Question

3. The nicotine content in cigarettes of a certain brand is normally distributed with mean (in milligrams) 4. The brand advertises that the mean nicotine content of their cigarettes is 1.5, but measurements on a random sample of 100 cigarettes of this brand gave a mean of r = 1.53 and standard deviation s=0.1. Is there sufficient evidence in the sample to suggest that the mean nicotine content is actually higher than advertised? Use a = 0.05. (Hint: follow the steps: 1) set Ho and Ha; 2) set the rejection region; 3) compute the test statistic; 4) make decision based on the test statistic.) (15 pts.)

Answer 1

x = 1.53, s = 0.1, s = 0.1,n = 100, μ = 1.5

H₀: μ = 1.5, Mean nicotine content is not higher than that advertised

H₁: μ > 1.5, Mean nicotine content is higher than that advertised

Test statistic: t = (x-μ)/(s/n)^0.5 = (1.53-1.5)(0.1/100)^0.5 = 0.0009

Degrees of freedom: df = n-1 = 100-1 = 99

Level of significance = 0.05

Test statistic (Using Excel function ABS(T.INV(probability, df))) = ABS(T.INV(0.05,99)) = 1.66

Since test statistic is higher than critical value, we reject the null hypothesis and conclude that μ > 1.5.

So, mean nicotine content is higher than that advertised.