IE533 - Industrial Applications of Statistics - HW-2 Ashwath Kumar

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IE533: Industrial Applications of Statistics Homework #2 Ashwath Kumar Purpose: Practice using software to perform paired t-test, ANOVA, multiple testing adjustments, General Linear Model, and Sample size calculation. Output: Please address the following items in a report (word doc or PDF) no longer than 4 pages. Succinctness and orderly formatting are prized and rewarded (0.5 Points). 1. Use the data set provided on Brightspace (HW 2 Data Problem 1) to perform a paired t-test to examine if the mean amount of methane produced in digestion is different between two baby food formulas. Include the estimated mean difference, 95% confidence interval for the mean difference, p-value from the t-test, and your conclusion about the findings in your output. Descriptive Statistics Sample N Mean StDev SE Mean C1 20 9.099 1.026 0.230 C2 20 6.658 1.074 0.240 Estimation for Paired Difference Mean StDev SE Mean 95% CI for μ_difference 2.441 1.525 0.341 (1.727, 3.155) μ_difference: mean of (C1 - C2) Test Null hypothesis H₀: μ_difference = 0 Alternative hypothesis H₁: μ_difference ≠ 0 T-Value P-Value 7.16 0.000 Inference: We reject the null hypothesis as the p-value is 0, which is below the 5% threshold. This indicates a difference in the amount of methane produced between the ginger formula and the new formula. 2. (2 Points) Generate a data set to mimic the baby food storage medium example from class (generate the mean taste score directly instead of separate taste tester scores for each food item). Use the following generating distributions to draw 25 observations for each medium: Jar ~ 𝑵(𝝁 = ?𝟓, 𝝈 = ?) Pouch ~ 𝑵(𝝁 = ?𝟎, 𝝈 = ?)

IE533: Industrial Applications of Statistics Homework #2 Ashwath Kumar Foil ~ 𝑵(𝝁 = ?𝟓, 𝝈 = ?) Perform a single factor ANOVA test for differences in the mean tastiness associated with the storage medium. Perform Tukey-Pairwise Comparisons among the means and check your model assumptions of Normality and Constant Variance. In your output include the ANOVA table, Tukey Output (table or plot), Normal P-P plot and constant variance plot (residuals by factor level). What are your conclusions from each of these four items? Analysis of Variance Source DF Adj SS Adj MS F-Value P-Value Factor 2 1299 649.54 21.13 0.000 Error 72 2213 30.74 Total 74 3512 Inference: P-value is less than alpha (=5%), therefore null is rejected in this case also. Tukey Pairwise Comparisons Grouping Information Using the Tukey Method and 95% Confidence Factor N Mean Grouping Jar 25 85.227 A Pouch 25 79.98 B Foil 25 75.034 C Means that do not share a letter are significantly different Inference : In Tukey's pairwise comparison, the differences observed for Foil – Pouch and Pouch - Jar are significantly smaller in comparison to the Foil – Jar pairwise comparison. This suggests that Pouch is closer to the mean of both foil and Jar. However, it is important to note that the foil-jar comparison exhibits a higher mean difference.

IE533: Industrial Applications of Statistics Homework #2 Ashwath Kumar Normal P-P Plot Conclusion: The proximity of mean data points to the reference line indicates that the data follows a normal distribution. Constant Variance Plot Conclusion: The symmetrical distribution observed in the Residual vs Fitted Value plot for all three storage mediums further confirms the normal distribution of the data. 3. (0.5 Point) Using the data from problem 2, refit the ANOVA using the Fisher LSD multiple comparison method. Include the output from the Fisher LSD test (table or plot), comment on how the result compares to the Tukey adjustment. Fisher Pairwise Comparisons Grouping Information Using the Fisher LSD Method and 95% Confidence Factor N Mean Grouping Jar 25 85.227 A Pouch 25 79.98 B Foil 25 75.034 C Means that do not share a letter are significantly different.

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IE533: Industrial Applications of Statistics Homework #2 Ashwath Kumar Inference: In Fischer, pairwise comparison plots difference for Foil – Pouch, and Pouch - Jar are much lesser compared to Foil – Jar pairwise comparison. It is much like that of Tukey. However, mean difference range is slightly lower compared to Tukey method for all three pairs 4. (0.5 Point) Using the data from problem 2, fit a general linear model with tastiness as the response and storage medium as the factor (you may have to transform the shape of your data to get it to work). Include the coefficient table in your output and interpret each coefficient . Regression Equation Total obs = 80.080 - 5.046 Medium_Foil + 5.147 Medium_Jar - 0.101 Medium_Pouch COEF: 80.0796 -5.0459 5.1470 -0.1012 Inference: The taste improves with extended food storage in the jar. Also, it diminishes with prolonged storage in foil and pouch. In contrast to foil, food stored in a pouch has superior taste. 5. (0.5 Point) Perform a sample size calculation assuming 3 levels, a maximum difference between means 8, power of 80%, and standard deviation of 3. Include a power curve or screen shot of computer output, how many replicates do you need for each factor level?

IE533: Industrial Applications of Statistics Homework #2 Ashwath Kumar No. of replicates for each factor level is 4 with the power of 80%. Results Maximum Difference Sample Size Target Power Actual Power 8 4 0.8 0.815048 6. Identify a published (0.25 Points – Optional) Identify a published experiment (journal article, white paper, blog, etc.) that uses a one-factor ANOVA to analyze their data. Give the hypothesis from their F-test and briefly describe their key finding (three lines or less). Journal Article: Understanding one-way ANOVA using conceptual figures - Tae Kyun Kim 1. Link F-test: 3.629 (bigger than 3.101). Hence, null hypothesis should be rejected. Insights: Further using ANOVA, various group pairings are compared through an additional step to determine specific group differences. This subsequent process is known as the post-hoc test.