A production process is considered to be out of control if the produced parts have lengths with a standard deviation greater than .5 mm. A sample of 41 parts yields a sample standard deviation of .58 mm. At the .05 significance level, does this sample indicate that the process should be adjusted in order to correct the standard deviation of the product? Assume the lengths have a normal distribution.
To determine whether the process should be adjusted, we need to conduct a hypothesis test. The null hypothesis (H₀) states that the process is in control and the standard deviation is not greater than 0.5 mm, while the alternative hypothesis (H₁) states that the process is out of control and the standard deviation is greater than 0.5 mm.
Here is how to perform the hypothesis test:
Step 1: Set up hypotheses:
- Null hypothesis (H₀): σ ≤ 0.5 mm
- Alternative hypothesis (H₁): σ > 0.5 mm
Step 2: Select the significance level (α):
- Given in the question as 0.05 (or 5%).
Step 3: Find the test statistic:
- Since we are testing the standard deviation, we need to use a chi-square test statistic.
- The test statistic is calculated as follows: χ² = (n - 1) * (s / σ₀)²
where n = sample size, s = sample standard deviation, and σ₀ = hypothesized standard deviation under the null hypothesis.
Step 4: Determine the critical value or p-value:
- Since the alternative hypothesis is "greater than," we need to find the critical value from the right-tailed chi-square distribution with (n - 1) degrees of freedom and a significance level of α.
- The critical value represents the cutoff point beyond which we can reject the null hypothesis.
- Alternatively, we can calculate the p-value associated with the test statistic, which represents the probability of observing a test statistic as extreme or more extreme than the one obtained, assuming the null hypothesis is true.
Step 5: Compare the test statistic with the critical value or p-value:
- If the test statistic is greater than the critical value (or the p-value is less than α), we reject the null hypothesis and conclude that the process should be adjusted.
- If the test statistic is less than or equal to the critical value (or the p-value is greater than or equal to α), we fail to reject the null hypothesis and do not conclude that the process should be adjusted.
Now let's calculate the test statistic and compare it to the critical value or p-value.
Given values:
Sample size (n) = 41
Sample standard deviation (s) = 0.58 mm
Hypothesized standard deviation under the null hypothesis (σ₀) = 0.5 mm
Significance level (α) = 0.05 (or 5%)
Step 3: Calculate the test statistic:
χ² = (n - 1) * (s / σ₀)²
= (41 - 1) * (0.58 / 0.5)²
Calculating this, we get χ² ≈ 30.360
Step 4: Determine the critical value or p-value:
Since we have a right-tailed test, we need to find the critical value from the chi-square distribution table with (n - 1) degrees of freedom and a significance level of α = 0.05.
With (41 - 1) = 40 degrees of freedom, the critical value is approximately 56.327.
Alternatively, we can calculate the p-value associated with the test statistic using a chi-square distribution calculator, given the test statistic (χ²), degrees of freedom (df), and type of test (right-tailed).
Step 5: Compare the test statistic with the critical value or p-value:
- χ² > critical value: 30.360 > 56.327
- p-value < α: p-value < 0.05
Since the test statistic is less than the critical value (30.360 ≤ 56.327) and the p-value is greater than the significance level (p-value > 0.05), we fail to reject the null hypothesis.
Therefore, based on this sample, there is not enough evidence to suggest that the process should be adjusted to correct the standard deviation of the product.