Six Sigma LSSMBB Exam Dumps & Practice Test Questions

Question 1:

If you find that data from one of two suppliers is not normally distributed and cannot be transformed, what should you do next before comparing shield thickness between suppliers?

A Use the Shapiro-Wilk test
B Continue with the t-test
C Apply a non-parametric test
D Stop the analysis

Correct answer: C

Explanation:

When conducting statistical tests like a t-test, it is crucial to ensure that the data meets certain assumptions, with normality being one of the most important. The t-test assumes that the data in both groups is normally distributed. If you discover that the data from one supplier's group violates this assumption and cannot be made normal through transformation, the results from a t-test could be invalid or misleading.

In this situation, the best alternative is to use a non-parametric test. Non-parametric tests do not require the data to follow a normal distribution and are therefore suitable when normality assumptions are violated. For comparing two independent groups, the Mann-Whitney U test (also called the Wilcoxon rank-sum test) is a widely accepted non-parametric alternative. Instead of comparing means, this test compares medians or the general distribution of ranks between groups.

Option A, the Shapiro-Wilk test, is actually a tool used to check for normality. While useful at the beginning to assess whether the data is normally distributed, it is not the next step after you have already determined that normality is violated.

Option B, proceeding with the t-test despite non-normal data, is inappropriate because the t-test’s validity hinges on the normality assumption. Ignoring this can lead to incorrect conclusions.

Option D, discontinuing the analysis, is unnecessary because alternative methods like non-parametric tests exist that allow you to proceed and obtain meaningful results.

In summary, when you confirm that data from one group is not normally distributed and cannot be transformed, switching to a non-parametric test like the Mann-Whitney U test is the correct course of action. This ensures your analysis remains statistically sound and reliable.

Question 2:

When comparing the average test results from three machines running the same test, with data that is normally distributed and has equal variances, which statistical test should you use?

A Kruskal-Wallis test
B Chi-Square test
C ANOVA
D Bartlett’s or Levene’s test

Correct answer: C

Explanation:

When you have data from three or more groups—in this case, three parallel machines performing the same test—the appropriate statistical test depends on the nature of the data and underlying assumptions. Here, the data is normally distributed, and the variances within each machine’s results are equal. These conditions satisfy the assumptions required to use Analysis of Variance (ANOVA).

ANOVA is specifically designed to test whether the means of three or more groups are statistically significantly different from one another. The null hypothesis states that all group means are equal, while the alternative hypothesis posits that at least one group mean differs. Since your data meets the assumptions of normality and homogeneity of variance, ANOVA provides a valid, powerful way to compare these machine averages simultaneously.

Option A, the Kruskal-Wallis test, is a non-parametric alternative used when normality or equal variance assumptions are violated. Since those assumptions hold here, Kruskal-Wallis is less appropriate.

Option B, the Chi-Square test, is designed for categorical data, such as frequencies or counts, and is not suitable for comparing means of continuous data.

Option D, Bartlett’s or Levene’s test, are tests used to assess whether variances across groups are equal—a prerequisite check before running ANOVA. They do not test for differences in means themselves.

In conclusion, since your data is normally distributed with equal variances across the three machines, ANOVA is the correct choice for determining if there are statistically significant differences in the average test values.

Question 3:

Which statistical test is commonly employed to analyze the relationship between two or more categorical variables?

A. Kruskal-Wallace Test
B. Shapiro-Wilkes Test
C. Student’s t-Test
D. Chi-Square Test

Correct Answer: D

Explanation:

When dealing with categorical data, determining whether two or more variables are associated or independent requires a test designed for discrete, non-numeric data. The Chi-Square Test is the most commonly used method for this purpose. It assesses whether the distribution of observed frequencies across categories deviates significantly from what would be expected if the variables were independent.

The Chi-Square Test operates by comparing actual observed counts of occurrences in each category against expected counts, which are calculated under the assumption that there is no relationship between the variables. The resulting Chi-Square statistic measures the magnitude of deviation between observed and expected frequencies. This statistic follows a Chi-Square distribution, allowing researchers to calculate a p-value and assess significance. If the p-value is below a predefined threshold (often 0.05), it indicates that the variables likely have a statistically significant association.

