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

Answer:

$t=\frac{26.8-29.9}{\frac{2.7}{\sqrt{9}}}=-3.44$    

$df=n-1=9-1=8$  

$p_v =P(t_{(8)}$  

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

$\bar X=26.8$ 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:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

Calculate the statistic

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

$t=\frac{26.8-29.9}{\frac{2.7}{\sqrt{9}}}=-3.44$    

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:  

$p_v =P(t_{(8)}$  

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.