Sampling Variability and Sampling Distributions

Chapter 8: Sampling Variability and Sampling Distributions

Introduction to Sampling

• We often want to estimate a population parameter (e.g., mean fat content $m$ of hamburgers).

• We take a sample of size $n$ (e.g., $n = 50$ hamburgers) and calculate a sample statistic (e.g., sample mean).

• Key questions:

  • Is the sample mean a good estimate of $m$?

  • How close is the sample mean to $m$?

  • Will other samples of $n = 50$ have the same sample mean?

Sampling Distribution

• The sampling distribution describes the long-run behavior of a sample statistic.

• The sample mean (denoted as $\bar{x}$) is a statistic.

Statistics and Sampling Variability _

• Statistic: A number computed from sample data.

• Examples of statistics:

  • $\bar{x}$ - sample mean

  • $s$ - sample standard deviation

  • $\hat{p}$ - sample proportion

• The value of a statistic depends on the specific sample selected.

• Sampling Variability: The variability of a statistic from sample to sample.

Illustrative Example: Fish Pond

• Consider a pond with 20 fish.

• Fish lengths (in inches): 4.5, 5.4, 10.3, 7.9, 8.5, 6.6, 11.7, 8.9, 2.2, 9.8, 6.3, 4.3, 9.6, 8.7, 13.3, 4.6, 10.7, 13.4, 7.7, 5.6

• True population mean: $\,mu = 8$

• Example Samples:

  • Sample 1: 6.3, 2.2, 13.3 inches; $\bar{x} = 7.27$ inches

  • Sample 2: 8.5, 4.6, 5.6 inches; $\bar{x} = 6.23$ inches

  • Sample 3: 10.3, 8.9, 13.4 inches; $\bar{x} = 10.87$ inches

• Demonstrates that sample means vary (sampling variability).

• Some sample means are closer to the true mean, some are farther; some are above, and some are below the true mean.

Sampling Distribution of $\bar{x}$ (Fish Pond Continued)

• There are 1140 ( $<em>{20}C</em>3$ ) possible samples of size 3 from the fish population.

• If we calculate the mean length of each possible sample, we would have the sampling distribution of $\bar{x}$.

Definition: Sampling Distributions of $\bar{x}$

• The distribution formed by considering the value of a sample statistic (like $\bar{x}$) for every possible sample of a given size from a population.

Fish Pond Revisited (Smaller Population)

• Simplified scenario: Only 5 fish in the pond.

• Lengths: 6.6, 11.7, 8.9, 2.2, 9.8

• Population mean: $\,mu_x = 7.84$

• Population standard deviation: $\,sigma_x = 3.262$

• We keep the population size small to find all possible samples.

Sampling Distribution with Samples of Size 2

• Consider all possible pairs (samples of size 2) from the 5 fish.

• Possible pairs and their means $\bar{x}$:

  • 6.6 & 11.7: $\bar{x} = 9.15$

  • 6.6 & 8.9: $\bar{x} = 7.75$

  • 6.6 & 2.2: $\bar{x} = 4.4$

  • 6.6 & 9.8: $\bar{x} = 8.2$

  • 11.7 & 8.9: $\bar{x} = 10.3$

  • 11.7 & 2.2: $\bar{x} = 6.95$

  • 11.7 & 9.8: $\bar{x} = 10.75$

  • 8.9 & 2.2: $\bar{x} = 5.55$

  • 8.9 & 9.8: $\bar{x} = 9.35$

  • 2.2 & 9.8: $\bar{x} = 6$

• There are 10 possible samples.

• Mean of sample means: $\,mu_{\bar{x}} = 7.84$

• Standard deviation of sample means: $\,sigma_{\bar{x}} = 1.998$

• These values define the sampling distribution of $\bar{x}$ for samples of size 2.

• The mean of the sampling distribution equals the population mean.

Sampling Distribution with Samples of Size 3

• Consider all possible triplets (samples of size 3) from the 5 fish.

• Possible triplets and their means $\bar{x}$:

  • 6.6, 11.7, 8.9: $\bar{x} = 9.067$

  • 6.6, 11.7, 2.2: $\bar{x} = 6.833$

  • 6.6, 11.7, 9.8: $\bar{x} = 9.367$

  • 6.6, 8.9, 2.2: $\bar{x} = 5.9$

  • 6.6, 8.9, 9.8: $\bar{x} = 8.433$

  • 6.6, 2.2, 9.8: $\bar{x} = 6.2$

  • 11.7, 8.9, 2.2: $\bar{x} = 7.6$

  • 11.7, 8.9, 9.8: $\bar{x} = 10.133$

  • 11.7, 2.2, 9.8: $\bar{x} = 7.9$

  • 8.9, 2.2, 9.8: $\bar{x} = 6.967$

• There are 10 possible samples.

• Mean of sample means: $\,mu_{\bar{x}} = 7.84$

• Standard deviation of sample means: $\,sigma_{\bar{x}} = 1.332$

• These values determine the sampling distribution of $\bar{x}$ for samples of size 3.

