Sampling and Sampling Distributions Study Notes
Sampling and Sampling Distributions
Introduction to Sampling Distributions
Sampling Distribution of \bar{x}
Point Estimation
Selecting a Sample
Other Sampling Methods
Sampling Distribution of \bar{p}
Selecting a Sample
Sampling from a Finite Population
Finite populations are often defined by lists such as:
Organization membership roster
Credit card account numbers
Inventory product numbers
A simple random sample of size n from a finite population of size N is a sample selected such that each possible sample of size n has the same probability of being selected.
In large sampling projects, computer-generated random numbers are often used to automate the sample selection process.
Sampling without replacement is the procedure used most often.
Replacing each sampled element before selecting subsequent elements is called sampling with replacement.
Example: St. Andrew’s College
St. Andrew’s College received 900 applications for admission in the upcoming year from prospective students.
The applicants were numbered from 1 to 900 as their applications arrived.
The Director of Admissions would like to select a simple random sample of 30 applicants.
Step 1: Assign a random number to each of the 900 applicants.
Step 2: Select the 30 applicants corresponding to the 30 smallest random numbers.
Sampling from an Infinite Population
Sometimes we want to select a sample, but find it is not possible to obtain a list of all elements in the population.
As a result, we cannot construct a frame for the population.
Hence, we cannot use the random number selection procedure.
Most often this situation occurs in infinite population cases.
Populations are often generated by an ongoing process where there is no upper limit on the number of units that can be generated.
Some examples of on-going processes, with infinite populations, are:
parts being manufactured on a production line
transactions occurring at a bank
telephone calls arriving at a technical help desk
customers entering a store
A random sample from an infinite population is a sample selected such that the following conditions are satisfied.
Each element selected comes from the population of interest.
Each element is selected independently.
In the case of an infinite population, we must select a random sample in order to make valid statistical inferences about the population from which the sample is taken.
Point Estimation
Example: St. Andrew’s College
Recall that St. Andrew’s College received 900 applications from prospective students.
The application form contains a variety of information including the individual’s Scholastic Aptitude Test (SAT) score and whether or not the individual desires on-campus housing.
At a meeting in a few hours, the Director of Admissions would like to announce the average SAT score and the proportion of applicants that want to live on campus, for the population of 900 applicants.
However, the necessary data on the applicants have not yet been entered in the college’s computerized database.
So, the Director decides to estimate the values of the population parameters of interest based on sample statistics.
The sample of 30 applicants is selected using computer-generated random numbers.
\bar{x} as Point Estimator of \mu
s as Point Estimator of \sigma
\bar{p} as Point Estimator of p
Note: Different random numbers would have identified a different sample which would have resulted in different point estimates.
Population Parameter
\mu = Population Mean SAT Score
\sigma = Population Standard Deviation for SAT Score
p = Population Proportion Wanting On-Campus Housing
Population Parameter | Point Estimator | Point Estimate | Parameter Value |
|---|---|---|---|
\mu = Population mean SAT score | \bar{x} | 1097 | 1090 |
\sigma = Population std. deviation for SAT score | s | 75.2 | 80 |
p = Population proportion wanting campus housing | \bar{p} | .68 | .72 |
Summary of Point Estimates Obtained from a Simple Random Sample
Sampling Distribution of \bar{x}
Process of Statistical Inference
A simple random sample of n elements is selected from the population.
The sample data provide a value for the sample mean \bar{x}.
The value of \bar{x} is used to make inferences about the value of \mu.
The sampling distribution of \bar{x} is the probability distribution of all possible values of the sample mean \bar{x}.
where:
\mu = the population mean
Expected Value of \bar{x}
E(\bar{x}) = \mu
When the expected value of the point estimator equals the population parameter, we say the point estimator is unbiased.
Standard Deviation of \bar{x}
We will use the following notation to define the standard deviation of the sampling distribution of \bar{x}.
\sigma_{\bar{x}} = the standard deviation of \bar{x}
\sigma = the standard deviation of the population
n = the sample size
N = the population size
Finite Population
\sigma_{\bar{x}} = \sqrt{\frac{N-n}{N-1}} * \frac{\sigma}{\sqrt{n}}
Infinite Population
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}
\sigma_{\bar{x}} is referred to as the standard error of the mean.
\sqrt{\frac{N-n}{N-1}}
is the finite population correction factor.A finite population is treated as being infinite if \frac{n}{N} < .05
When the population has a normal distribution, the sampling distribution of \bar{x} is normally distributed for any sample size.
In cases where the population is highly skewed or outliers are present, samples of size 50 may be needed.
In most applications, the sampling distribution of \bar{x} can be approximated by a normal distribution whenever the sample is size 30 or more.
The sampling distribution of \bar{x} can be used to provide probability information about how close the sample mean \bar{x} is to the population mean \mu.
