Trial Test Answers - T-tests & Probability

TRIAL TEST ANSWERS Part I — Multiple-choice questions T-tests & Probability (1–20) 1. B. Student’s t 11. A. 𝑃(𝐴 ∣ 𝐵) = 2. A. Two means 3. A. Form the difference values 4. A. Probability of an event given another event has occurred 5. A. Chance of rain given cloudy weather 6. A. 𝑃(𝐵 ∣ 𝐴) = 𝑃(𝐴∩𝐵) 𝑃(𝐴) 7. A. 𝑃(𝐴) > 0 8. D. All of the above 9. A. 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴) 𝑃(𝐵 ∣ 𝐴) 10. A. Update probabilities when new information becomes available 𝑃(𝐵∣𝐴) 𝑃(𝐴) 𝑃(𝐵) 12. A. Determining the probability a defective item came from a specific supplier 13. A. Visualizing conditional probabilities 14. A. (3/6) 15. A. 0.03 16. A. Bayes’ theorem 17. A. 0 ≤ 𝑃(𝐴) ≤ 1 18. A. Mutually exclusive events 19. A. 0.5 20. A. Mean difference ÷ standard error Regression & Correlation (21–40) 21. A. Strength and direction of a linear relationship 29. A. The product of the standard deviations 22. A. They tend to increase together 30. A. Strong negative correlation 23. A. Higher x values correspond to lower y values 31. A. Predict y from x 24. A. –1 to +1 25. A. No linear association 26. A. Whether two variables move together 27. A. x and y tend to rise together 28. A. x increases while y decreases 32. A. 𝑦 = 𝑎𝑥 + 𝑏 33. A. Change in y for each unit change in x 34. A. Predicted value of y when x = 0 35. A. 𝑒𝑖 = 𝑦𝑖 − 𝑦̂𝑖 36. A. Sum of squared residuals 37. A. Proportion of variance in y explained by x 39. A. The correlation coefficient 40. A. 𝑅 2 = 1 38. A. 85% of variation in y is explained by x Part II — Assignment questions Assignment 1: Customer service response time (Paired t-test) Goal: Test whether the software upgrade reduces average response time. 1. Hypotheses Let 𝑑 = Before − After (positive (d) means time decreased after upgrade). • Null hypothesis 𝐻0 : 𝜇𝑑 = 0 (no change in mean response time) • Alternative hypothesis 𝐻1 : 𝜇𝑑 = 0 (mean response time decreased after upgrade) (You could also phrase it as After < Before.) 2. Differences, mean difference, and SD of differences Compute 𝑑𝑖 = Before𝑖 − After𝑖 : Agent 1 2 3 4 5 6 7 8 9 10 • Number of pairs: (n = 10) • Sum of differences: Before 12 15 14 16 13 17 11 18 14 16 After 10 13 11 14 12 15 10 15 12 14 di = Before – After 2 2 3 2 1 2 1 3 2 2 ∑𝑑𝑖 = 2 + 2 + 3 + 2 + 1 + 2 + 1 + 3 + 2 + 2 = 20 • Mean difference: 𝑑ˉ = ∑𝑑𝑖 20 = =2 𝑛 10 Now compute the sample standard deviation of differences: Differences from mean (2): (0, 0, 1, 0, −1, 0, −1, 1, 0, 0) Squares: (0, 0, 1, 0, 1, 0, 1, 1, 0, 0) Sum of squared deviations: ∑(𝑑𝑖 − 𝑑ˉ )2 = 4 • Sample variance: 𝑠𝑑2 • ∑(𝑑𝑖 − 𝑑ˉ )2 4 = = ≈ 0.4444 𝑛−1 9 Sample standard deviation: 𝑠𝑑 = √0.4444 ≈ 0.667 3. t-statistic and standard error • Standard error of mean difference: 𝑆𝐸𝑑ˉ = • 𝑠𝑑 √𝑛 0.667 √10 ≈ 0.667 ≈ 0.211 3.162 t-statistic: 𝑡= • = 𝑑ˉ − 0 2 = ≈ 9.49 𝑆𝐸𝑑ˉ 0.211 Degrees of freedom: 𝑑𝑓 = 𝑛 − 1 = 9 4. p-value For 𝑡 ≈ 9.49 with 𝑑𝑓 = 9: • The p-value (one-sided) is much smaller than 0.001. • Even a two-sided test would give 𝑝 ≪ 0.001. So (p) is effectively ≈ 0 at normal significance levels like 0.05 or 0.01. 5. Interpretation • Since (p) is extremely small, we reject 𝐻0 . • There is strong statistical evidence that the software upgrade reduced average response time. • In practical terms, average response time decreased by about 2 minutes per inquiry. Assignment 2: Shipment error analysis (Conditional probability + Bayes) Given: • P(X) = 0.55 (Center X) • P(Y) = 0.45 (Center Y) • Error rates: o 𝑃(Error ∣ 𝑋) = 0.06 o 𝑃(Error ∣ 𝑌) = 0.12 1. Tree diagram structure (in words) First level: source • Branch 1: Center X with probability 0.55 • Branch 2: Center Y with probability 0.45 Second level: outcome for each source • • From X: o Error: (0.06) o No Error: (0.94) From Y: o Error: (0.12) o No Error: (0.88) Joint probabilities (multiply along branches): • • • • 𝑃(𝑋 ∩ Error) = 0.55 × 0.06 = 0.033 𝑃(𝑋 ∩ No Error) = 0.55 × 0.94 = 0.517 𝑃(𝑌 ∩ Error) = 0.45 × 0.12 = 0.054 𝑃(𝑌 ∩ No Error) = 0.45 × 0.88 = 0.396 2. Probability a randomly selected shipment is erroneous 𝑃(Error) = 𝑃(𝑋 ∩ Error) + 𝑃(𝑌 ∩ Error) = 0.033 + 0.054 = 0.087 So 8.7% of shipments are erroneous overall. 