Sampling Distributions Homework Answers
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1 Sampling Distributions Homework Answers John Fox Soc. 6Z03, Fall % of registered voters are Democrats: parameter 36% of registered voters are Republicans: parameter 34% of voters contacted are Democrats: statistic No: The histogram of the sample values will resemble the histogram of the population; if the population is non-normal, then so will be the distribution of the sample. It s the sampling distribution of the sample means that becomes more normal as the sample size grows when we sample from a non-normal population State This is simply the problem, as stated: We want to find the probability that the average loss for 10,000 policies will be greater than $85 when the long-run average loss is $75. Plan Use the central limit theorem to approximate this probability; this is justified because for = the sampling distribution of sample means will be very close to normal. Solve The sample distribution of is ( )= ( ) = (75 3). Then, = = = and ( 85) = ( 3 33) = = 0004 Conclude The company can be 99.96% certain that its average loss won t exceed $85 per policy % is (c) a parameter (the percentage of all registered voters who voted) (a) (b) 1 = = = = = (20 30) = ( ) = = 5646 mean( ) = =25 SD( ) = = 6 4 = ( ) 1
2 (c) 1 = = = = = (20 30) = ( ) = 9996 (The normal table doesn t go beyond = ±3 49.) 2
3 Soc 6Z03: Sampling Distributions Homework Answers, Supplemental Questions John Fox ## Loading required package: splines ## Loading required package: RcmdrMisc ## Loading required package: car ## Loading required package: sandwich ## The Commander GUI is launched only in interactive sessions (a) Setting the seed for the R random number generator: > set.seed(12345) > NormalSamples < as.data.frame(matrix(rnorm(1000*10, mean=100, sd=15), + ncol=10)) > rownames(normalsamples) < paste("sample", 1:1000, sep="") > colnames(normalsamples) < paste("obs", 1:10, sep="") > NormalSamples$mean < rowmeans(normalsamples[,1:10]) n = 10 Histogram of 1000 sample means, each sample of size : > with(normalsamples, Hist(mean, scale="frequency", breaks="sturges", + col="darkgray")) file:///d:/courses/ /Soc6Z03/Homework/R sampling dist homework.html 1/9
![> library(abind, pos=15) > library(e1071, pos=16) > numsummary(normalsamples[,"mean"], statistics=c("mean", "sd"), +](/docs-images/77/75537318/images/4-0.jpg "quantiles=c(0,.25,.5,.75,1)) mean sd n 99.98965 4.")
4 > library(abind, pos=15) > library(e1071, pos=16) > numsummary(normalsamples[,"mean"], statistics=c("mean", "sd"), + quantiles=c(0,.25,.5,.75,1)) mean sd n file:///d:/courses/ /Soc6Z03/Homework/R sampling dist homework.html 2/9
5 The distribution of the sample means looks normal (as far as one can tell from a histogram): it is unimodal and symmetric with centre near 100 and standard deviation These values are close to the theoretical values, and n. μ = 100 σ/ = 15/ 10 = 4.74 Notes: Because the data are generated randomly, your results won t be exactly the same as mine. I suppressed the display of the interquartile range and quantiles in the numerical summaries computed for the 1000 sample means, leaving only the boxes for the mean and standard deviation checked. (b) > NormalSamples < as.data.frame(matrix(rnorm(1000*50, mean=100, sd=15), + ncol=50)) > rownames(normalsamples) < paste("sample", 1:1000, sep="") > colnames(normalsamples) < paste("obs", 1:50, sep="") > NormalSamples$mean < rowmeans(normalsamples[,1:50]) n = 50 For samples of size : > with(normalsamples, Hist(mean, scale="frequency", breaks="sturges", + col="darkgray")) file:///d:/courses/ /Soc6Z03/Homework/R sampling dist homework.html 3/9
![> numsummary(normalsamples[,"mean"], statistics=c("mean", "sd"), + quantiles=c(0,.25,.5,.75,1)) mean sd n 100.0186 2.](/docs-images/77/75537318/images/6-0.jpg "195417 1000 The distribution of sample means looks normal, as before; the mean of the sample means is again very close to μ = 100 (in fact, it rounds to 100); and the standard deviation of the sample")