Looking at the other options clarifies why they are less suitable:

• The Kruskal-Wallace Test is a non-parametric method used to compare medians among three or more independent groups. It is not designed to analyze associations between categorical variables but rather to compare distributions of continuous or ordinal data.

• The Shapiro-Wilkes Test is used exclusively to test whether a dataset is normally distributed. It does not measure relationships between variables, especially categorical ones.

• The Student’s t-Test compares the means of two groups and assumes normally distributed continuous data. It is not applicable for testing relationships among categorical variables.

Thus, the Chi-Square Test is the correct choice because it is specifically designed to examine the association or independence of multiple discrete or categorical variables. Its widespread use in fields like social sciences, biology, and market research underscores its importance in analyzing categorical data.

Question 4:

An engineer aims to increase the average measurement of a product characteristic from 850 to more than 855. The standard deviation for both the current and proposed processes is assumed to be 7.7. The engineer wants to test if the new process's average is statistically significantly greater by more than 5 units compared to the old process. 

What are the appropriate null and alternative hypotheses?

A. Ho: μ New - μ Old ≤ 5, Ha: μ New - μ Old > 5
B. Ho: μ New - μ Old = 5, Ha: μ New - μ Old ≠ 5
C. Ho: μ New = 850, Ha: μ New > 850
D. Ho: σ New ≥ 7.7, Ha: σ New > 7.7

Correct Answer: A

Explanation:

When testing for improvements in a process mean, hypothesis testing typically involves setting a null hypothesis that represents the status quo or no significant improvement, and an alternative hypothesis representing the expected improvement. Here, the engineer’s goal is to verify that the new process mean exceeds the old mean by more than 5 units.

The null hypothesis (Ho) usually embodies the assumption that any observed difference is not statistically significant or that the difference does not surpass the threshold of interest. Therefore, Ho states that the difference between the new and old means is less than or equal to 5 (μ New - μ Old ≤ 5). This means the new process does not show a meaningful improvement beyond this margin.

The alternative hypothesis (Ha) reflects the engineer’s intent—to prove that the new process mean exceeds the old mean by more than 5 (μ New - μ Old > 5). Since the focus is on testing for an increase, a one-tailed test is appropriate here.

Examining the other options highlights why they do not fit:

• Option B posits the null hypothesis that the difference equals exactly 5 and the alternative that it is different (either higher or lower). This represents a two-tailed test, which does not align with the engineer’s interest in only increases above 5.
  
  

• Option C tests whether the new mean is greater than 850 but does not specify a difference of more than 5 units compared to the old process. This misses the specific target difference the engineer wants to test.
  
  

• Option D focuses on testing changes in the standard deviation rather than the mean, which is not relevant here since the engineer is concerned about average improvements.

In conclusion, option A correctly formulates the hypotheses to test whether the new process achieves a statistically significant improvement greater than 5 units, matching the engineer’s objective and appropriate statistical methodology.

Question 5:

Which statistical method is suitable for analyzing the relationship between one continuous independent variable (X) and one continuous dependent variable (Y)?

A. T-test
B. Chi-Square test
C. One-Way ANOVA
D. Correlation

Correct Answer: D

Explanation:

When studying how two continuous variables relate to each other, the best statistical approach is correlation analysis. This method quantifies the strength and direction of the linear association between a continuous input variable (X) and a continuous output variable (Y). The most commonly used correlation measure is the Pearson correlation coefficient, which ranges from -1 to 1. A value close to 1 indicates a strong positive linear relationship, near -1 indicates a strong negative linear relationship, and around 0 indicates little to no linear association.

Correlation is ideal because it directly measures the degree to which changes in one variable predict changes in another when both variables are measured on continuous scales, such as height, temperature, or test scores.

Let’s review why the other options are not suitable here:

• T-test (A) compares the means between two groups, typically involving one categorical independent variable and one continuous dependent variable. Since both variables here are continuous, the T-test is not applicable.

• Chi-Square test (B) is used for testing relationships between categorical variables, especially in contingency tables. Since both variables in this scenario are continuous, this test does not fit.

• One-Way ANOVA (C) compares means across more than two groups and requires a categorical independent variable with multiple levels and a continuous dependent variable. This question involves no categorical variables, so ANOVA is inappropriate.

