Question

To help consumers assess the risks they are taking, the Food and Drug administration publishes the amount of nicotine found in all commerical brands of cigs. A new cig has recently been marketed. The FDA tests on this cigarette gave a mean nicotine content of 28.4 milligrams and standard deviation of 2.6 milligrams for a sample of n =9 cigarettes. Assume the amounts of nicotine are normally distributed.
A) construct a 90% confidence interval that estimates the mean nicotine content for this brand of cigarette. Make sure to check the requirements, find the point estimate, the confidence interval and interpret.
B) the FDA claims that the mean nicotine content exceeds 31.0 milligrams for this brand of cigarette, and their stated reliability is 90%. Do you agree? Hint: Think about interval you found in part a and remember to justify.

Answer 1

a)

sample mean, xbar = 28.4
sample standard deviation, s = 2.6
sample size, n = 9
degrees of freedom, df = n - 1 = 8

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.86

CI = (28.4 - 1.86 * 2.6/sqrt(9) , 28.4 + 1.86 * 2.6/sqrt(9))
CI = (26.79 , 30.01)

CI = (x-te x 8 +te x )

b)

yes, we agree because As the confidence interval does not contain 31 reject H0