Purpose: This activity is intended to illustrate properties of the sampling distribution of sample proportions pˆ.

Transcription

1 Rectangularity: Part I a. Sampling Distribution of Sample Proportions pˆ Purpose: This activity is intended to illustrate properties of the sampling distribution of sample proportions pˆ. Categorical Variable of Interest = Big rectangles (defined as area >=0) vs Small rectangles (areas < 9) What is the true percentage of Big rectangles? The Population of Rectangles Sheet shows a population of size 00 consisting of rectangles of varying areas. There are 24 Big rectangles, so the true percentage of Big rectangles is p = 24% Pretend we do not know p and wish to estimate it through random sampling, we could draw a simple random sample of rectangles from the population and use the sample proportion pˆ to estimate it. The sample proportion pˆ will vary from sample to sample. The distribution of the pˆ values for many simple random samples of size n is called the sampling distribution of the statistic pˆ Instructions:. Select MANY different simple random samples of size from the population (sample with replacement -- so that it is possible to select the same rectangle more than once) using R as follows: areas=c(3,0,,6,,4,,,8,3,,22,,6,3,2,,,2,0,,3,8,2,3, 2,8,,6,8,2,8,,6,6,4,2,4,9,,,,2,,2,,3,4,,2,3, 8,2,7,24,,6,3,,4,8,8,6,3,,2,7,,0,,24,,4,20,8,0, 2,0,2,0,,,2,2,9,3,2,6,7,2,,,3,9,,,4,0,7,8) length(areas) repeat.time=0 set.seed(7) X=sample(areas,,replace=T) ph.=sum(x>=0)/ set.seed(7) # This will allow you to have the same random sequences # for sample size n= ph. = c() X=sample(areas,,replace=T) ph.[i] = sum(x>=0)/}

2 For each sample of size n=, list the areas, and then calculate the value of pˆ. Repeat it MANY times for the same n= and plot the distribution of values of pˆ. Do the same for n=. Do the same for n=2. Questions:. For each sample size n =,, and 2 construct a histogram of the sample proportion values. 2. For each sample size, describe the shape of the distribution of pˆ values. 3. Compare the shape of the distributions of the pˆ values to the shape of the distribution of the population. Which looks more normal? 4. Based on your histograms, what do you think is the relationship between the sample size and the shape of the distribution of the sample proportion?. (a) For each sample size, calculate the standard deviation and the mean of the sample proportions. (b) For which sample size is the standard deviation the largest and for which sample size is the standard deviation the smallest? Why do you suppose this happens? 6. How does the standard deviation of the pˆ values compare for n =,, and? 7. Checking: the mean of the sample proportions = p (the true proportion). 8. Checking: the standard deviation of the sample proportions = 9. What is the likelihood (probability) that pˆ >= 0.4 when n=? When n=? When n=2? 2

3 Population of Rectangles: (The population of rectangles sheet is adapted from Scheaffer et al. 996.) 3

4 Histogram and Table of the Areas of the Rectangles in the Population: Histogram of the Areas of the Rectangles in the Population: Area Table of the Areas of the Rectangles in the Population: AREA Total

5 Histogram of ph. Histogram of ph. Histogram of ph ph ph ph.2 Histogram of ph. Histogram of ph. Histogram of ph ph ph ph.2 Note that increasing repeat.time will NOT change the mean and spread of the sampling distribution The first row of three histograms are from repeat.time=0000, and the second row are from repeat.time= The R code to generate the above histogram is as follows: # Rectangularity Part I # Sampling Distribution of the sample means areas=c(3,0,,6,,4,,,8,3,,22,,6,3,2,,,2,0,,3,8,2,3, 2,8,,6,8,2,8,,6,6,4,2,4,9,,,,2,,2,,3,4,,2,3, 8,2,7,24,,6,3,,4,8,8,6,3,,2,7,,0,,24,,4,20,8,0, 2,0,2,0,,,2,2,9,3,2,6,7,2,,,3,9,,,4,0,7,8) length(areas) p= sum(areas>=0)/00 repeat.time=0000

6 # for sample size n= ph. = c() X=sample(areas,,replace=T) ph.[i] = sum(x>=0)/} # for sample size n= ph. = c() X=sample(areas,,replace=T) ph.[i] = sum(x>=0)/} # for sample size n=2 ph.2 = c() X=sample(areas,2,replace=T) ph.2[i] = sum(x>=0)/2} # Plots par(mfrow=c(2,3)) hist(ph.,nc=00,xlim=range(ph.)) abline(v=mean(ph.),col='red') hist(ph.,nc=00,xlim=range(ph.)) abline(v=mean(ph.),col='red') hist(ph.2,nc=00,xlim=range(ph.)) abline(v=mean(ph.2),col='red') # Compare the center and spread with the theoretical values rbind(c(mean(ph.),sd(ph.)), c(mean(ph.),sd(ph.)), c(mean(ph.2),sd(ph.2))) p c(sqrt(p*(-p)/), sqrt(p*(-p)/), sqrt(p*(-p)/2)) # Increasing repeat.time will NOT change the mean and spread # of the sampling dustribution repeat.time=