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 yielded mean nicotine content of 25.325.3 milligrams and standard deviation of 2.72.7 milligrams for a sample of n equals 9n=9 cigarettes. Construct a 9090​% confidence interval for the mean nicotine content of this brand of cigarette.

Answer 1

The 90​% confidence interval for the mean nicotine content of this brand of cigarette is between 20.3 milligrams and 30.3 milligrams.

Explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 9 - 1 = 8

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 8 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.9)/(2) = 0.95`. So we have T = 1.8595

The margin of error is:

M = T*s = 1.8595*2.7 = 5

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 25.3 - 5 = 20.3 milligrams

The upper end of the interval is the sample mean added to M. So it is 25.3 + 5 = 30.3 milligrams.

The 90​% confidence interval for the mean nicotine content of this brand of cigarette is between 20.3 milligrams and 30.3 milligrams.