In summary, when your data consist of one continuous predictor and one continuous outcome variable, correlation provides the most meaningful insight into their relationship. Hence, D is the correct choice.

Question 6:

In the context of statistical hypothesis testing, what does beta risk (β) represent?

A. The chance of rejecting the null hypothesis when it is true
B. Always fixed at 0.10
C. Influenced by the cost of sampling
D. The chance of failing to reject the null hypothesis when it is false

Correct Answer: D

Explanation:

Beta risk, symbolized by β, is the probability of making a Type II error in hypothesis testing. This type of error occurs when a test fails to reject the null hypothesis even though the null hypothesis is actually false. Essentially, it’s the risk that the test misses detecting a real effect or difference that truly exists.

When conducting hypothesis tests, the null hypothesis is usually presumed true unless there is sufficient evidence against it. However, sometimes the test might not be powerful enough to detect a real difference or relationship, causing a Type II error. The value of β quantifies how likely this error is to occur.

The complement of beta risk (1 - β) is called the power of the test, which is the probability of correctly rejecting a false null hypothesis.

Now, examining the other options:

• Option A confuses beta risk with alpha risk (α), which is the probability of rejecting the null hypothesis when it is actually true (Type I error).

• Option B is incorrect because beta risk is not a fixed value like 0.10; it varies based on factors such as sample size, effect size, and significance level.

• Option C is also inaccurate since beta risk is not directly influenced by sampling costs but by statistical considerations such as sample size and test sensitivity.

To conclude, beta risk (β) specifically measures the likelihood of failing to reject a false null hypothesis, making D the correct and precise answer.

Question 7:

Sigma Saving and Loans wants to determine if their average loan processing cycle time is less than 9.5 days. Which statistical test should they use to evaluate this claim?

A. A one-sample t-test
B. A two-sample t-test
C. A one-way ANOVA
D. A chi-square test of means

Correct answer: A

Explanation:

Choosing the correct statistical test depends on the nature of the data and the research question. Here, Sigma Saving and Loans aims to test if the average cycle time for processing loans is less than 9.5 days. This is a comparison of a single sample mean against a known or hypothesized value (9.5 days).

In this scenario, the one-sample t-test is the appropriate method because it is designed to compare the mean of one sample to a specific number or population mean. This test is especially useful when the sample size is relatively small and when the population standard deviation is unknown—both common situations in practical business data analysis.

The other options are not suitable for this case:

• A two-sample t-test compares the means between two independent groups or samples, such as two different branches or loan types. Since the question only involves one sample compared to a fixed number, this test does not apply.

• A one-way ANOVA is used to compare means across three or more groups to see if at least one group mean differs significantly. There is no indication of multiple groups here.

• The chi-square test of means is not appropriate because chi-square tests apply to categorical data to examine frequencies or proportions, not continuous data like cycle time.

Since the goal is to determine if the current average cycle time is significantly less than 9.5 days, the one-sample t-test (option A) is the correct choice. This test will measure whether the observed sample mean is statistically lower than the hypothesized average, providing the CEO with a meaningful decision-making tool based on data.

Question 8:

Two random samples from the same population are taken, one with size n=10 and the other with size n=100. Separate two-sided confidence intervals for the mean are calculated. How will these intervals compare?

A. The confidence interval for n=10 will be smaller.
B. The confidence interval for n=10 will be larger.
C. Both confidence intervals will be the same size.
D. There is insufficient information to determine.

Correct answer: B

Explanation:

The width of a confidence interval (CI) for a population mean depends heavily on the sample size and variability in the data. The confidence interval formula is generally:

CI=xˉ±zα/2×sn\text{CI} = \bar{x} \pm z_{\alpha/2} \times \frac{s}{\sqrt{n}}CI=xˉ±zα/2​×n​s​

Where:

• xˉ\bar{x}xˉ is the sample mean,

• zα/2z_{\alpha/2}zα/2​ is the critical value from the standard normal distribution (based on the confidence level),

• sss is the sample standard deviation, and

• nnn is the sample size.

The key factor influencing the interval width is the standard error, calculated as sn\frac{s}{\sqrt{n}}n​s​. Since the denominator involves the square root of the sample size, larger sample sizes lead to smaller standard errors.

