Math Problem Statement

 A research engineer for a tire manufacturer is investigating tire life for a new rubber com-

pound and has built 16 tires and tested them to end-of-life in a road test. The sample mean and

standard deviation are 60,139.7 and 3645.94 kilometers. Can you conclude, using α= 0.05, that the

standard deviation of tire life is less than 4000 kilometers?

Solution

This problem involves testing the standard deviation of a population based on a sample, which can be done using a chi-squared test for the population variance.

Step-by-Step Solution:

  1. Formulate the Hypotheses:

    • Null Hypothesis (H0H_0): σ240002=16,000,000\sigma^2 \geq 4000^2 = 16,000,000 (the standard deviation is at least 4000 km).
    • Alternative Hypothesis (HaH_a): σ2<40002=16,000,000\sigma^2 < 4000^2 = 16,000,000 (the standard deviation is less than 4000 km).
  2. Test Statistic: The test statistic for a population variance is: χ2=(n1)s2σ02\chi^2 = \frac{(n-1)s^2}{\sigma_0^2} where:

    • n=16n = 16 (sample size),
    • s=3645.94s = 3645.94 (sample standard deviation),
    • σ02=40002=16,000,000\sigma_0^2 = 4000^2 = 16,000,000 (hypothesized population variance).

    Compute s2s^2: s2=3645.942=13,299,854.44s^2 = 3645.94^2 = 13,299,854.44

    Substituting values: χ2=(161)(13,299,854.44)16,000,000=15×13,299,854.4416,000,00012.47\chi^2 = \frac{(16-1)(13,299,854.44)}{16,000,000} = \frac{15 \times 13,299,854.44}{16,000,000} \approx 12.47

  3. Degrees of Freedom: The degrees of freedom for this test are df=n1=161=15df = n - 1 = 16 - 1 = 15.

  4. Critical Value: This is a one-tailed test with α=0.05\alpha = 0.05. From a chi-squared table or calculator, the critical value for χ0.05,152\chi^2_{0.05, 15} is: χcritical2=24.996\chi^2_{\text{critical}} = 24.996

  5. Decision Rule:

    • If χ2χcritical2\chi^2 \leq \chi^2_{\text{critical}}, reject H0H_0.
    • If χ2>χcritical2\chi^2 > \chi^2_{\text{critical}}, fail to reject H0H_0.
  6. Compare the Test Statistic to the Critical Value: χ2=12.47<24.996\chi^2 = 12.47 < 24.996

  7. Conclusion: Since the test statistic is less than the critical value, we reject the null hypothesis. At α=0.05\alpha = 0.05, there is sufficient evidence to conclude that the standard deviation of tire life is less than 4000 kilometers.


Do you have any questions or would you like me to explain any part in more detail?

Related Questions:

  1. What assumptions must be satisfied for the chi-squared test to be valid?
  2. How does increasing the sample size affect the power of this test?
  3. What would happen if the significance level α\alpha was changed to 0.01?
  4. Can you explain how to use a chi-squared distribution table?
  5. What are other applications of the chi-squared test for variance in engineering?

Tip:

For chi-squared tests, always ensure that the sample data comes from a normally distributed population, as the test is sensitive to deviations from normality.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Chi-Squared Test
Statistics
Variance

Formulas

\chi^2 = \frac{(n-1)s^2}{\sigma_0^2}
H_0: \sigma^2 \geq 4000^2 = 16,000,000
H_a: \sigma^2 < 4000^2 = 16,000,000

Theorems

Chi-Squared Distribution
One-Tailed Test

Suitable Grade Level

Undergraduate (Statistics or Engineering)