Question

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?

Answer 1

`t=(26.8-29.9)/((2.7)/(√(9)))=-3.44`

`df=n-1=9-1=8`

`p_v =P(t_((8))<-3.44)=0.0044`

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.

Explanation:

Data given

`\bar X=26.8` represent the sample mean

`s=2.7` represent the sample standard deviation

`n=9` sample size

`\mu_o =29.9` represent the value that we want to test

`\alpha=0.05` represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

`p_v` 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:
`\mu \geq 29.9`

Alternative hypothesis:
`\mu < 29.9`

The statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Calculate the statistic

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

`t=(26.8-29.9)/((2.7)/(√(9)))=-3.44`

P-value

The degrees of freedom are given by:

`df=n-1=9-1=8`

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

`p_v =P(t_((8))<-3.44)=0.0044`

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.