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 a mean nicotine content of 26.2 milligrams and standard deviation of 2.9 milligrams for a sample of n = 95 cigarettes. Estimate the mean nicotine content with a 95% confidence interval.

Answer 1

The mean nicotine content with a 95% confidence interval is (25.52mg, 26.88mg).

Explanation:

The sample size is 95.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

`df = 95-1 = 94`

Then, we need to subtract one by the confidence level
`\alpha` and divide by 2. So:

`(1-0.95)/(2) = (0.05)/(2) = 0.025`

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 94 and 0.025 in the t-distribution table, we have
`T = 2.28`.

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

`s = (2.9)/(√(95)) = 0.2975`

Now, we multiply T and s

`M = T*s = 2.28*0.2975 = 0.6783`

For the lower end of the interval, we subtract the mean by M. So 26.2 - 0.6783 = 25.52mg

For the upper end of the interval, we add the mean to M. So 26.2 + 0.6783 = 26.88mg

The mean nicotine content with a 95% confidence interval is (25.52mg, 26.88mg).