Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.958 g and a standard deviation of 0.298 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 36 cigarettes with a mean nicotine amount of 0.859 g.

Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly seleting 36 cigarettes with a mean of 0.859 g or less.
P(M < 0.859 g) =
Enter your answer as a number accurate to 4 decimal places. NOTE: Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Assume that any probability below 5% is sufficient to conclude that the new product really have lower nicotine. 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

Here it is given that distribution is normal with mean=0.958 and standard deviation=0.298

For n=36, we need to find

Here population is normal so sample mean is also normal, hence we can convert M to z

As probability is less than 5% we conclude

• Yes. The probability of this data is unlikely to have occurred by chance alone.