Key Observations

• The mean of the sampling distribution EQUALS the mean of the population: $\,mu_{\bar{x}} = \mu$

• As the sample size ($n$) increases, the standard deviation of the sampling distribution decreases: $\,sigma_{\bar{x}}$ decreases as $n$ increases.

General Properties of Sampling Distributions of $\bar{x}$

• Rule 1: The mean of the sampling distribution of $\bar{x}$ is equal to the population mean: $\,mu_{\bar{x}} = \mu$

• Rule 2: The standard deviation of the sampling distribution of $\bar{x}$ is given by:

  • $\,sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

  • This is exact for infinite populations.

  • Approximately correct if the population is finite and the sample size is no more than 10% of the population.

  • (The fish pond examples didn't satisfy this condition, so the formula wasn't accurate there.)

Example: Platelet Volume

• Study: “Mean Platelet Volume in Patients with Metabolic Syndrome and Its Relationship with Coronary Artery Disease” (Thrombosis Research, 2007).

• Platelet volume of patients without metabolic syndrome is approximately normal.

• Population mean: $\,mu = 8.25$

• Population standard deviation: $\,sigma = 0.75$

• Minitab was used to generate 500 random samples for each of the following sample sizes: $n = 5, 10, 20, 30$.

• Density histograms of the 500 sample means were created.

• Observations:

  • The means of the histograms are approximately the population mean $\,mu = 8.25$.

  • The standard deviation of the histograms decreases as $n$ increases (consistent with Rule 2).

  • The shape of the histograms is approximately normal (consistent with Rule 3).

Rule 3: Normality

• When the population distribution is normal, the sampling distribution of $\bar{x}$ is also normal for any sample size $n$.

Example: NHL Overtime Game Length

• Study: “Is the Overtime Period in an NHL Game Long Enough?” (American Statistician, 2008).

• Data: Time (in minutes) from the start of the game to the first goal scored in overtime for 281 games.

• The distribution is strongly positively skewed.

• Population mean: $\,mu = 13$ minutes.

• Population median: 10 minutes.

• Minitab was used to generate 500 samples of sizes $n = 5, 10, 20, 30$.

• Observations:

  • Histograms are centered approximately at $\,mu = 13$.

  • Standard deviations decrease as $n$ increases.

  • Shapes of histograms become more normal as $n$ increases, even though the population is skewed.

Rule 4: Central Limit Theorem (CLT)

• When $n$ is sufficiently large, the sampling distribution of $\bar{x}$ is well approximated by a normal curve, even when the population distribution is not normal.

• A common guideline: CLT can be applied if n > 30.

Example: Soft-Drink Bottler

• Claim: Cans contain an average of 12 oz of soda.

• Let $x$ = actual volume of soda in a randomly selected can.

• $x$ is normally distributed with $\,sigma = 0.16$ oz.

• $n = 16$ cans are randomly selected, and $\bar{x}$ is calculated.

• If the claim is correct, the sampling distribution of $\bar{x}$ is normal with:

  • Mean: $\,mu_{\bar{x}} = \mu = 12$ oz

  • Standard Deviation: $\,sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{0.16}{\sqrt{16}} = 0.04$ oz

Soda Problem Continued

• What is P(11.96 < \bar{x} < 12.08)?

• Standardize the endpoints:

  • $z_1 = \frac{11.96 - 12}{0.04} = -1$

  • $z_2 = \frac{12.08 - 12}{0.04} = 2$

• P(-1 < Z < 2) = P(Z < 2) - P(Z < -1) = 0.9772 - 0.1587 = 0.8185

Example: Hot Dog Manufacturer

• Claim: Average fat content of 18 grams per hot dog with $\,sigma = 1$ gram.

• Consumers would be unhappy if the mean exceeded 18 grams.

• An independent testing organization analyzes a random sample of $n = 36$ hot dogs.

• Suppose the sample mean is $\bar{x} = 18.4$ grams. Does this suggest the claim is incorrect?

• Because n > 30, the Central Limit Theorem applies, and the distribution of $\bar{x}$ is approximately normal.

Hot Dogs Continued

• The mean is $\,mu = 18$

• The standard deviation = $\,sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{1}{\sqrt{36}} = \frac{1}{6} = 0.167$

• P(\bar{x} > 18.4) = P(Z > \frac{18.4 - 18}{0.167}) = P(Z > 2.40) = 1 - 0.9918 = 0.0082

• Values of $\bar{x}$ at least as large as 18.4 would be observed only about 0.82% of the time if the claim were true.

• The sample mean of 18.4 is large enough the claim might be incorrect.

Sampling Distribution of $\hat{p}$

• Illustrative experiment: Toss a penny 20 times, record the number of heads, and calculate the sample proportion of heads.

• Mark the proportion on a dot plot.

