The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.953 g and a standard deviation of 0.285 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 39 cigarettes with a mean nicotine amount of 0.857 g.

Assuming that the given mean and standard deviation **have NOT changed**, find the probability of randomly selecting 39 cigarettes with a mean of 0.857 g **or less**.
P(*M* < 0.857 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.

Based on the result above, ¿is it valid to claim that the amount of nicotine is lower?  (Let’s use a 5% cut-off for our definition of unusual.)

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.

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.

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.953 g and a standard deviation of 0.285 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 39 cigarettes with a mean nicotine amount of 0.857 g. Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly selecting 39 cigarettes with a mean of 0.857 g or less. Is it valid to claim that the amount of nicotine is lower?

Determine the probability of selecting 39 cigarettes with a mean nicotine content of 0.857 g or less, using the Central Limit Theorem and Z-scores. This example explains how to calculate the standard error and Z-score, and interpret the result in the context of significance testing.

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