For the two samples:

• When n=10n=10n=10, the square root is approximately 3.16, making the standard error relatively larger. A larger standard error means more uncertainty around the estimate, thus a wider confidence interval.
  
  

• When n=100n=100n=100, the square root is 10, so the standard error is much smaller, producing a narrower confidence interval that provides a more precise estimate of the population mean.

Since the critical value and confidence level are the same for both intervals, the difference in width comes solely from the sample size. Therefore, the confidence interval based on the smaller sample n=10n=10n=10 will be larger (wider) due to higher variability and less precision, while the interval for n=100n=100n=100 will be narrower and more reliable.

Hence, the correct answer is B—the confidence interval for the smaller sample size will be larger.

Question 9:

What is the main objective of conducting a screening experiment within the Design of Experiments (DOE) framework?

A. To determine the optimal input factor levels to maximize the response
B. To identify and separate the few critical factors from many insignificant ones
C. To compare different levels of a single input factor
D. To find input settings that yield a product with robust performance

Correct answer: B

Explanation:

In the context of Design of Experiments (DOE), a screening experiment is primarily designed to identify which factors significantly affect the outcome and to distinguish these from those that have minimal or no effect. The goal is to separate the “vital few” from the “trivial many.” This step is crucial for efficiently focusing further experimentation and resources on the variables that truly influence the response.

Screening experiments typically occur early in the DOE process when many potential factors may influence the response, but there is uncertainty about which are most important. These experiments help reduce the complexity by eliminating factors that don’t impact the process or product meaningfully, enabling researchers to focus on optimizing and modeling the critical few factors later.

For example, in a manufacturing process, you may start by testing factors such as temperature, pressure, and speed to identify which of these has the most influence on product quality. The factors deemed insignificant can be excluded from more complex and resource-intensive experiments.

To clarify why the other options are incorrect:
Option A refers to optimization, which is typically the focus of response surface or optimization experiments after the vital factors are identified. Screening is not aimed at finding the best factor levels but at finding which factors matter.
Option C focuses on examining levels of a single factor. Screening experiments usually test multiple factors simultaneously to determine their overall influence rather than isolate one factor.
Option D relates to robust design experiments aimed at minimizing variability and ensuring consistent product quality, which comes after identifying important factors.

Therefore, Option B correctly states the purpose of a screening experiment—to identify the most influential factors out of many candidates and narrow down the scope for subsequent study.

Question 10:

Referring to the DOE results and using the Hierarchy of Effects model at an alpha level of 0.10, which factors and interactions should be retained in the experimental model?

A. Temperature, Time, and the Temperature × Pressure interaction
B. Temperature, Time, Pressure, and the Temperature × Pressure interaction
C. Time and the main effects of Temperature and Pressure
D. Temperature and Time only

Correct answer: A

Explanation:

When interpreting Design of Experiments (DOE) results, the Hierarchy of Effects principle guides which terms—main effects and interactions—should be included in the final model. This principle recommends keeping significant main effects and important interactions to ensure the model accurately reflects the factors influencing the response.

The alpha level of 0.10 means that any factor or interaction with a p-value less than 0.10 is considered statistically significant enough to remain in the model. This threshold is slightly more lenient than the conventional 0.05, allowing some borderline significant terms to be included.

Given this, the terms Temperature (Temp), Time, and the interaction Temperature × Pressure (TempPressure) are retained because their p-values fall below 0.10, indicating they meaningfully affect the response variable.

Let’s consider why the other options are less appropriate:
Option B includes the main effect Pressure even though it is not statistically significant at the 0.10 level, leading to a more complex model without improving explanatory power.
Option C mixes main effects and excludes the important interaction, which may lead to missing key combined effects.
Option D leaves out the interaction term, potentially oversimplifying the model and ignoring important factor interactions.

By selecting Option A, the model retains the most influential terms that meet the statistical significance criteria, balancing model accuracy and simplicity. This approach optimizes predictive performance while avoiding unnecessary complexity.

In summary, following the Hierarchy of Effects with alpha = 0.10, the terms in Option A are the statistically supported and logically consistent factors to keep in the DOE model.