Commit cb60fc16 by Robert Ricci

### R Commands to be used during the lecture

parent c1ee67ff
 Determining probababilities for different execution times: Execution time is normally distributed with mean 10s, stddev 2s Evaluate probability of having a runtime of 3s: dnorm(3, mean = 10, sd = 2) Plot the probability curve over 20s: x <- seq(0,20, by = .1) plot(x,dnorm(x, mean = 10, sd = 2)) Plot the PDF over 20s: plot(x,pnorm(x, mean = 10, sd = 2)) Part A: Probability of time being more than 12s 1 - probability of less than 12s 1 - pnorm(12, mean = 10, sd = 2) Part B: Probability of time being less than 9s pnorm(9, mean = 10, sd = 2) Part C: Probability of time being between 8 and 12s pnorm(12, mean = 10, sd = 2) - pnorm(8, mean = 10, sd = 2) Part D: 95% of exection time Invert the PDF: x2 = seq(0,1,by = 0.01) plot(x2,qnorm(x2, mean = 10, sd = 2)) qnorm(0.95, mean = 10, sd = 2) ---------------------------------------------------------------------- Basic confidence intervals: Mean number of Disk I/Os: I <- c(28, 31, 12, 18, 42, 29, 33, 45, 53, 34, 42, 21, 36, 23, 36, 36, 27, 9, 11, 19, 35, 24, 31, 29, 16, 23, 34, 24, 38, 15, 18, 35, 27) Show quantiles: quantile(I) Part A: 10th and 90th percentiles quantile(I,c(.10,.90)) Part B: Mean mean(I) Part C: 90% CI for the mean Note: for double-sided 90% CI, qnorm() parameter is half as wide (95%) sd(I) length(I) mean(I) - (qnorm(.95)*sd(I))/sqrt(length(I)) mean(I) + (qnorm(.95)*sd(I))/sqrt(length(I)) Part D: What fraction make less than 25 IOs, and what is 90% CI for this fraction? Ismall <- c(12, 18, 21, 23, 9, 11, 19, 24, 16, 23, 24, 15, 18) p = length(Ismall) / length(I) Proportion: p + qnorm(0.95) * sqrt((p*(1-p))/length(I)) p - qnorm(0.95) * sqrt((p*(1-p))/length(I)) Part E: What is the one-sided 90% confidence interval for the mean? Using formula from Secion 13.7 mean(I) - qt(.90,length(I)-1) * (sd(I)/sqrt(length(I))) ---------------------------------------------------------------------- Testing for differences: System A: A <- c(140, 120, 176, 288, 992, 144, 2736, 2796, 752, 17720, 96) System B: B <- c(98, 120, 141, 317, 893, 86, 1642, 1678, 376, 8860, 67) Difference between the systems: (paired) diff <- A - B 90% confidence interval for differences: mean(diff) - (qnorm(.95)*sd(diff))/sqrt(length(diff)) mean(diff) + (qnorm(.95)*sd(diff))/sqrt(length(diff)) Part A: Graph the CIs x3 <- seq(.80,.99, by = .01) plot(x3,mean(diff) - (qnorm(x3)*sd(diff))/sqrt(length(diff)), type='l', ylim=c(-1000,3000)) lines(x3,mean(diff) + (qnorm(x3)*sd(diff))/sqrt(length(diff))) Part B: How many measurements does it take to get a difference at 90% confidence? Bottom end of 90% confidence interval for A: mean(A) - (qnorm(.95)*sd(A))/sqrt(length(A)) Upper end of CI for B: mean(B) + (qnorm(.95)*sd(B))/sqrt(length(B)) So, we need: mean(A) - (qnorm(.95)*sd(A))/sqrt(n)) >= mean(B) + (qnorm(.95)*sd(B))/sqrt(n) Which becomes: ((qnorm(.95)*(sd(A) + sd(B)))/(mean(A) + mean(B)))^2
Markdown is supported
0% or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!