Stat 422: Takehome

1 From the population of size N=5, given by {0,1,3,5,7}

• Calculate hte values of mu and sigma squared.

# 1A
pop = c(0, 1, 3, 5, 7)
N = length(pop)
mu = mean(pop)
mu

## [1] 3.2

sigma = mean((pop - mu)^2)
sigma

## [1] 6.56

• Calculate the expected value of ybar and the variance of ybar for an SRS of size n=2, without replacement and with equal probaiblities.

suppressMessages(library(combinat))

# 1B: Without Replacement list all possible subsets
n = 2

dat = combn((pop), n)
c1 = t(dat)
c2 = 1/nrow(t(dat))
c3 = N * apply(c1, 1, mean)
c4 = apply(c1, 1, mean)
c5 = apply(c1, 1, var)
c6 = (1 - (n/N)) * (c5/n)

paste("The possible samples are")

## [1] "The possible samples are"

print(c1)

##       [,1] [,2]
##  [1,]    0    1
##  [2,]    0    3
##  [3,]    0    5
##  [4,]    0    7
##  [5,]    1    3
##  [6,]    1    5
##  [7,]    1    7
##  [8,]    3    5
##  [9,]    3    7
## [10,]    5    7

paste("The replication of the table with N is:")

## [1] "The replication of the table with N is:"

cbind(c1, Prob = c2, Tauhat = c3, Ybar = c4, S2 = c5, VhatYbar = c6)

##           Prob Tauhat Ybar   S2 VhatYbar
##  [1,] 0 1  0.1    2.5  0.5  0.5     0.15
##  [2,] 0 3  0.1    7.5  1.5  4.5     1.35
##  [3,] 0 5  0.1   12.5  2.5 12.5     3.75
##  [4,] 0 7  0.1   17.5  3.5 24.5     7.35
##  [5,] 1 3  0.1   10.0  2.0  2.0     0.60
##  [6,] 1 5  0.1   15.0  3.0  8.0     2.40
##  [7,] 1 7  0.1   20.0  4.0 18.0     5.40
##  [8,] 3 5  0.1   20.0  4.0  2.0     0.60
##  [9,] 3 7  0.1   25.0  5.0  8.0     2.40
## [10,] 5 7  0.1   30.0  6.0  2.0     0.60

# The expected value of ybar equals the average of all possible Ybars.
mean(c4)

## [1] 3.2

# note that it is equal to mu since it is unbiased.
all.equal(mean(c4), mu)

## [1] TRUE

# The variance of ybar is equal to (written 2 ways)
(sigma/n) * ((N - n)/(N - 1))

## [1] 2.46

mean((c4 - mu)^2)

## [1] 2.46

# the expected value of the variance of ybar (written 2 ways)
sum(c6 * c2)

## [1] 2.46

mean(c6)

## [1] 2.46

• Calculate the expected value of ybar and the variance of ybar for an SRS of size n=2, with replacement and with equal probaiblities.

##     [,1] [,2]
##        1    2
##        1    3
##        1    4
##        1    5
##        2    3
##        2    4
##        2    5
##        3    4
##        3    5
##        4    5
## r11    0    0
## r12    1    1
## r13    3    3
## r14    5    5
## r15    7    7

##            Prob Tauhat Ybar  S2 VhatYbar
##     1 2 0.06667    7.5  1.5 0.5     0.15
##     1 3 0.06667   10.0  2.0 2.0     0.60
##     1 4 0.06667   12.5  2.5 4.5     1.35
##     1 5 0.06667   15.0  3.0 8.0     2.40
##     2 3 0.06667   12.5  2.5 0.5     0.15
##     2 4 0.06667   15.0  3.0 2.0     0.60
##     2 5 0.06667   17.5  3.5 4.5     1.35
##     3 4 0.06667   17.5  3.5 0.5     0.15
##     3 5 0.06667   20.0  4.0 2.0     0.60
##     4 5 0.06667   22.5  4.5 0.5     0.15
## r11 0 0 0.06667    0.0  0.0 0.0     0.00
## r12 1 1 0.06667    5.0  1.0 0.0     0.00
## r13 3 3 0.06667   15.0  3.0 0.0     0.00
## r14 5 5 0.06667   25.0  5.0 0.0     0.00
## r15 7 7 0.06667   35.0  7.0 0.0     0.00

## [1] 3.067

## [1] 2.713

## [1] 0.5

2 The owner of a movie rental business wants to estimate the proportion of DVD's that were rented in the past month. She randomly samples 15 of the 85 racks of DVD's in her store, then for each DVD in the rack, she records whether or not it was rented. In this sample, identify each of the following: the population, the frame, the sampling units, and the elements. Of the types of probability sampling methods that we reviewed in Chapter 2, which one does this seem most like? Briefly explain.

population = a collection of elements about which we wish to make an inference = all dvd's in the store. We wish to infer the proportion rented in the past month.

the frame = a list of sampling units = racks
sampling units = nonoverlapping collections of elements from the population that cover the entire population = the racks are nonoverlapping collections of DVD's.
element = an object on which a measurement is taken = the DVD's are measured as watched or not watched, making them elements.

