Lab4a
Umais Siddiqui
September 30, 2017
Foundations for statistical inference - Sampling distributions
load("more/ames.RData")
area <- ames$Gr.Liv.Area
price <- ames$SalePrice
summary(area)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 334 1126 1442 1500 1743 5642hist(area)
Exercise 1 : Describe this population distribution.
Answer:
This population data is unimodal and skewed to the right.
samp1 <- sample(area, 50)Exercise 2: Describe the distribution of this sample. How does it compare to the distribution of the population?
summary(samp1)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 848 1187 1429 1460 1616 2522hist(samp1)
The results of the distribution of the sample are variable sometimes they are unimodal somtimes bimodal. The Skewness is right however and it does sometime come pretty close the population
Exercise 3:Take a second sample, also of size 50, and call it samp2. How does the mean of samp2 compare with the mean of samp1? Suppose we took two more samples, one of size 100 and one of size 1000. Which would you think would provide a more accurate estimate of the population mean?
samp2 <- sample(area, 50)
hist(samp2)
mean(samp1)## [1] 1459.9mean(samp2)## [1] 1532.3samp3 <- sample(area, 1000)
hist(samp3)
mean(samp3)## [1] 1511.559mean(area)## [1] 1499.69Answer:
The mean of sample 1 and sample 2 are pretty close but some times one is greater than the other as it is variable. I think the one with size of 100 would provide a more accurate estimate of the population but it may also vary because it is a random selection of the population that is used for sampling.
Exercise 4 How many elements are there in sample_means50? Describe the sampling distribution, and be sure to specifically note its center. Would you expect the distribution to change if we instead collected 50,000 sample means?
sample_means50 <- rep(NA, 5000)
for(i in 1:5000){
samp <- sample(area, 50)
sample_means50[i] <- mean(samp)
}
hist(sample_means50)
length(sample_means50)## [1] 5000summary(sample_means50)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 1291 1452 1496 1499 1544 1885sample_means50_a <- rep(NA, 50000)
for(i in 1:50000){
samp <- sample(area, 50)
sample_means50_a[i] <- mean(samp)
}
hist(sample_means50_a)
length(sample_means50_a)## [1] 50000summary(sample_means50_a)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 1239 1452 1498 1500 1546 1864Answer:
The sample size is 5000. The centre is close to 1500 and I do not expect the distribution to change if we collected 50,000 sample means.
Exercise 5 To make sure you understand what you’ve done in this loop, try running a smaller version. Initialize a vector of 100 zeros called sample_means_small. Run a loop that takes a sample of size 50 from area and stores the sample mean in sample_means_small, but only iterate from 1 to 100. Print the output to your screen (type sample_means_small into the console and press enter). How many elements are there in this object called sample_means_small? What does each element represent?
Answer:
sample_means_small <- rep(NA, 100)
for(i in 1:100){
samp <- sample(area, 50)
sample_means_small[i] <- mean(samp)
}
sample_means_small## [1] 1508.94 1676.46 1434.06 1692.28 1506.04 1425.04 1483.60 1333.28
## [9] 1445.00 1397.16 1472.88 1549.58 1440.58 1507.10 1474.96 1579.32
## [17] 1484.16 1586.44 1606.06 1476.98 1501.82 1553.12 1545.10 1272.40
## [25] 1637.56 1481.98 1510.40 1457.36 1450.46 1460.62 1500.20 1497.20
## [33] 1480.46 1503.52 1535.56 1521.98 1583.00 1354.18 1417.96 1454.10
## [41] 1442.80 1464.92 1606.00 1463.42 1660.50 1496.38 1562.28 1493.98
## [49] 1514.78 1548.34 1434.18 1482.94 1681.24 1422.42 1507.52 1490.34
## [57] 1395.72 1624.94 1486.02 1326.86 1481.58 1457.46 1520.78 1469.44
## [65] 1377.20 1384.80 1532.08 1512.82 1449.24 1628.68 1485.06 1493.12
## [73] 1358.94 1502.62 1542.14 1463.78 1383.26 1572.48 1495.26 1477.94
## [81] 1399.20 1506.10 1520.58 1405.72 1479.60 1543.74 1464.54 1482.98
## [89] 1478.32 1498.26 1481.70 1529.84 1560.94 1559.82 1487.08 1491.48
## [97] 1390.48 1521.62 1445.10 1519.82There are 100 elements in sample_means_small. The 100 elemenets represent the sample mean for the 100 iterations done to get a ‘sample’ of the area.
Exercise 6 When the sample size is larger, what happens to the center? What about the spread?
hist(sample_means50)
sample_means10 <- rep(NA, 5000)
sample_means100 <- rep(NA, 5000)
for(i in 1:5000){
samp <- sample(area, 10)
sample_means10[i] <- mean(samp)
samp <- sample(area, 100)
sample_means100[i] <- mean(samp)
}
par(mfrow = c(3, 1))
xlimits <- range(sample_means10)
hist(sample_means10, breaks = 20, xlim = xlimits)
hist(sample_means50, breaks = 20, xlim = xlimits)
hist(sample_means100, breaks = 20, xlim = xlimits)
Answer :
As the sample mean Increases the center becomes larger and the spread gets narrower.
On YOur Own
Question 1:Take a random sample of size 50 from price. Using this sample, what is your best point estimate of the population mean?
Answer:
sample_meansprice50 <- sample(price, 50)
mean(sample_meansprice50)## [1] 179060.5Question 2: Since you have access to the population, simulate the sampling distribution for x price by taking 5000 samples from the population of size 50 and computing 5000 sample means. Store these means in a vector called sample_means50. Plot the data, then describe the shape of this sampling distribution. Based on this sampling distribution, what would you guess the mean home price of the population to be? Finally, calculate and report the population mean.
Answer:
sample_means50 <- rep(NA, 5000)
for(i in 1:5000){
samp <- sample(price, 50)
sample_means50[i] <- mean(samp)
}
hist(sample_means50, breaks = 25)
summary(sample_means50)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 146700 172900 180300 180700 187800 226400mean(sample_means50)## [1] 180693.8mean(price)## [1] 180796.1Question 3 Change your sample size from 50 to 150, then compute the sampling distribution using the same method as above, and store these means in a new vector called sample_means150. Describe the shape of this sampling distribution, and compare it to the sampling distribution for a sample size of 50. Based on this sampling distribution, what would you guess to be the mean sale price of homes in Ames?
Answer:
sample_means150 <- rep(NA, 5000)
for(i in 1:5000){
samp <- sample(price, 50)
sample_means150[i] <- mean(samp)
}
hist(sample_means150, breaks = 25)
summary(sample_means150)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 147100 173200 180300 180800 188000 239300mean(sample_means150)## [1] 180797.9mean(price)## [1] 180796.1