From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant at around 2.1 years. A survey of 37 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level?

Note: If you are using a Student's *t*-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.) 
*H*0:

 𝜇 ≥ 19 *Ha*:

 𝜇 < 19 
*X*

 represents the average age when smokers first start to smoke.  State the distribution to use for the test. (Round your answers to four decimal places.)

*X*

 ~   

   ,   

To analyze whether the data support the claim that the mean starting age of smokers is at least 19 years, we can conduct a one-sample \(t\)-test. Here’s how to go through the steps:

### Step 1: State the Hypotheses
- Null Hypothesis (\(H_0\)): \(\mu \geq 19\)
- Alternative Hypothesis (\(H_a\)): \(\mu < 19\)

### Step 2: Identify the Distribution
Since the sample size is less than 30 and the population standard deviation is unknown, we will use the Student's \(t\)-distribution.

### Step 3: Sample Information
- Sample mean (\(\bar{X}\)): 18.1
- Sample standard deviation (s): 1.3
- Sample size (n): 37

### Step 4: Calculate the Test Statistic
The test statistic for a one-sample \(t\)-test is calculated as follows:

\[
t = \frac{\bar{X} - \mu_0}{s / \sqrt{n}}
\]

Where:
- \(\mu_0\) is the hypothesized population mean (19 in this case).

Plugging in the values:

\[
t = \frac{18.1 - 19}{1.3 / \sqrt{37}} = \frac{-0.9}{1.3 / 6.0828} \approx \frac{-0.9}{0.2133} \approx -4.22
\]

### Step 5: Determine the Critical Value and Decision Rule
For a one-tailed test at the 5% significance level (\(\alpha = 0.05\)) with \(n - 1 = 36\) degrees of freedom, we can look up the critical value in the \(t\)-distribution table.

The critical value for \(t_{0.05, 36}\) is approximately \(-1.688\).

### Step 6: Compare Test Statistic to Critical Value
- Calculated \(t\): \(-4.22\)
- Critical value: \(-1.688\)

Since \(-4.22 < -1.688\), we reject the null hypothesis.

### Conclusion
The data provide sufficient evidence to support the claim that the mean starting age of smokers is less than 19 years at the 5% significance level.

### Summary of Results
- Test Statistic \(t \approx -4.22\)
- Distribution: \(X \sim t_{36}\)

Would you like further details or have any questions? Here are some follow-up questions to consider:

1. How would the result change if the sample mean were higher?
2. What are the implications of rejecting the null hypothesis in this context?
3. How would you interpret the p-value for this test?
4. What assumptions must hold true for the \(t\)-test to be valid?
5. How might the results differ if the standard deviation were not constant?

**Tip:** Always ensure that your sample size is adequate to draw reliable conclusions from statistical tests.

From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant at around 2.1 years. A survey of 37 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level?

Explore the results of a statistical analysis to determine if the mean starting age of smokers is less than 19 years, based on a survey of 37 smokers. The sample mean was 18.1 with a standard deviation of 1.3, analyzed using a t-test at the 5% significance level.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation