To find the probability of randomly selecting 37 cigarettes with a mean nicotine amount of 8.078 mg or less, we can use the properties of the normal distribution.

1. Calculate the standard error (SE) of the sample mean: $SE = \frac{\sigma}{\sqrt{n}} = \frac{1.4 \text{ mg}}{\sqrt{37}} \approx \frac{1.4}{6.08} \approx 0.2305 \text{ mg}$

2. Calculate the z-score for the sample mean of 8.078 mg: $z = \frac{\bar{x} - \mu}{SE} = \frac{8.078 - 8.4}{0.2305} \approx \frac{-0.322}{0.2305} \approx -1.396$

3. Find the probability associated with this z-score using the standard normal distribution table:

   • A z-score of -1.396 corresponds to a cumulative probability of approximately 0.0811.

Thus, the probability of randomly selecting 37 cigarettes with a mean nicotine amount of 8.078 mg or less is approximately: $P(M < 8.078 \text{ mg}) \approx 0.0811$

Conclusion

Since the probability of obtaining a sample mean of 8.078 mg or less is about 0.0811 (or 8.11%), which is relatively high, we would conclude:

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

Additional Questions

1. How does the sample size affect the standard error of the mean?
2. What is the significance level typically used to determine statistical significance?
3. How would the result change if the mean nicotine level were even lower?
4. What assumptions are made when using the normal distribution for this analysis?
5. How can you conduct a hypothesis test for this situation?

Tip

Always ensure the conditions for using a normal distribution are met, especially regarding sample size and population distribution.