[tex]\bar X=26.8[/tex] represent the sample mean

[tex]s=2.7[/tex] represent the sample standard deviation

[tex]n=9[/tex] sample size

[tex]\mu_o =29.9[/tex] represent the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value for the test (variable of interest)

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean exceed 29.9 or no, the system of hypothesis would be:

Null hypothesis:[tex]\mu \geq 29.9[/tex]

Alternative hypothesis:[tex]\mu < 29.9[/tex]

The statistic is given by:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)

Calculate the statistic

We can replace in formula (1) the info given like this:

[tex]t=\frac{26.8-29.9}{\frac{2.7}{\sqrt{9}}}=-3.44[/tex]

P-value

The degrees of freedom are given by:

[tex]df=n-1=9-1=8[/tex]

Since is a one sided test the p value would be:

[tex]p_v =P(t_{(8)}<-3.44)=0.0044[/tex]

Conclusion

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and is not enough evidence to conclude that the claim is true.