In: Statistics and Probability
The nicotine content in cigarettes of a certain brand is normally distributed with mean (in milligrams) ? μ and standard deviation ?=0.1 σ = 0.1 . The brand advertises that the mean nicotine content of their cigarettes is 1.5 mg. Now, suppose a reporter wants to test whether the mean nicotine content is actually higher than advertised. He takes measurements from a SRS of 15 cigarettes of this brand. The sample yields an average of 1.4 mg of nicotine. Conduct a test using a significance level of ?=0.01 α = 0.01 . (a) The test statistic (b) The critical value, z*
Solution :
= 1.5
= 1.4
=0.1
n = 15
This is the right tailed test .
The null and alternative hypothesis is
H0 : =1.5
Ha : > 1.5
Test statistic = z
= ( - ) / / n
= (1.4 -1.5) /0.1 / 15
= -3.87
p(Z >-3.87 ) = 1-P (Z < -3.87) =0.9999
P-value = 0.9999
= 0.01
The critical value = zc = 2.326
0.9999 > 0.01
Do not reject the null hypothesis .
There is insufficient evidence to suggest that