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Improve Phase

Regression analysis and developing solutions for process improvement.

The Improve phase focuses on identifying and implementing solutions using regression analysis techniques. It covers simple linear regression, correlation analysis, multiple regression, residual analysis, data transformation including Box-Cox, and confidence/prediction intervals to optimize processes and reduce variation.
5 minutes 5 Questions

The Improve Phase is the fourth stage in the DMAIC (Define, Measure, Analyze, Improve, Control) methodology of Lean Six Sigma. This critical phase focuses on developing, testing, and implementing solutions to address the root causes identified during the Analyze Phase. The primary objective is to o…

Concepts covered: Correlation, Correlation Coefficient (r), Coefficient of Determination (R-squared), Simple Linear Regression, Regression Equations, Fitted Line Plot, Residuals Analysis, Residual Patterns, Non-Linear Regression, Multiple Linear Regression, Multiple Regression Coefficients, Adjusted R-Squared, Confidence Intervals in Regression, Prediction Intervals, Data Transformation, Box-Cox Transformation, Multicollinearity, Variance Inflation Factor (VIF)

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LSSGB - Improve Phase Example Questions

Test your knowledge of Improve Phase

Question 1

A Green Belt is investigating response time data from a call center that exhibits both positive skewness and increasing variance as mean values rise. The team applied a Box-Cox transformation with an optimal lambda of 0.25, but stakeholders are concerned about the interpretability of results when communicating findings to management. Which consideration should guide the decision between using the exact optimal lambda versus rounding to a more conventional transformation power?

Correct Answer: The confidence interval width around the optimal lambda determines whether conventional alternatives like log or fourth root provide statistically equivalent normalization

The correct answer addresses a fundamental principle in Box-Cox transformation application: the confidence interval around the optimal lambda value.

When performing a Box-Cox transformation, statistical software provides not just a point estimate for lambda, but also a confidence interval. If this confidence interval includes conventional transformation values (like 0 for log transformation, 0.5 for square root, or 0.25 for fourth root), then using the conventional value provides essentially equivalent statistical performance while offering significant practical advantages:

  1. Interpretability: Conventional transformations (log, square root, fourth root) are easier to explain to stakeholders and management
  2. Reversibility: Back-transformation to original units is more intuitive with conventional powers
  3. Comparability: Results can be more easily compared with other studies or benchmarks that use standard transformations

In this case, with an optimal lambda of 0.25, the fourth root transformation (which is exactly 0.25) may be within the confidence interval anyway, making it the natural choice.

The second option is incorrect because it overstates the rigidity required. Box-Cox transformations have uncertainty, and slight variations in lambda within the confidence interval do not meaningfully bias subsequent analyses.

The third option incorrectly suggests sample size should drive the choice between exact and conventional lambdas. Sample size affects the precision of the lambda estimate but doesn't create a rule about preferring conventional powers above certain thresholds.

The fourth option incorrectly implies that application type (control charting vs. regression) should be the primary decision factor. While different applications may have varying sensitivity to transformation choice, this is not the primary consideration when evaluating optimal vs. conventional lambda values.

Question 2

In multiple linear regression, what does a p-value of 0.03 for a predictor variable typically indicate?

Correct Answer: The predictor has a statistically significant relationship with the response variable at the 0.05 significance level

The correct answer states that the predictor has a statistically significant relationship with the response variable at the 0.05 significance level.

In multiple linear regression, the p-value for a predictor variable tests the null hypothesis that the coefficient for that predictor is equal to zero (meaning no relationship exists). A p-value of 0.03 means there is only a 3% probability of observing such a relationship by chance alone if the null hypothesis were true.

Since 0.03 is less than the commonly used significance level of 0.05 (α = 0.05), we reject the null hypothesis and conclude that the predictor has a statistically significant relationship with the response variable.

The other answers are incorrect for the following reasons:

  • The answer suggesting the predictor explains exactly 3% of variance is wrong because p-values do not indicate the percentage of variance explained. That would be measured by metrics like R-squared or partial R-squared values, not p-values.

  • The answer claiming a positive correlation coefficient of 0.97 is incorrect because p-values are not correlation coefficients. A p-value of 0.03 has no mathematical relationship to a correlation of 0.97.

  • The answer stating the predictor contributes 3% to the overall R-squared is also wrong because p-values do not measure contribution to R-squared. The p-value tests statistical significance, not the magnitude of contribution to explained variance.

Question 3

A quality engineer performs regression analysis between incoming material temperature and final product hardness. The resulting R² value is 0.45. A team member asks what percentage of the variation in hardness remains unexplained by the temperature variable. What is the correct response?

Correct Answer: 55% of the variation in hardness is not explained by the temperature predictor

The R² (coefficient of determination) value represents the proportion of variance in the dependent variable (hardness) that is explained by the independent variable (temperature).

When R² = 0.45, this means that 45% of the variation in hardness IS explained by temperature.

To find the unexplained variation, we calculate: 1 - R² = 1 - 0.45 = 0.55 or 55%.

Therefore, 55% of the variation in hardness is not explained by the temperature predictor. This unexplained portion represents other factors affecting hardness that are not captured by the temperature variable in the regression model.

The second option incorrectly states that 45% is unexplained, when actually 45% is the explained portion (the R² value itself).

The third option confuses the concept of R² with confidence intervals. R² measures explained variance, not the percentage of data points falling within confidence bounds.

The fourth option incorrectly interprets R² as relating to measurement error and correlation percentage. R² does not measure total measurement error; it measures the proportion of variance explained by the regression model.

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