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Let \(N\) denote the number of “observational units” in a population of observations. This can be infinite or finite.
In an experiment the number of observational units is (theoretically) infinite.
In an survey the number of observational units is finite.
When the number of observational units is finite we need to be concerned with if we are sampling with or without replacement.
When sampling with replacement, it is possible to observe the same unit more than once.
Example: Suppose we had a very small forest with \(N\) = 3 trees with volumes of 10, 20, and 30 cubic feet. Assume we will obtain a sample of \(n\) = 2 observations of tree volumes using sampling with replacement. We select trees one at a time, and each time we do every tree in the population has the same probability of being selected.| Sample | Probability | \(\bar{x}\) |
|---|---|---|
| 10,10 | 1/9 | 10 |
| 20,10 | 1/9 | 15 |
| 30,10 | 1/9 | 20 |
| 10,20 | 1/9 | 15 |
| 20,20 | 1/9 | 20 |
| 30,20 | 1/9 | 25 |
| 10,30 | 1/9 | 20 |
| 20,30 | 1/9 | 25 |
| 30,30 | 1/9 | 30 |
| \(\bar{x}\) | Probability |
|---|---|
| 10 | 1/9 |
| 15 | 2/9 |
| 20 | 3/9 |
| 25 | 2/9 |
| 30 | 1/9 |
Here \(\bar{x}\) has mean 20 and standard deviation (i.e., standard error) of about 5.77.
When sampling without replacement, it is not possible to observe the same unit more than once.
Example: Suppose we had a very small forest with \(N\) = 3 trees with volumes of 10, 20, and 30 cubic feet. Assume we will obtain a sample of \(n\) = 2 observations of tree volumes using sampling without replacement. We select trees one at a time, and each time we do every tree in the population that was not previously selected has the same probability of being selected.| Sample | Probability | \(\bar{x}\) |
|---|---|---|
| 10,20 | 1/6 | 15 |
| 20,10 | 1/6 | 15 |
| 10,30 | 1/6 | 20 |
| 30,10 | 1/6 | 20 |
| 20,30 | 1/6 | 25 |
| 30,20 | 1/6 | 25 |
| \(\bar{x}\) | Probability |
|---|---|
| 15 | 1/3 |
| 20 | 1/3 |
| 25 | 1/3 |
The sampling distribution shows that \(\bar{x}\) has mean 20 and standard deviation (i.e., standard error) of about 4.08.
Example: Now consider another forest with \(N\) = 100 trees, of which 60 have a volume of 10 cubic feet, and 40 have a volume of 20 cubic feet, and suppose we select a random sample of \(n\) = 5 trees by using sampling without replacement.| Volume | Frequency | Probability |
|---|---|---|
| 10 | 60 | 0.6 |
| 20 | 40 | 0.4 |
| Observation | 10 | 20 | Sample |
|---|---|---|---|
| 1 | 60/100 | 40/100 | 10 |
| 2 | 59/99 | 40/99 | 10, 20 |
| 3 | 59/98 | 39/98 | 10, 20, 10 |
| 4 | 58/97 | 39/97 | 10, 20, 10, 10 |
| 5 | 58/96 | 38/96 | 10, 20, 10, 10, 20 |
If we were sampling with replacement, the probability of the sample 10, 20, 10, 10, 20 would be \[ 0.6 \times 0.4 \times 0.6 \times 0.6 \times 0.4 = 0.03456, \] but because we were sampling without replacement, the probability of the sample is \[ 0.6 \times 40/99 \times 59/98 \times 58/97 \times 39/96 \approx 0.0347294. \] When sampling without replacement, the observations are not independent. Observations are independent only if the probabilities for one observation do not depend on the other observations.
Suppose we are estimating \(\mu\) with \(\bar{x}\). The standard error formula \(s/\sqrt{n}\) assumes sampling with replacement. When sampling without replacement the correct formula for the standard error of \(\bar{x}\) is \[ \frac{s}{\sqrt{n}}\sqrt{1 - \frac{n}{N}}, \] where \(N\) is the population size. The margin of error is then \[ t\frac{s}{\sqrt{n}}\sqrt{1 - \frac{n}{N}}, \] and the confidence interval for estimating \(\mu\) is \[ \bar{x} \pm t\frac{s}{\sqrt{n}}\sqrt{1 - \frac{n}{N}}. \] The term \(1-n/N\) is sometimes called the finite population correction (FPC).
Example: The mean test score of a simple random sample of 50 students from a school of 200 is 70 points, with a standard deviation of 10 points. What is the margin of error and confidence interval for the mean test score for all 200 students?
Example: Conservation researchers conducted a survey to estimate the density and abundance of a species of “larkspur” (a flowering plant of the genus Delphinium) in a rectangular region. To survey the larkspur they divided the region into 150 one-meter wide transects from north to south. The researchers selected a sample of 20 transects using random sampling. The mean density of larkspur in these 20 transects was 0.83 larkspur per square meter, and the standard deviation was 0.72 larkspur per square meter. What is the point estimate, margin of error, and confidence interval for the mean density of larkspur in the region?
Question: In some cases we might want to omit the finite population correction term \(1-n/N\) by effectively setting \(1-n/N = 1\) so that \(\sqrt{1-n/N}\) “disappears” from our equations. Sometimes we do this to simplify calculations, but also we may need to do this if we do not know \(N\). When would it be reasonable to omit the finite population correction term?
In survey sampling, the parameter \(\mu\) is the mean of all units in the population, so that \[ \mu = \frac{x_1 + x_2 + \cdots + x_N}{N}. \] The parameter \(\tau\) is the total of all units in the population, so that \[ \tau = x_1 + x_2 + \cdots + x_N. \] Note that \(\mu = \tau/N\) and \(\tau = N\mu\).
Examples of population totals that we might want to estimate:
The point estimate of \(\tau\) is \(N\bar{x}\) since \(\tau = N\mu\) and \(\bar{x}\) estimates \(\mu\). The standard error of \(N\bar{x}\) (assuming sampling without replacement) is \[ N\frac{s}{\sqrt{n}}\sqrt{1 - \frac{n}{N}}, \] so the margin of error for estimating \(\tau\) is \[ tN\frac{s}{\sqrt{n}}\sqrt{1 - \frac{n}{N}}, \] and the confidence interval for \(\tau\) is \[ N\bar{x} \pm tN\frac{s}{\sqrt{n}}\sqrt{1 - \frac{n}{N}}. \] Note: We will always assume sampling without replacement when estimating \(\tau\).
Example: Based on a random sample (without replacement) of \(n\) = 100 students at a university of \(N\) = 1000 students, the mean expenditure per year on phone apps was \(\bar{x}\) = 20 dollars with a standard deviation of \(s\) = 10 dollars. What is the estimate of the total expenditure on phone apps per year for all of the students at the university?
Example: Conservation researchers conducted a survey to estimate the density and abundance of a species of “larkspur” (a flowering plant of the genus Delphinium) in a rectangular region. To survey the larkspur they divided the region into 150 one-meter wide transects from north to south. The researchers selected a sample of 20 transects using random sampling. The mean number of larkspur in these 20 transects was 21.45 larkspur, and the standard deviation was 21.52 larkspur. What is the point estimate, margin of error, and confidence interval for the total number of larkspur in the region?