to help consumers assess the risks they are taking, the food and drug administration (FDA) publishes the amount of nicotine found in all commerical brands of ciagrettes. a new ciagrette has recently been marketed. the FDA tests on this ciagrette yielded mean nicotine content of 24.3 milligrams and standard deviation of 2.1 milligrams for a sample of 9 ciagrettes. construct a 99% confidence interval for the mean nicotine content of this brand of ciagrette.

The 99% confidence interval for the mean nicotine content of this brand of cigarette is between 21.95 and 26.65 milligrams.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used 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

99% 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 . So we have T = 3.3554

The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 24.3 - 2.35 = 21.95 milligrams

The upper end of the interval is the sample mean added to M. So it is 24.3 + 2.35 = 26.65 milligrams

The 99% confidence interval for the mean nicotine content of this brand of cigarette is between 21.95 and 26.65 milligrams.

230/f=f-253

Step-by-step explanation: i took the quiz