Math Problem Statement

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.) H0:

 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

 ~   

   ,   

Solution

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 tt-test. Here’s how to go through the steps:

Step 1: State the Hypotheses

  • Null Hypothesis (H0H_0): μ19\mu \geq 19
  • Alternative Hypothesis (HaH_a): μ<19\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 tt-distribution.

Step 3: Sample Information

  • Sample mean (Xˉ\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 tt-test is calculated as follows:

t=Xˉμ0s/nt = \frac{\bar{X} - \mu_0}{s / \sqrt{n}}

Where:

  • μ0\mu_0 is the hypothesized population mean (19 in this case).

Plugging in the values:

t=18.1191.3/37=0.91.3/6.08280.90.21334.22t = \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 (α=0.05\alpha = 0.05) with n1=36n - 1 = 36 degrees of freedom, we can look up the critical value in the tt-distribution table.

The critical value for t0.05,36t_{0.05, 36} is approximately 1.688-1.688.

Step 6: Compare Test Statistic to Critical Value

  • Calculated tt: 4.22-4.22
  • Critical value: 1.688-1.688

Since 4.22<1.688-4.22 < -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 t4.22t \approx -4.22
  • Distribution: Xt36X \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 tt-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.

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Math Problem Analysis

Mathematical Concepts

Statistics
Hypothesis Testing
t-Distribution

Formulas

t = (X̄ - μ₀) / (s / √n)

Theorems

Central Limit Theorem
t-Test for Means

Suitable Grade Level

Grades 11-12