Certified Six Sigma Black Belt Hypothesis Testing Questions and Answers — Questions and Answers
Question 1: A Six Sigma Black Belt is analyzing a process improvement aimed at reducing the average processing time. The null hypothesis (H₀) is that the mean time is unchanged (μ = 30 minutes), and the alternative hypothesis (H₁) is that the mean time has been reduced (μ < 30 minutes). After conducting the test, the calculated p-value is 0.04. Using a significance level (α) of 0.05, what is the correct conclusion?
- Fail to reject the null hypothesis; there is not enough evidence to conclude the mean time has been reduced.
- Reject the null hypothesis; there is sufficient evidence to conclude the mean time has been reduced. (Correct answer)
- Reject the null hypothesis; there is sufficient evidence to conclude the mean time has increased.
- The p-value is greater than the significance level, so the alternative hypothesis is accepted.
Correct answer: Reject the null hypothesis; there is sufficient evidence to conclude the mean time has been reduced.
The rule for making a decision using a p-value is: If the p-value is less than or equal to the significance level (α), we reject the null hypothesis (H₀). In this case, the p-value (0.04) is less than α (0.05), leading to the rejection of H₀ in favor of the alternative hypothesis (H₁), which states that the mean time has been reduced.
Question 2: In the context of a hypothesis test for a process improvement project, what is the consequence of committing a Type I error?
- Correctly concluding that the process has not improved.
- Concluding that the process has not improved when, in fact, it has.
- Incorrectly failing to reject a false null hypothesis.
- Concluding that the process has improved when, in fact, it has not. (Correct answer)
Correct answer: Concluding that the process has improved when, in fact, it has not.
A Type I error (also known as alpha risk or a false positive) occurs when we incorrectly reject a true null hypothesis. In a process improvement context, the null hypothesis often represents the 'status quo' or 'no change.' Therefore, rejecting a true null hypothesis means concluding there was an improvement or change when, in reality, no change occurred. This can lead to implementing an ineffective solution.
Question 3: A project team wants to determine if there is a statistically significant difference in the average customer satisfaction scores (measured on a continuous scale from 1 to 10) among four different service centers. Which of the following statistical tests is the most appropriate to use?
- Analysis of Variance (ANOVA) (Correct answer)
- Paired t-test
- 2-Sample t-test
- Chi-Square test
Correct answer: Analysis of Variance (ANOVA)
Analysis of Variance (ANOVA) is used to compare the means of three or more groups to determine if at least one group mean is statistically different from the others. A 2-Sample t-test is used for two groups, a paired t-test is for comparing two related samples, and a Chi-Square test is used for categorical data, not continuous data like satisfaction scores.
Question 4: A Black Belt is investigating a claim that a new supplier's raw material leads to a different proportion of defective products compared to the old supplier. To determine if there is a statistically significant difference in the defect rates between the two suppliers, which hypothesis test is most suitable?
- F-test for variances
- One-way ANOVA
- 2-Proportion test (Correct answer)
- 2-Sample t-test
Correct answer: 2-Proportion test
The goal is to compare the proportions (or percentages) of defective products from two independent populations (old supplier vs. new supplier). The 2-Proportion test is specifically designed for this purpose. A 2-Sample t-test compares means of continuous data, an F-test compares variances, and ANOVA compares the means of more than two groups.
Question 5: Which of the following statements correctly describes the relationship between statistical power, sample size, and Type II error (beta risk)?
- Decreasing the sample size increases power and decreases the risk of a Type II error.
- Increasing the sample size increases power and decreases the risk of a Type II error. (Correct answer)
- Increasing the sample size decreases power and increases the risk of a Type II error.
- Sample size has no direct effect on power or the risk of a Type II error.
Correct answer: Increasing the sample size increases power and decreases the risk of a Type II error.
The power of a test (1 - β) is the probability of correctly detecting a real effect. A larger sample size provides more information and reduces sampling error, making it easier to detect a true effect. This increases the power of the test and, by definition, decreases the probability of a Type II error (β), which is the error of failing to detect a real effect.
Question 6: A quality manager wants to verify a machine manufacturer's claim that the variance of their machine's output is no more than 0.5 mm². The manager will conduct a hypothesis test to challenge this claim. What is the correct formulation for the null (H₀) and alternative (H₁) hypotheses?
- H₀: μ ≤ 0.5, H₁: μ > 0.5
- H₀: σ² = 0.5, H₁: σ² ≠ 0.5
- H₀: σ² > 0.5, H₁: σ² ≤ 0.5
- H₀: σ² ≤ 0.5, H₁: σ² > 0.5 (Correct answer)
Correct answer: H₀: σ² ≤ 0.5, H₁: σ² > 0.5
The null hypothesis (H₀) always contains the condition of equality and represents the status quo or the claim being tested. In this case, the claim is that the variance (σ²) is no more than 0.5 (σ² ≤ 0.5). The alternative hypothesis (H₁) is what the quality manager wants to prove, which is that the variance is actually greater than the claimed value (σ² > 0.5). The test is for variance (σ²), not the mean (μ).
Certified Six Sigma Black Belt Hypothesis Testing Questions and Answers
A Six Sigma Black Belt is analyzing a process improvement aimed at reducing the average processing time.
The null hypothesis (H₀) is that the mean time is unchanged (μ = 30 minutes), and the alternative hypothesis (H₁) is that the mean time has been reduced (μ < 30 minutes).
After conducting the test, the calculated p-value is 0.04.
Using a significance level (α) of 0.05, what is the correct conclusion?