To solve this problem, we need to determine the probability of obtaining a sample mean of 0.857 g or less, assuming that the original population mean is 0.953 g and the standard deviation is 0.285 g. We'll use the Central Limit Theorem, which states that the sampling distribution of the sample mean is approximately normally distributed if the sample size is sufficiently large (in this case, $n = 39$).

Step 1: Calculate the Standard Error (SE)

The standard error of the sample mean is given by:

$\text{Standard Error} = \frac{\sigma}{\sqrt{n}}$

where:

• $\sigma = 0.285$ (standard deviation of the population),
• $n = 39$ (sample size).

$\text{Standard Error} = \frac{0.285}{\sqrt{39}}$

Step 2: Calculate the Z-Score

The $z$-score formula for a sample mean is:

$z = \frac{\bar{X} - \mu}{\text{Standard Error}}$

where:

• $\bar{X} = 0.857$ (sample mean),
• $\mu = 0.953$ (population mean).

Step 3: Find the Probability

After calculating the $z$-score, we'll look up the corresponding probability $P(M < 0.857)$ using the standard normal distribution table.

I'll calculate each step now.### Results:

• Z-Score: -2.104 (rounded to three decimal places)
• Probability: $P(M < 0.857) = 0.0177$ (rounded to four decimal places)

Interpretation:

The probability of randomly selecting 39 cigarettes with a mean nicotine content of 0.857 g or less, assuming the population mean is still 0.953 g, is approximately 0.0177 (1.77%).

Conclusion:

Since this probability is lower than the 5% significance level (0.05), it is unlikely to have occurred by chance alone. Thus, there is enough evidence to support the claim that the amount of nicotine has been reduced.

Answer:

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

Do you want further details or have any questions?

Here are 5 related questions to extend your understanding:

1. How does increasing the sample size affect the standard error?
2. What would happen if the standard deviation of the population were higher?
3. How would the conclusion change if we used a 1% significance level instead?
4. How would you interpret a $z$-score of 0?
5. What is the probability of obtaining a sample mean of exactly 0.953 g?

Tip: When working with probabilities, always remember to compare the computed probability to the chosen significance level to draw conclusions.