c:ms:2020:schedule
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c:ms:2020:schedule [2020/06/26 15:37] – [Week16 (June 22, 25)] hkimscil | c:ms:2020:schedule [2020/06/27 13:28] – [Week15 (June 22, 25)] hkimscil | ||
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<WRAP half column> | <WRAP half column> | ||
===== Concepts and ideas ===== | ===== Concepts and ideas ===== | ||
- | Do the following | ||
- | < | ||
- | 98, 87, 94, 77, 87, 98, 85, 89, 95, 85, 93, 93, | ||
- | 97, 71, 97, 93, 75, 68, 98, 95, 79, 94, 98, 95) | ||
- | S2 <- c(60, 98, 94, 95, 99, 97, 100, 73, 93, 91, 98, | ||
- | 86, 66, 83, 77, 97, 91, 93, 71, 91, 95, 100, | ||
- | 72, 96, 91, 76, 100, 97, 99, 95, 97, 77, 94, | ||
- | 99, 88, 100, 94, 93, 86) | ||
- | S3 <- c(95, 86, 90, 90, 75, 83, 96, 85, 83, 84, 81, 98, | ||
- | 77, 94, 84, 89, 93, 99, 91, 77, 95, 90, 91, 87, | ||
- | 85, 76, 99, 99, 97, 97, 97, 77, 93, 96, 90, 87, | ||
- | 97, 88) | ||
- | S4 <- c(67, 93, 63, 83, 87, 97, 96, 92, 93, 96, 87, 90, | ||
- | 94, 90, 82, 91, 85, 93, 83, 90, 87, 99, 94, 88, | ||
- | 90, 72, 81, 93, 93, 94, 97, 89, 96, 95, 82, 97) | ||
- | |||
- | scores <- list(S1=S1, | ||
- | |||
- | * find means for each element in " | ||
- | * find standard deviation for each element in " | ||
- | * find variance | ||
- | |||
- | < | ||
- | | ||
- | | ||
- | | ||
- | * make " | ||
- | * name columns " | ||
- | * name rows " | ||
- | * get means for each subject | ||
- | * attach the above data to the matrix data and name it " | ||
- | * get standard deviation for each trial | ||
- | * attach the above data to the matrix data, " | ||
- | |||
- | |||
- | < | ||
- | * get subrubs data as the above | ||
- | * get population means by each state (listed in the data, suburbs) | ||
- | * use aggregate and refer to the below e.g. | ||
- | < | ||
- | aggregate(MPG.city ~ Origin, Cars93, mean)</ | ||
- | * get population sum by each county with tapply function. | ||
- | * tapply(number, | ||
- | * how many counties are there? | ||
- | * Use Cars93 data, get MPG.city mean by Origin. | ||
- | |||
- | __Using pnorm, qnorm__ | ||
- | pnorm : get proportion out of normal distribution whose characteristics are mean and sd | ||
- | < | ||
- | * What is the value of the below? | ||
- | < | ||
- | * How would you get 68, 95, 99% from pnorm | ||
- | * use ?pnorm and see the default option | ||
- | |||
- | * generate 10 random numbers with runif function | ||
- | |||
- | < | ||
- | world.series <- data.frame(year)</ | ||
- | * get 10 year samples out of world.series data with " | ||
- | * how would you get the sample sample again latter? | ||
- | |||
- | < | ||
- | * What would be the result from the above? | ||
- | |||
- | < | ||
- | tbl = table(survey$Smoke, | ||
- | tbl # the contingency table</ | ||
- | |||
- | < | ||
- | </ | ||
- | * read the above output and interpret | ||
- | * what about the below one? | ||
- | < | ||
- | </ | ||
- | |||
- | see first [[: | ||
- | see [[:b:r cookbook: | ||
- | |||
- | < | ||
- | | ||
- | | ||
- | </ | ||
- | * Can you say the types of cars are different by the Origins? | ||
- | |||
- | < | ||
- | dur</ | ||
- | * make the above data into z-score (zdur). | ||
- | * get mean of the zdur | ||
- | * get sd of the zdur | ||
- | |||
- | < | ||
- | set.seed(1123) | ||
- | x <- rnorm(50, mean=100, sd=15) | ||
- | </ | ||
- | * test x against population | ||
- | * test x against population | ||
- | * are they different from each other? | ||
- | * what would you do if you want to see the different result from the second one? | ||
- | |||
- | < | ||
- | |||
- | > t.test(a, mu=60) | ||
- | |||
- | One Sample t-test | ||
- | |||
- | data: a | ||
- | t = 2.3079, df = 9, p-value = 0.0464 | ||
- | alternative hypothesis: true mean is not equal to 60 | ||
- | 95 percent confidence interval: | ||
- | | ||
- | sample estimates: | ||
- | mean of x | ||
- | | ||
- | </ | ||
- | * find the t critical value with function qt. | ||
- | * explain what happens in the next code | ||
- | * read (or remind) what pnorm and qnorm do. | ||
- | < | ||
- | > m <- mean(x) | ||
- | > n <- length(x) | ||
- | > n | ||
- | [1] 50 | ||
- | > m | ||
- | [1] 96.00386 | ||
- | > s | ||
- | [1] 17.38321 | ||
- | > SE <- s / sqrt(n) | ||
- | > SE | ||
- | [1] 2.458358 | ||
- | > E <- qt(.975, df=n-1)*SE | ||
- | > E | ||
- | [1] 4.940254 | ||
- | > m + c(-E, E) | ||
- | [1] 91.0636 100.9441 | ||
- | > </ | ||
- | |||
- | |||
- | * what's wrong with the below? | ||
- | < | ||
- | |||
- | < | ||
- | * using aggregate, get mean for each trnas. type. | ||
- | * compare the difference of mileage between auto and manual cars. | ||
- | * use t.test (two sample) | ||
- | * "use var.equal=T" | ||
- | |||
- | < | ||
- | b = c(185, 169, 173, 173, 188, 186, 175, 174, 179, 180) | ||
- | </ | ||
- | * stack them into data c | ||
- | * convert colnames into score and trans | ||
- | * t.test score by trans with var.equal option true. | ||
- | * aov test | ||
- | * see t.test t value, t = -0.9474 and F value, | ||
</ | </ | ||
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</ | </ | ||
<WRAP half column> | <WRAP half column> | ||
+ | ===== Assignment ===== | ||
+ | 그룹과제 내용수정 ---- | ||
+ | 그룹과제 내용을 아래와 같이 수정합니다. | ||
+ | |||
</ | </ | ||
====== Week16 (June 22, 25) ====== | ====== Week16 (June 22, 25) ====== |
c/ms/2020/schedule.txt · Last modified: 2021/03/02 09:42 by hkimscil