Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.902 g and a standard deviation of 0.287 g. The company that produces these cigarettes claims that it has now reduced the amount of nicotine. The supporting evidence consists of a sample of 33 cigarettes with a mean nicotine amount of 0.832 g.

Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly seleting 33 cigarettes with a mean of 0.832 g or less.
P(x-bar < 0.832 g) =  
Enter your answer as a number accurate to 4 decimal places.

Based on the result above, is it valid to claim that the amount of nicotine is lower?

• No. The probability of obtaining this data is high enough to have been a chance occurrence.
• Yes. The probability of this data is unlikely to have occurred by chance alone.

Answer 1

µ =    0.902                                      
σ =    0.287                                      
n=   33                                      
                                          
X =   0.832                                      
                                          
Z =   (X - µ )/(σ/√n) = (   0.832   -   0.902   ) / (   0.287   / √   33   ) =   -1.401  
                                          
P(X ≤   0.832   ) = P(Z ≤   -1.401   ) =   0.0806                       (answer)

excel formula for probability from z score is =NORMSDIST(Z)  

-------------------------

Based on the result above, it is not valid to claim that the amount of nicotine is lower

because probability is greater than 0.05.

• No. The probability of obtaining this data is high enough to have been a chance occurrence.