Lean Six Sigma Black Belt Analyze Phase: Hypothesis Testing 1 Flashcards
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A Black Belt runs a hypothesis test and obtains a p-value of 0.03 with alpha set at 0.05. What is the correct conclusion?
Answer: Reject the null hypothesis because the p-value is less than alpha
When the p-value is less than the chosen significance level (alpha), the evidence against the null hypothesis is sufficient to reject it. Here, 0.03 < 0.05, so H₀ is rejected. Note that we 'fail to reject' or 'reject' H₀ — we never formally 'accept' it.
A Black Belt measures cycle time on the same 20 workstations before and after a process change. Which hypothesis test is most appropriate?
Answer: Paired t-test
A paired t-test is used when the same subjects or units are measured twice (before and after). It accounts for the natural correlation between paired observations, reducing variability and increasing statistical power compared to treating the groups as independent.
Before applying a parametric hypothesis test such as a t-test, a Black Belt should use the Anderson-Darling test to verify which assumption?
Answer: That the data is approximately normally distributed
Parametric tests assume the underlying data (or residuals) are approximately normally distributed. The Anderson-Darling test is one of the most common normality tests used in Minitab and Six Sigma workflows to verify this assumption before proceeding.
A Black Belt increases the significance level (alpha) from 0.05 to 0.10. What is the direct statistical consequence?
Answer: The probability of a Type I error increases
Alpha is by definition the probability of committing a Type I error (rejecting a true null hypothesis). Raising alpha from 0.05 to 0.10 directly raises the Type I error rate. As a side effect, power increases and the Type II error rate decreases, but the primary direct consequence is an increased Type I error probability.
A Black Belt wants to determine whether defect type (scratch, dent, discoloration) is independent of production shift (day, evening, night). Which test is most appropriate?
Answer: Chi-square test of independence
The chi-square test of independence evaluates whether two categorical variables are associated by comparing observed cell frequencies to expected frequencies under the assumption of independence. Both defect type and production shift are categorical, making this the correct choice.
A dataset severely violates the normality assumption and the sample size is too small to invoke the Central Limit Theorem. Which non-parametric test is the appropriate alternative to the independent two-sample t-test?
Answer: Mann-Whitney U test
The Mann-Whitney U test (also called the Wilcoxon rank-sum test) is the non-parametric equivalent of the independent two-sample t-test. It compares the rank-ordered distributions of two independent groups without requiring normally distributed data, making it suitable when normality is violated.