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To solve this problem, we need to compute the probability \( P(M < 0.855 \, g) \), where \( M \) represents the mean nicotine content of 47 cigarettes. Given that the population nicotine content is normally distributed with a mean \( \mu = 0.899 \, g \) and standard deviation \( \sigma = 0.304 \, g \), and we are considering a sample size of 47 cigarettes, the sample mean \( M \) is normally distributed as well, with the same population mean \( \mu \) but a smaller standard error \( \sigma_M \) given by:

\[
\sigma_M = \frac{\sigma}{\sqrt{n}} = \frac{0.304}{\sqrt{47}} \approx 0.0443
\]

### Step 1: Calculate the z-score
The z-score for the sample mean \( M = 0.855 \, g \) is calculated as:

\[
z = \frac{M - \mu}{\sigma_M} = \frac{0.855 - 0.899}{0.0443} \approx \frac{-0.044}{0.0443} \approx -0.993
\]

### Step 2: Find the probability
Using a standard normal distribution table or a calculator, we now find the probability corresponding to a z-score of -0.993.

\[
P(M < 0.855) = P(Z < -0.993)
\]

From the standard normal table, the cumulative probability for \( z = -0.993 \) is approximately **0.1608**.

### Step 3: Conclusion
The probability of randomly selecting 47 cigarettes with a mean nicotine amount of 0.855 g or less is approximately **0.1608**.

#### Interpretation of the result:
Since the probability is about 16.08%, which is greater than the 5% significance level, we **cannot** conclude that the new product has lower nicotine. Thus, the answer to the second question is:

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

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Would you like further details or clarification? Here are five related questions:

1. What is the significance of the sample size in this problem?
2. How is the standard error of the mean calculated, and why is it important?
3. What assumptions do we make when using the z-score method in hypothesis testing?
4. How would the conclusion change if the probability was less than 5%?
5. What are the implications of using a higher or lower significance level in hypothesis testing?

**Tip:** Always check the assumptions of normality and independence when performing z-tests, especially for small samples.

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.899 g and a standard deviation of 0.304 g. The company claims to have reduced the nicotine. A sample of 47 cigarettes has a mean nicotine amount of 0.855 g. Assuming the mean and standard deviation have not changed, find the probability of randomly selecting 47 cigarettes with a mean of 0.855 g or less. Is the difference significant at a 5% level?

Learn how to calculate the probability of obtaining a sample mean of nicotine content below 0.855 g based on a population mean of 0.899 g and standard deviation of 0.304 g. This step-by-step guide covers z-score calculation and hypothesis testing for statistical significance. Suitable for college-level or AP Statistics students.

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