6 > numsummary(normalsamples[,"mean"], statistics=c("mean", "sd"), + quantiles=c(0,.25,.5,.75,1)) mean sd n The distribution of sample means looks normal, as before; the mean of the sample means is again very close to μ = 100 (in fact, it rounds to 100); and the standard deviation of the sample means, 2.19, is close to the theoretical value, 15/ 50 = The variability of the sample means is therefore smaller for samples of size than for samples of size. n = 50 n = 10 ( c ) (optional) file:///d:/courses/ /Soc6Z03/Homework/R sampling dist homework.html 4/9
7 > ExponentialSamples < as.data.frame(matrix(rexp(1000*2, rate=1), ncol=2)) > rownames(exponentialsamples) < paste("sample", 1:1000, sep="") > colnames(exponentialsamples) < paste("obs", 1:2, sep="") > ExponentialSamples$mean < rowmeans(exponentialsamples[,1:2]) Here are the histogram, mean, and standard deviation for the distribution of the means of 1000 samples drawn from the exponential distribution with for : μ = σ = 1 n = 2 > with(exponentialsamples, Hist(mean, scale="frequency", breaks="sturges", + col="darkgray")) file:///d:/courses/ /Soc6Z03/Homework/R sampling dist homework.html 5/9
8 > numsummary(exponentialsamples[,"mean"], statistics=c("mean", "sd"), + quantiles=c(0,.25,.5,.75,1)) mean sd n ( x ) = μ = 1 ( x ) = σ/ = 1/ 2 = 0.71 The theoretical values are mean and SD n. > ExponentialSamples < as.data.frame(matrix(rexp(1000*5, rate=1), ncol=5)) > rownames(exponentialsamples) < paste("sample", 1:1000, sep="") > colnames(exponentialsamples) < paste("obs", 1:5, sep="") > ExponentialSamples$mean < rowmeans(exponentialsamples[,1:5]) Here are the histogram, mean, and standard deviation for the distribution of the means of 1000 samples drawn from the exponential distribution with for : μ = σ = 1 n = 5 > with(exponentialsamples, Hist(mean, scale="frequency", breaks="sturges", + col="darkgray")) file:///d:/courses/ /Soc6Z03/Homework/R sampling dist homework.html 6/9
![> numsummary(exponentialsamples[,"mean"], statistics=c("mean", "sd"), + quantiles=c(0,.25,.5,.75,1)) mean sd n 1.004399 0.4465361 1000 ( x ) = μ = 1 ( x ) = σ/ = 1/ 5 = 0.](/docs-images/77/75537318/images/9-0.jpg "45 The theoretical values are mean and SD n. > ExponentialSamples < as.data.")
9 > numsummary(exponentialsamples[,"mean"], statistics=c("mean", "sd"), + quantiles=c(0,.25,.5,.75,1)) mean sd n ( x ) = μ = 1 ( x ) = σ/ = 1/ 5 = 0.45 The theoretical values are mean and SD n. > ExponentialSamples < as.data.frame(matrix(rexp(1000*25, rate=1), ncol=25)) > rownames(exponentialsamples) < paste("sample", 1:1000, sep="") > colnames(exponentialsamples) < paste("obs", 1:25, sep="") > ExponentialSamples$mean < rowmeans(exponentialsamples[,1:25]) file:///d:/courses/ /Soc6Z03/Homework/R sampling dist homework.html 7/9
10 And here are the histogram, mean, and standard deviation for the distribution of the means of 1000 samples drawn from the exponential distribution with for : μ = σ = 1 n = 25 > with(exponentialsamples, Hist(mean, scale="frequency", breaks="sturges", + col="darkgray")) > numsummary(exponentialsamples[,"mean"], statistics=c("mean", "sd"), + quantiles=c(0,.25,.5,.75,1)) mean sd n file:///d:/courses/ /Soc6Z03/Homework/R sampling dist homework.html 8/9
11 ( x ) = μ = 1 ( x ) = σ/ = 1/ 25 = 0.2 The theoretical values are mean and SD n. As the sample size grows, the sampling distribution of the sample means becomes more symmetric, looking increasingly like a normal distribution, and thus illustrating the central limit theorem. The distribution of sample means is always centred near, but its spread decreases as the sample size grows. μ = 1 The means and standard deviations of the sample means are similar to the theoretical values. file:///d:/courses/ /Soc6Z03/Homework/R sampling dist homework.html 9/9
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