Questions: The piston diameter of a certain hand pump is 0.6 inch. The manager determines that the diameters are normally distributed, with a mean of 0.6 inch and a standard deviation of 0.005 inch. After recalibrating the production machine, the manager randomly selects 22 pistons and determines that the standard deviation is 0.0044 inch. Is there significant evidence for the manager to conclude that the standard deviation has decreased at the α=0.10 level of significance? What are the correct hypotheses for this test? The null hypothesis is H0 : The alternative hypothesis is H1 :

Solution Steps

We set up the hypotheses for the test as follows:

• Null Hypothesis: \( H_0: \sigma = 0.005 \)
• Alternative Hypothesis: \( H_1: \sigma < 0.005 \)

The test statistic for the chi-square test is calculated using the formula:

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

Substituting the values:

\[ \chi^2 = \frac{(22 - 1) \cdot (0.0044)^2}{(0.005)^2} = 16.2624 \]

To find the P-value, we calculate the cumulative distribution function (CDF) of the chi-square distribution for the test statistic with \( n-1 = 21 \) degrees of freedom:

\[ P = P(\chi^2(21) \leq 16.2624) = 0.2453 \]

The critical value for a left-tailed test at the significance level \( \alpha = 0.10 \) with 21 degrees of freedom is:

\[ \text{Critical Value} = 13.2396 \]

Since the test statistic \( \chi^2 = 16.2624 \) is greater than the critical value \( 13.2396 \) and the P-value \( 0.2453 \) is greater than \( \alpha = 0.10 \), we fail to reject the null hypothesis.

Final Answer