Central Limit Theorem
When the population from which we are selecting a random sample does not have a normal distribution, the central limit theorem is helpful in identifying the shape of the sampling distribution of \bar{x}.
CENTRAL LIMIT THEOREM: In selecting random samples of size n from a population, the sampling distribution of the sample mean \bar{x} can be approximated by a normal distribution as the sample size becomes large.
Example: St. Andrew’s College Sampling Distribution of \bar{x} for SAT Scores
What is the probability that a simple random sample of 30 applicants will provide an estimate of the population mean SAT score that is within +/- 10 of the actual population mean \mu?
In other words, what is the probability that \bar{x} will be between 1080 and 1100?
Step 1: Calculate the z-value at the upper endpoint of the interval.
z = \frac{1100-1090}{14.6} = .68
Step 2: Find the area under the curve to the left of the upper endpoint.
P(z < .68) = .7517
Step 3: Calculate the z-value at the lower endpoint of the interval.
z = \frac{1080-1090}{14.6} = -.68
Step 4: Find the area under the curve to the left of the lower endpoint.
P(z < -.68) = .2483
Step 5: Calculate the area under the curve between the lower and upper endpoints of the interval.
P(-.68 < z < .68) = P(z < .68) - P(z < -.68) = .7517 - .2483 = .5034
The probability that the sample mean SAT score will be between 1080 and 1100 is:
P(1080 < \bar{x} < 1100) = .5034
Example: St. Andrew’s College Relationship Between the Sample Size and the Sampling Distribution of \bar{x}
Suppose we select a simple random sample of 100 applicants instead of the 30 originally considered.
E(\bar{x}) = \mu
regardless of the sample size. In our example, E(\bar{x}) remains at 1090.
Whenever the sample size is increased, the standard error of the mean \sigma_{\bar{x}} is decreased. With the increase in the sample size to n = 100, the standard error of the mean is decreased from 14.6 to:
With n = 30,
\sigma_{\bar{x}} = 14.6
With n = 100,
\sigma_{\bar{x}} = 8.04
Recall that when n = 30, P(1080 < \bar{x} < 1100) = .5034.
We follow the same steps to solve for P(1080 < \bar{x} < 1100) when n = 100 as we showed earlier when n = 30.
Now, with n = 100, P(1080 < \bar{x} < 1100) = .7888.
Because the sampling distribution with n = 100 has a smaller standard error, the values of \bar{x} have less variability and tend to be closer to the population mean than the values of \bar{x} with n = 30.
Sampling Distribution of \bar{p}
Process of Statistical Inference
A simple random sample of n elements is selected from the population.
The sample data provide a value for the sample proportion \bar{p}.
The value of \bar{p} is used to make inferences about the value of p.
The sampling distribution of \bar{p} is the probability distribution of all possible values of the sample proportion \bar{p}.
where: p = the population proportion
Expected Value of \bar{p}
E(\bar{p}) = p
Standard Deviation of \bar{p}
Finite Population
\sigma_{\bar{p}} = \sqrt{\frac{N-n}{N-1}} * \sqrt{\frac{p(1-p)}{n}}
Infinite Population
\sigma_{\bar{p}} = \sqrt{\frac{p(1-p)}{n}}
\sigma_{\bar{p}} is referred to as the standard error of the proportion.
\sqrt{\frac{N-n}{N-1}}
is the finite population correction factor.
Form of the Sampling Distribution of \bar{p}
The sampling distribution of \bar{p} can be approximated by a normal distribution whenever np > 5 and n(1 – p) > 5
Example: St. Andrew’s College Sampling Distribution of \bar{p}
Recall that 72% of the prospective students applying to St. Andrew’s College desire on-campus housing.
What is the probability that a simple random sample of 30 applicants will provide an estimate of the population proportion of applicant desiring on-campus housing that is within plus or minus .05 of the actual population proportion?
For our example, with n = 30 and p = .72, the normal distribution is an acceptable approximation because:
n*p* = 30(.72) = 21.6 > 5 and n(1 - p) = 30(.28) = 8.4 > 5
Step 1: Calculate the z-value at the upper endpoint of the interval.
z = \frac{.77-.72}{.082} = .61
Step 2: Find the area under the curve to the left of the upper endpoint.
P(z < .61) = .7291
Step 3: Calculate the z-value at the lower endpoint of the interval.
z = \frac{.67-.72}{.082} = -.61
Step 4: Find the area under the curve to the left of the lower endpoint.
P(z < -.61) = .2709
Step 5: Calculate the area under the curve between the lower and upper endpoints of the interval.
P(-.61 < z < .61) = P(z < .61) - P(z < -.61) = .7291 - .2709 = .4582
The probability that the sample proportion of applicants wanting on-campus housing will be within +/- .05 of the actual population proportion :
P(.67 < \bar{p} < .77) = .4582
Other Sampling Methods
Stratified Random Sampling
Cluster Sampling
Systematic Sampling
Convenience Sampling
Judgment Sampling