Math Problem Statement

The piston diameter of a certain hand pump is 0.5 inch. The manager determines that the diameters are normally​ distributed, with a mean of 0.5 inch and a standard deviation of 0.006 inch. After recalibrating the production​ machine, the manager randomly selects 27 pistons and determines that the standard deviation is 0.0052 inch. Is there significant evidence for the manager to conclude that the standard deviation has decreased at the α=0.05 level of​ significance? Calculate the value of the test statistic.

Solution

To determine if there is significant evidence that the standard deviation has decreased, we can perform a hypothesis test for the variance using the Chi-square distribution.

Step 1: Set Up Hypotheses

  • Null Hypothesis (H0H_0): The standard deviation has not decreased (i.e., σ=0.006\sigma = 0.006 inch).
  • Alternative Hypothesis (HaH_a): The standard deviation has decreased (i.e., σ<0.006\sigma < 0.006 inch).

Step 2: Calculate the Test Statistic

The test statistic for the variance is calculated using the formula:

χ2=(n1)s2σ02\chi^2 = \frac{(n-1)s^2}{\sigma_0^2}

Where:

  • nn = sample size = 27
  • ss = sample standard deviation = 0.0052 inch
  • σ0\sigma_0 = population standard deviation under the null hypothesis = 0.006 inch

First, we need to compute s2s^2 and σ02\sigma_0^2:

  • s2=(0.0052)2=0.00002704s^2 = (0.0052)^2 = 0.00002704
  • σ02=(0.006)2=0.000036\sigma_0^2 = (0.006)^2 = 0.000036

Now, substituting the values into the Chi-square formula:

χ2=(271)(0.00002704)0.000036=26×0.000027040.000036\chi^2 = \frac{(27-1)(0.00002704)}{0.000036} = \frac{26 \times 0.00002704}{0.000036}

Calculating this:

χ2=0.000703040.00003619.5\chi^2 = \frac{0.00070304}{0.000036} \approx 19.5

Step 3: Determine the Critical Value

For a one-tailed test at the α=0.05\alpha = 0.05 level of significance with n1=26n-1 = 26 degrees of freedom, we can find the critical value from the Chi-square distribution table.

Using a Chi-square table or calculator, we find:

  • Critical value at df=26df = 26 and α=0.05\alpha = 0.05 is approximately 14.Margins14. Margins.

Step 4: Conclusion

  • Since our calculated χ2\chi^2 (approximately 19.5) is greater than the critical value (approximately 14. Margins), we do not reject the null hypothesis.

Summary

  • Test statistic: χ219.5\chi^2 \approx 19.5
  • Decision: There is insufficient evidence at the α=0.05\alpha = 0.05 level to conclude that the standard deviation has decreased.

Would you like any more details or have further questions? Here are some related topics you might consider exploring:

  1. What are the assumptions of the Chi-square test?
  2. How to interpret the p-value in this context?
  3. What would happen if the sample size were larger?
  4. How can we improve the reliability of the test results?
  5. What other statistical tests could be used to analyze this data?

Tip: Always check the assumptions of normality and independence before conducting hypothesis tests for variance.

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Math Problem Analysis

Mathematical Concepts

Statistics
Hypothesis Testing
Chi-Square Distribution

Formulas

\(\chi^2 = \frac{(n-1)s^2}{\sigma_0^2}\)

Theorems

Chi-Square Test for Variance

Suitable Grade Level

Grades 11-12