3. Bayes’ theorem: 𝑃(𝑋 ∣ Error) and 𝑃(𝑌 ∣ Error) Using Bayes’ theorem: 𝑃(𝑋 ∣ Error) = 𝑃(𝑋 ∩ Error) 0.033 = ≈ 0.3793 𝑃(Error) 0.087 𝑃(𝑌 ∣ Error) = 𝑃(𝑌 ∩ Error) 0.054 = ≈ 0.6207 𝑃(Error) 0.087 So: • • 𝑃(𝑋 ∣ Error) ≈ 0.38 𝑃(𝑌 ∣ Error) ≈ 0.62 4. Interpretation for operations • Even though Center X sends more shipments (55%), most errors (about 62%) come from Center Y. • Center Y has the higher error rate, and it contributes more to total defective shipments. • Management should: o Focus quality-improvement efforts on Center Y (process checks, training, stricter inspections). o Possibly renegotiate standards or volumes with Center Y, or benchmark it against Center X. Assignment 3: Advertising budget vs weekly sales (Correlation + Regression) 1. Split into training and testing sets Dataset: Week 1 2 3 4 5 6 7 X (Ad Budget, $1000s) 4 5 3 6 4 7 5 Y (Sales, $1000s) 52 60 45 72 55 80 63 8 9 10 • Training set: Weeks 1–7 • Testing set: Weeks 8–10 6 3 7 70 47 82 Training X: 4, 5, 3, 6, 4, 7, 5 Training Y: 52, 60, 45, 72, 55, 80, 63 2. Correlation between advertising and sales (training set) First compute means: • 𝑛=7 ∑𝑋 = 4 + 5 + 3 + 6 + 4 + 7 + 5 = 34 ⇒ 𝑋ˉ = 34/7 ≈ 4.857 ∑𝑌 = 52 + 60 + 45 + 72 + 55 + 80 + 63 = 427 ⇒ 𝑌ˉ = 427/7 = 61 Now compute: 𝑆𝑥𝑥 = ∑(𝑥𝑖 − 𝑋ˉ)2 ≈ 10.856 𝑆𝑦𝑦 = ∑(𝑦𝑖 − 𝑌ˉ)2 = 860 𝑆𝑥𝑦 = ∑(𝑥𝑖 − 𝑋ˉ)(𝑦𝑖 − 𝑌ˉ) ≈ 96 Then the correlation coefficient is: 𝑟= 𝑆𝑥𝑦 √𝑆𝑥𝑥 𝑆𝑦𝑦 √𝑆𝑥𝑥 𝑆𝑦𝑦 ≈ √10.856 × 860 ≈ √9336.16 ≈ 96.6 𝑟≈ 96 ≈ 0.994 96.6 So there is a very strong positive correlation between advertising budget and sales in the training data. 3. Fit a simple linear regression model (training set) Model form: 𝑌 = 𝑎𝑋 + 𝑏 • Slope: 𝑎= • 𝑆𝑥𝑦 96 ≈ ≈ 8.84 𝑆𝑥𝑥 10.856 Intercept: 𝑏 = 𝑌ˉ − 𝑎𝑋ˉ ≈ 61 − 8.84 × 4.857 ≈ 61 − 42.94 ≈ 18.06 So the regression equation is approximately: 𝑌̂ = 8.84𝑋 + 18.06 4. Interpret the slope in business terms • The slope 𝑎 ≈ 8.84 means: For each additional $1000 spent on advertising, weekly sales increase by about $8,840 (in this scaled unit system, 8.84 in thousands). • This suggests a strong and substantial positive effect of advertising on sales in the observed range. 5. Use the testing set to predict and compute errors Testing data: Week 8 9 10 X 6 3 7 Actual Y 70 47 82 Use 𝑌̂ = 8.84𝑋 + 18.06. • Week 8 (X = 6): 𝑌̂8 = 8.84 × 6 + 18.06 = 53.04 + 18.06 = 71.10 Error: Error8 = Actual − Predicted = 70 − 71.10 = −1.10 • Week 9 (X = 3): 𝑌̂9 = 8.84 × 3 + 18.06 = 26.52 + 18.06 = 44.58 Error: Error9 = 47 − 44.58 = 2.42 • Week 10 (X = 7): 𝑌̂10 = 8.84 × 7 + 18.06 = 61.88 + 18.06 = 79.94 Error: Error10 = 82 − 79.94 = 2.06 Summary of errors: • Week 8: –1.10 • Week 9: +2.42 • Week 10: +2.06 On a sales scale from ~45 to 80, these are quite small. 6. Evaluate whether the model generalizes well • The model trained on Weeks 1–7 produces accurate predictions for Weeks 8–10, with errors of only about 1–2.5 units. • The direction and magnitude of the relationship are consistent across training and testing. • This suggests the model generalizes well within this data range. • Management can reasonably use this model for short-term planning and for understanding how changes in advertising budget affect sales, while remembering it is still a simple linear approximation.