Simple random sampling = consists of selecting a group of n sampling units in such a way that each sample of size n has the same chance of being selected. In this case, we select 15

3 You have been hired to help estimate the total corn production (in bushels) for a particular state for last year. From N=98 counties in the state, a simple random sample of n = 15 was taken, and the corn production (in bushels) was recorded. The data is available in the “Corn data” file.

• Calculate an estimate of the corn production in this state for last year, and calculate a bound for your estimate.
• If it is desired to achieve an error bound of B=15,000,000 bushels for future estimates, how many counties should be sampled?

# part A
dat = read.csv("H:\\STAT422\\corndata.csv", header = TRUE)
N = 98
n = 15
### total is N times ybar.
ybar = mean(dat$bushels)
total = ybar * N
total

## [1] 52479653

Var.total = N^2 * (1 - n/N) * (var(dat$bushels/n))
Var.total

## [1] 1.427e+13

Bound = 2 * sqrt(Var.total)
Bound

## [1] 7554215

# part B
B = 1.5e+07
N = 98
sig2 = var(dat$bushels)
D = B^2/(4 * N^2)
n = (N * sig2)/((N - 1) * D + sig2)
n

## [1] 40.17

# so round up
n = ceiling(n)
n

## [1] 41

4 In a state with approximately 20,000 registered voters, a sample of 500 is taken to estimate what proportion support each of 4 candidates for governor. The results are that 192 support candidate Johnson, 160 support candidate Smith, 97 support candidate Brown, adn 51 support candidate Jones.

• a. Estimate hte proportion p that support candidate Johnson in the population, and calculate a bound on the error of estimate.
• b. Is there a true difference in the proportion of voters supporting candidates Johnson and Smith?

N = 20000
n = 500
phat1 = 192/n
phat2 = 160/n
phat3 = 97/n
phat4 = 51/n
paste(c("The estimated proportion in support of Johnson is", phat1))

## [1] "The estimated proportion in support of Johnson is"
## [2] "0.384"                                            

Var.phat1 = (1 - n/N) * ((phat1 * (1 - phat1))/(n - 1))
Bound.phat1 = 2 * sqrt(Var.phat1)
Bound.phat1

## [1] 0.043

Var.phat2 = (1 - n/N) * ((phat2 * (1 - phat2))/(n - 1))
Bound.phat2 = 2 * sqrt(Var.phat2)
Bound.phat2

## [1] 0.04124

Var.phat3 = (1 - n/N) * ((phat3 * (1 - phat3))/(n - 1))
Bound.phat3 = 2 * sqrt(Var.phat3)
Bound.phat3

## [1] 0.03496

Var.phat4 = (1 - n/N) * ((phat4 * (1 - phat4))/(n - 1))
Bound.phat4 = 2 * sqrt(Var.phat4)
Bound.phat4

## [1] 0.02676

rbind(c(phat1, phat2, phat3, phat4), c(Bound.phat1, Bound.phat2, 
    Bound.phat3, Bound.phat4))

##       [,1]    [,2]    [,3]    [,4]
## [1,] 0.384 0.32000 0.19400 0.10200
## [2,] 0.043 0.04124 0.03496 0.02676

# Part B the point estimator for the difference in proportions:
p.diff = phat1 - phat2
var.p.diff = Var.phat1 + Var.phat2 + (2 * phat1 * phat2)/n
var.p.diff

## [1] 0.001379

Bound.p.diff = 2 * sqrt(var.p.diff)
Bound.p.diff

## [1] 0.07427

5 For the cars93 dataset with N=92 types of cars, a simple random sample was taken of n=10 cars. The miles per gallon on the highway for these 10 carss were: 31, 26, 20, 30, 36, 25, 26, 30, 23, and 30. Estimate the mean of highway miles per gallon for the population of N=92 cars, and calculate a bound on the error of estimation.

# just for kicks, cuz we dont need dataset?
cardat = read.table("H:\\STAT422\\test1\\cars93.txt", sep = "")
N = 92
n = 10
mpg = c(31, 26, 20, 30, 36, 25, 26, 30, 23, 30)

# the point estimate of mu is ybar:
ybar = mean(mpg)
ybar

## [1] 27.7

Var.ybar = (1 - n/N) * (var(mpg)/n)
Bound.ybar = 2 * sqrt(Var.ybar)
Bound.ybar

## [1] 2.744