1 / 29

Statistics

Statistics. continued. CPS 807. 2 k r Factorial Designs with Replications. r replications of 2 k Experiments 2 k r observations. Allows estimation of experimental errors Model: y = q 0 + q A x A + q B x B +q AB x A x B +e e= Experimental error. Computation of Effects.

nhu
Download Presentation

Statistics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Statistics continued CPS 807

  2. 2kr Factorial Designs with Replications • r replications of 2k Experiments • 2kr observations. • Allows estimation of experimental errors • Model: y = q0 + qAxA + qBxB +qABxAxB+e e=Experimental error

  3. Computation of Effects • Simply use means of r measurements Effects: q0 = 41, qA = 21.5, qB = 9.5, qAB = 5

  4. Estimated Response: yi = q0 + qAxAi + qBxBi +qABxAixBi • Experimental Error = Estimated-Measured eij = yij - yi = yij - q0 - qAxAi - qBxBi - qABxAixBi ^ ^ • Sum of Squared Errors: SSE = Estimation of Experimental Errors

  5. Estimated Response: y1 = q0 - qA - qB + qAB = 41 -21.5 -9.5 +5 = 15 Experimental errors: e11 = y11 - y1 = 15 - 15 = 0 ^ ^ Estimated Measured Effect ^ SSE = 02 + 32 + (-3)2 + (-3)2 + ... + 42 = 102 Experimental Errors: Example

  6. Total variation or total sum of squares: SST = SST = SSA + SSB + SSAB + SSE Allocation of Variation

  7. Model: Since x’s, their products, and all errors add to zero Derivation

  8. Mean response: Squaring both sides of the model and ignoring cross product terms: SSY = SS0 + SSA + SSB + SSAB + SSE Derivation (cont’d)

  9. Total Variation: One way to compute SSE: Derivation (cont’d)

  10. Example: Memory-Cache Study

  11. Example: Memory-Cache Study(cont’d) SSA + SSB + SSAB + SSE =5547 + 1083 + 300 + 102 = 7032 = SST Factor A explains 5547/7032 or 78.88% Factor B explains 15.40% Interaction AB explains 4.27% 1.45% is unexplained and is attributed to errors.

  12. Review: Confidence Interval for the Mean • Problem: How to get a single estimate of the population mean from k sample estimates? • Answer: Get probabilistic bounds. • Eg., 2 bounds, C1 & C2 There is a high probability, 1-, that the mean is in the interval (C1, C2 ): • Pr {C1   C2} = 1 -  • Confidence interval  (C1, C2 ) •   Significance Level • 100 (1-)  Confidence Level • 1-  Confidence Coefficient.

  13. Confidence Interval for the Mean (cont’d) Note: Confidence Level is traditionally expressed as a percentage (near 100%); whereas, significance level , is expressed as a fraction & is typically near zero; e.g., 0.05 or 0.01.

  14. Example: • Given sample with: • mean = x = 3.90 • SD = s = 0.95 • n = 32 • A 90 % CI for the mean = 3.90 + (1.645)(0.95)/ • = (3.62, 4.17), used the central limit theorem. • Note: A 90 % CI => We can state with 90 % confidence that the population mean is between 3.62 & 4.17. The chance of error in this statement is 10 % Confidence Interval for the Mean (cont’d)

  15. Testing for a Zero Mean • Difference in processor times of two different implementations of the same algorithms was measured on 7 similar workloads. The differences are: • {1.5, 2.6, -1.8, 1.3, -0.5, 1.7, 2.4} • Can we say with 99 % confidence that one implementation is superior to the other

  16. Sample size = n = 7 • mean = x = 1.03 • sample variance = s2 = 2.57 • sample deviation = s = 1.60 • CI = 1.03  tx 1.60/ = 1.03  0.605t • 100 (1- ) = 99,  = 0.01, 1- /2 = 0.995 • From Table, the t value at six degrees of freedom is: • t[0.995; 6] = 3.707 & the 99% CI = (-1.21, 3.27). • Since the CI includes zero, we can not say with 99% confidence that the mean difference is significantly different from Zero. Testing for a Zero Mean (cont’d)

  17. In testing a NULL, hypothesis, the level of significance is the probability of rejecting a true hypothesis. HYPOTHESIS Actually True Actually False DECISION  Error (Type II) To Accept Correct  Error (Type I) To Reject Correct Note: The letters  &  denote the probability related to these errors Type I & Type II Errors

  18. Confidence Intervals For Effects Effects are random variables. Errors ~ N(0,σe) => y ~ N( y.., σe) Since q0 = Linear combination of normal variables => q0 is normal with variance Variance of errors:

  19. Confidence Intervals For Effects (cont’d) Denominator = 22(r - 1) = # of independent terms in SSE => SSE has 22(r - 1) degrees of freedom. Estimated variance of q0 : Similarly, Confidence intervals (CI) for the effects: CI does not include a zero => significant

  20. Example For Memory-cache study: Standard deviation of errors: Standard deviation of effects: For 90% Confidence :

  21. Example (cont’d) Confidence intervals: No zero crossing => All effects are significant.

  22. Confidence Intervals for Contrasts Contrast  Linear combination with  coefficients = 0 Variance of  hiqi: For 100 ( 1 -  ) % confidence interval, use

  23. Example: Memory-cache study u = qA + qB -2qAB Coefficients = 0,1,1, and -2 => Contrast Mean u = 21.5 + 9.5 - 2 x 5 = 21 Variance Standard deviation t[0.95;8] = 1.86 90% Confidence interval for u : (16.31, 25.69)

  24. CI for Predicted Response ^ Mean response y : y = q0 + qA xA + qB xB + qABxA xB The standard deviation of the mean of m response: ^ neff = Effective deg of freedom = Total number of runs ^ 1 + Sum of DFs of params used in y

  25. 100 ( 1 -  ) % confidence interval: A single run (m = 1) : Population mean CI for Predicted Response (cont’d)

  26. Example: Memory-cache Study • For xA = -1 and xB = -1: • A single confirmation experiment: • y1 = q0 - qA - qB + qAB • = 41 - 21.5 - 9.5 + 5 = 15 • Standard deviation of the prediction: ^ Using t[0.95;8]=1.86, the 90% confidence interval is:

  27. Example: Memory-cache Study (cont’d) • Mean response for 5 experiments in future: The 90% confidence interval is: • Mean response for a large number of experiments in future: The 90% confidence interval is:

  28. Example: Memory-cache Study (cont’d) • Current mean response: Not for future. • (Use the formula for contrasts): 90% confidence interval: Notice: Confidence intervals become narrower.

  29. Assumptions 1. Errors are statistically independent. 2. Errors are additive. 3. Errors are normally distributed 4. Errors have a constant standard deviation e. 5. Effects of factors are additive. => observations are independent and normally distributed with constant variance.

More Related