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zhang-routledge, Water treatment
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Week 2 Day 2 Quan Sam
1/10/2020 Definitions:Inferential situation is when we take data from samples and make generalizations about a population allowing us to make predictions. We use 2 indicators for the estimation: Indicator of position or the average (Xbar): is calculated as Indicator of dispersion or the variance(^{2}): is the expectation of the squared deviation of a random variable from its mean, in other words, measures how far a set of numbers is spread out from their average value. It is calculated as Examples:First, we generate 11 different samples, and means_simulated=vector(mode = "numeric", length = 11) a=rnorm(20,85,2) b=rep(1,20) means_simulated[1]=mean(a) plot(a,b,cex = .5, col = "dark red") points(x=mean(a),y=0.8,pch=24,cex=0.5,col = "dark red") a=rnorm(20,85,2) b=rep(1.025,20) means_simulated[2]=mean(a) points(a,b,cex = .5, col = "pink") points(x=mean(a),y=0.8,pch=24,cex=0.5,col = "pink") a=rnorm(20,85,2) b=rep(1.05,20) means_simulated[3]=mean(a) points(a,b,cex = .5) points(x=mean(a),y=0.8,pch=24,cex=0.5) a=rnorm(20,85,2) b=rep(1.075,20) means_simulated[4]=mean(a) points(a,b,cex = .5, col = "green") points(x=mean(a),y=0.8,pch=24,cex=0.5,col = "green") a=rnorm(20,85,2) b=rep(1.1,20) means_simulated[5]=mean(a) points(a,b,cex = .5, col = "purple") points(x=mean(a),y=0.8,pch=24,cex=0.5,col = "purple") a=rnorm(20,85,2) b=rep(1.125,20) means_simulated[6]=mean(a) points(a,b,cex = .5, col = "tomato") points(x=mean(a),y=0.8,pch=24,cex=0.5,col = "tomato") a=rnorm(20,85,2) b=rep(1.15,20) means_simulated[7]=mean(a) points(a,b,cex = .5, col = "plum") points(x=mean(a),y=0.8,pch=24,cex=0.5,col = "plum") a=rnorm(20,85,2) b=rep(1.175,20) means_simulated[8]=mean(a) points(a,b,cex = .5, col = "gold") points(x=mean(a),y=0.8,pch=24,cex=0.5,col = "gold") a=rnorm(20,85,2) b=rep(1.2,20) means_simulated[9]=mean(a) points(a,b,cex = .5, col = "dimgrey") points(x=mean(a),y=0.8,pch=24,cex=0.5,col = "dimgrey") a=rnorm(20,85,2) b=rep(1.225,20) means_simulated[10]=mean(a) points(a,b,cex = .5, col = "aquamarine") points(x=mean(a),y=0.8,pch=24,cex=0.5,col = "aquamarine") a=rnorm(20,85,2) b=rep(1.25,20) means_simulated[11]=mean(a) points(a,b,cex = .5, col = "firebrick1") points(x=mean(a),y=0.8,pch=24,cex=0.5,col = "firebrick1") mean(means_simulated) ## [1] 85.15239 mean(means_simulated) ## [1] 85.15239 abline(v = mean(means_simulated), col = "red") We can conclude that although we do not have the mean values of the whole population, we can estimate it based on the mean values of 11 samples as they are approaching to the mean values of the population. Now, we generate 30 values. weight = c(56.6,54.8,59.0,60.4,61.8,62.6,65.0,58.1,61.4,60.8,59.2,58.1,57.5,55.2,54.6,61.6,56.9,61.3,67.2,53.9,54.1,62.0,63.5,58.1,56.0,51.5,63.8,58.1,58.2,61.3) mean(weight) ## [1] 59.08667 sd(weight) ## [1] 3.655203 If we only have a sample of 30 values, we still can estimate the mean value of the population if the mean value of the sample lies in the confidence interval of . The confidence interval is spanning from mean(weight) - 2*sd(weight)/sqrt(30) to mean(weight) + 2*sd(weight)/sqrt(30). Download 15.28 Kb. Share with your friends:
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