Simple Random Sampling

Goal

Goal: to know the mean number of boreal toad egg masses per pond in RMNP

• different egg masses per pond is meaningful why?

SRS

Conceptual Walkthrough

• We have a known population of ponds, N = 6

• We have enough to money for n = 2

• Will use SRS

Pond	egg.mass
A	2
B	6
C	8
D	10
E	10
F	12

SRS

Population Parameters:

• \(\mu = 8\)
• \(N = 6\)
• \(\sigma^2_{y} = 12.8\)

SRS

How many possible unique samples are there (w/o replacement)?

\[ {N}\choose{n} \]

\[ \frac{N!}{n!(N-n)!} \]

R

choose(6,2)

[1] 15

SRS

What is the probability of any one particular sample?

SRS: all samples have the same probability!

Convenient Sampling: What is the probability of one particular sample?

SRS

What is the probability pond “A” will be sampled”? Pond “B”?

Look at all possible combinations:

utils::combn(LETTERS[1:6],2)

     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14]
[1,] "A"  "A"  "A"  "A"  "A"  "B"  "B"  "B"  "B"  "C"   "C"   "C"   "D"   "D"  
[2,] "B"  "C"  "D"  "E"  "F"  "C"  "D"  "E"  "F"  "D"   "E"   "F"   "E"   "F"  
     [,15]
[1,] "E"  
[2,] "F"  

SRS

Sample.Number	First Pond	Second Pond	First Value	Second Value
1	A	B	2	6
2	A	C	2	8
3	A	D	2	10
4	A	E	2	10
5	A	F	2	12
6	B	C	6	8
7	B	D	6	10
8	B	E	6	10
9	B	F	6	12
10	C	D	8	10
11	C	E	8	10
12	C	F	8	12
13	D	E	10	10
14	D	F	10	12
15	E	F	10	12

What is the probability pond “A” will be sampled”? Pond “B”?

SRS

Is it okay to not like your random sample and resample?

sample(LETTERS[1:6],2)

[1] "A" "D"

sample(LETTERS[1:6],2)

[1] "C" "B"

sample(LETTERS[1:6],2)

[1] "B" "A"

SRS

Consider the sample mean \(\hat{\mu}\) for each sample

Sample.Number	First Pond	Second Pond	First Value	Second Value	Sample.Mean	Sample.Error
1	A	B	2	6	4	-4
2	A	C	2	8	5	-3
3	A	D	2	10	6	-2
4	A	E	2	10	6	-2
5	A	F	2	12	7	-1
6	B	C	6	8	7	-1
7	B	D	6	10	8	0
8	B	E	6	10	8	0
9	B	F	6	12	9	1
10	C	D	8	10	9	1
11	C	E	8	10	9	1
12	C	F	8	12	10	2
13	D	E	10	10	10	2
14	D	F	10	12	11	3
15	E	F	10	12	11	3

SRS

Sample.Number	Sample.Mean	Sample.Error
1	4	-4
2	5	-3
3	6	-2
4	6	-2
5	7	-1
6	7	-1
7	8	0
8	8	0
9	9	1
10	9	1
11	9	1
12	10	2
13	10	2
14	11	3
15	11	3

\[ \frac{1}{15}\times\sum_{i=1}^{15} (\hat{\mu}_{i}) = 8\]

\[\sum_{i=1}^N(\hat{\mu}_{i}-\mu) = 0 \]

SRS

Sample.Mean	Frequency	Relative.Freq	Mean.times.Rel.Freq
4	1	0.067	0.267
5	1	0.067	0.333
6	2	0.133	0.800
7	2	0.133	0.933
8	2	0.133	1.067
9	3	0.200	1.800
10	2	0.133	1.333
11	2	0.133	1.467
Sum	15	1.000	8.000

\[ E[\mu] = \sum_{q=1}^{Q} p_i \times \hat{\mu}_{i} = 8 \]

• \(Q\) = number of unique sample means

• \(p_i\) = probability of obtaining a given sample / relative frequency

SRS

Sampling Disribution

• Sample mean formula is an estimator of the population mean (parameter )
• Sample mean is a random variable with a sampling distribution
  • sample mean varies from sample-to-sample becasue of the sampling process
• The sampling distribution is specific to an estimator - has known outcomes and relative frequencies of values

Sampling Disribution

• Judge an estimator by its sampling distribution
• What properties do we want?

