SOLUTION: 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
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Question 1169147: 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 0.004 inch. After recalibrating the production machine, the manager randomly selects 21 pistons and determines that the standard deviation is 0.0035 inch. Is there significant evidence for the manager to conclude that the standard deviation has decreased at the a-0.01 level of significance?
ho: o= 0.0004
h1: o<0.004
Need to find x to the second power
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Here's how to solve this problem step-by-step:
**1. Define the Hypotheses and Significance Level:**
* Null Hypothesis (H₀): σ = 0.004 inch (The standard deviation has not decreased)
* Alternative Hypothesis (H₁): σ < 0.004 inch (The standard deviation has decreased)
* Significance level (α) = 0.01
**2. Gather the Data:**
* Sample size (n) = 21
* Sample standard deviation (s) = 0.0035 inch
* Population standard deviation (σ₀) = 0.004 inch
**3. Calculate the Chi-Square Test Statistic:**
* The formula for the chi-square test statistic is:
χ² = (n - 1) * s² / σ₀²
* Plug in the values:
χ² = (21 - 1) * (0.0035)² / (0.004)²
χ² = 20 * (0.00001225) / (0.000016)
χ² = 20 * (1225 / 1600)
χ² = 20 * 0.765625
χ² = 15.3125
**4. Determine the Degrees of Freedom:**
* Degrees of freedom (df) = n - 1 = 21 - 1 = 20
**5. Find the Critical Chi-Square Value:**
* Since this is a left-tailed test (H₁: σ < 0.004), we need to find the critical chi-square value from a chi-square distribution table with 20 degrees of freedom and an alpha level of 0.01.
* Using a chi-square table or calculator, the critical chi-square value is approximately 8.260.
**6. Compare the Test Statistic to the Critical Value:**
* Our calculated chi-square test statistic (15.3125) is greater than the critical chi-square value (8.260).
**7. Make a Decision:**
* Because our calculated chi-square value (15.3125) is *not* less than the critical value (8.260), we fail to reject the null hypothesis.
**8. Conclusion:**
* There is *not* significant evidence at the α = 0.01 level to conclude that the standard deviation of the piston diameters has decreased after recalibrating the production machine.
**Therefore, x to the second power is 15.3125**
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