• Repeat many times to create a partial graph of the sampling distribution of sample proportions ( $\,hat{p}$ ).

• This is a statistic!

• What would happen if we flipped the penny 50 times?

Definition: Sampling Distribution of $\hat{p}$

• The distribution formed by considering the value of the sample statistic $\,hat{p}$ for every possible sample of a given size from a population.

• $\,hat{p}$ represents the sample proportion.

Example: Students

• Population: Six students (Alice, Ben, Charles, Denise, Edward, & Frank).

• Parameter of interest: Proportion of females.

• Population proportion of females: $\,frac{1}{3}$

• Select samples of two from this population.

• Number of possible samples: $<em>{6}C</em>2=15$

Finding All Possible Samples and Sample Proportions

• List all 15 possible samples and the sample proportion of females in each:

  • Alice & Ben: $\,hat{p} = 0.5$

  • Alice & Charles: $\,hat{p} = 0.5$

  • Alice & Denise: $\,hat{p} = 1$

  • Alice & Edward: $\,hat{p} = 0.5$

  • Alice & Frank: $\,hat{p} = 0.5$

  • Ben & Charles: $\,hat{p} = 0$

  • Ben & Denise: $\,hat{p} = 0.5$

  • Ben & Edward: $\,hat{p} = 0$

  • Ben & Frank: $\,hat{p} = 0$

  • Charles & Denise: $\,hat{p} = 0.5$

  • Charles & Edward: $\,hat{p} = 0$

  • Charles & Frank: $\,hat{p} = 0$

  • Denise & Edward: $\,hat{p} = 0.5$

  • Denise & Frank: $\,hat{p} = 0.5$

  • Edward & Frank: $\,hat{p} = 0$

• Calculate the mean and standard deviation of these sample proportions.

• How does the mean of the sampling distribution compare to the population parameter ( $p$ )?

General Properties for Sampling Distributions of $\hat{p}$

• Rule 1: $\,mu_{\hat{p}} = p$

• Rule 2: $\,sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$

  • This rule is exact if the population is infinite.

  • This rule is approximately correct if the population is finite and no more than 10% of the population is included in the sample.

Example: Cal Poly Students

• Fall 2008: 18,516 students enrolled at Cal Poly SLO.

• 8091 (43.7%) were female ( $p = 0.437$ ).

• Statistical software simulates sampling from this population.

• 500 samples are generated for each of the following sample sizes: $n = 10, 25, 50, 100$.

• Histograms display the distribution of sample proportions for each sample size.

Cal Poly Students Continued

• The histograms are centered around the true proportion ( $p = 0.437$ ).

• What do you notice about the standard deviation of these distributions?

• What about the shape of these distributions?

Example: Viral Hepatitis after Blood Transfusions

• Study reported that hepatitis occurs in 7% of patients who receive blood transfusions during heart surgery ( $p = 0.07$ ).

• Simulate sampling from a population of blood recipients.

• Generate 500 samples for each of the following sample sizes: $n = 10, 25, 50, 100$.

• Histograms show sample proportion distributions for each sample size.

Blood Transfusions Continued

• The histograms are centered around the true proportion ( $p = 0.07$ ).

• What happens to the shape of these histograms as the sample size increases?

Rule 3: Normality for Proportions

• When $n$ is large and $p$ is not too near 0 or 1, the sampling distribution of $\,hat{p}$ is approximately normal.

• The further $p$ is from 0.5, the larger $n$ must be for the sampling distribution of $\,hat{p}$ to be approximately normal.

• A conservative rule of thumb: If np > 10 and n(1 - p) > 10, then a normal distribution provides a reasonable approximation to the sampling distribution of $\,hat{p}$.

Blood Transfusions Revisited

• $p$ = proportion of patients who contract hepatitis after a blood transfusion = 0.07

• Suppose a new blood screening procedure is believed to reduce the incidence rate of hepatitis.

• Blood screened using this procedure is given to $n = 200$ blood recipients.

• Only 6 of the 200 patients contract hepatitis ($\hat{p} = \frac{6}{200} = 0.03$).

• Does this indicate that the true proportion is less than 7%?

• To answer this, consider the sampling distribution of $\,hat{p}$.

Checking Conditions for Normality

• First, is the sampling distribution approximately normal?

• Check the conditions:

  • np = 200(0.07) = 14 > 10

  • n(1 - p) = 200(0.93) = 186 > 10

• Yes, we can use a normal approximation.

Calculating the Probability

• Assume the screening procedure is not effective and $p = 0.07$.

• Calculate P(\,hat{p} < 0.03).

• The Standard deviation is $\,sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.07(1-0.07)}{200}} = 0.018$

• The Z score is $\frac{0.03-0.07}{0.018} = -2.22$

• Then P(Z<-2.22) = 0.0132

• This small probability tells us that it is unlikely that a sample proportion of 0.03 or smaller would be observed if the screening procedure was ineffective.

• This screening procedure appears to yield a smaller incidence rate for hepatitis.