Estimator Properties

• Precise and unbiased estimator
• Is our estimator precise?

Variance of Sampling Distribution

Sample.Number	Sample.Mean	Sample.Error
1	4	-4
2	5	-3
3	6	-2
4	6	-2
5	7	-1
6	7	-1
7	8	0
8	8	0
9	9	1
10	9	1
11	9	1
12	10	2
13	10	2
14	11	3
15	11	3

var(ponds.simple$Sample.Mean)

[1] 4.571429

Population Variance

The variance of all sample units

\[ \sigma^{2} = \frac{1}{N-1}\sum_{i=1}^{N} \left(y_{i}-\mu\right)^{2} \]

\[ \sigma^{2} = \frac{1}{6-1}\sum_{i=1}^{6} \left(y_{i}-8\right)^{2} \]

(1/(6-1))*sum((ponds$egg.mass-8)^2)

[1] 12.8

Sample Variance

Estimate population variance from each sample

\[ \hat{\sigma}^{2} = \frac{1}{n-1}\sum_{i=1}^{n} \left(y_{i}-\hat{\mu}\right)^{2} \]

var.per.sample = apply(
                       cbind(ponds.all$`First Value`,ponds.all$`Second Value`),
                       1,
                       var
                       )
# Expected value of population variance
  mean(var.per.sample)

[1] 12.8

Var Sample Means: 4.2666667

Sample Distribution of Variance

• Unbiased estimate of the population variance.
• Individual values will deviate from the population variance.

Connect these two

• Population variance - variation among sample units; estimate from sample variance
• Var. Sampling Distribution - variance of mean values from each possible sample

Connect these two

Variance of all units vs variance of all sample means

Connect these two

\[ E[\text{Sampling Distribution Variance}] = \\\frac{1}{n} \left(\frac{N-n}{N}\right) \times \sigma^2 \]

• As the sample size (n) increases the sample variance declines by 1/n
• Finite-population correction factor (N-n/N)

Connect these two

• Sample size (n) = 2
• Total units (N) = 6
• E[\(\sigma^2\)] = 12.8
• E[sample dist. var] = 4.27
• 1/n = 0.5
• Finite correction factor = 0.67

• E[\(\sigma^2\)] \(\times\) 1/n \(\times\) FCF = 4.27

Connect these two

\[ E[\text{Sampling Distribution Variance}] = \\\frac{1}{n} \left(\frac{N-n}{N}\right) \times \sigma^2 \]

• As n–> N, (N-n)/(N) approaches zero.
• n = N then no sampling distribution variance
• n << N then correction factor ~1 and expected sampling distribution variance is related to the population variance by 1/n

Connect these two

Why does this matter?

Connect these two

If you know the E[\(\sigma^2\)] …

• you have a mechanism to understand the variation of sample distribution of the means
• the more variation in \(y\), the more variation in sampling distrubution of means (\(\hat{\mu}\)).

Sample Size

\[ E[\hat{\sigma}^2] \]

• n = 2 –> 4.27
• n = 3 –> 2.13
• n = 4 –> 1.06
• n = 5 –> 0.43
• n = 6 –> 0

Sample Size

Not pond example; When N is large enough . . .

Sample Size

• Sampling distribution is less variable
• Sampling distribution centers on population mean
• As n increases, the distribution becomes ‘Normal’

This is because of the Central limit theorem (CLT)

• CLT relies on random/prob. based sampling
• leads to parameter unbiasedness
• Does not depend on distribution of samples!
• allows estimation of precision of parameters
• allows estimation of confidence intervals

The TOTAL

Estimate the total population using the sample

\[ \hat{\tau} = N \times \hat{\mu} = \frac{N}{n}\sum_{i=1}^{n}(y_i) \]

The TOTAL

Back to ponds example

The TOTAL

Variance of the total

\[ \text{var}(\hat{\tau}) = N\times(N-1) \times \frac{\hat{\sigma}^2}{n} \]

Probabilty of

Whenever you have the sampling distribution, frame precision in terms of what matters to you.

Probabilty of

\[ P(\hat{\tau}\leq 2 \times \tau) \]

length(which( tau.est <=  2*48))/length(tau.est)

[1] 1

Sample Size

But what \(n\)?

• \(n\) impacts variation in estimates
• variation in \(y\) impacts the influence of \(n\)
• research objective probably impacts \(n\) the most
• higher \(n\) is not always better