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 28.5 milligrams and standard deviation of 2.8 milligrams for a sample of n = 9 cigarettes. The FDA claims that the mean nicotine content exceeds 31.7 milligrams for this brand of cigarette, and their stated reliability is 95%. Do you agree?

Answer 1

Answer:

We conclude that  the mean nicotine content is less than 31.7 milligrams for this brand of cigarette.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 31.7 milligrams

Sample mean, $\bar{x}$ = 28.5 milligrams

Sample size, n = 9

Alpha, α = 0.05

Sample standard deviation, s =  2.8 milligrams

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 31.7\text{ milligrams}\\H_A: \mu < 31.7\text{ milligrams}$

We use One-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{28.5 - 31.7}{\frac{2.8}{\sqrt{9}} } = -3.429$

Now, $t_{critical} \text{ at 0.05 level of significance, 8 degree of freedom } = -1.860$

Since,                  

$t_{stat} < t_{critical}$

We fail to accept the null hypothesis and accept the alternate hypothesis. We conclude that  the mean nicotine content is less than 31.7 milligrams for this brand of cigarette.