Sampling Distributions for Sample Means - AP Statistics Study Guide

Quick Summary

This guide will enable you to describe the sampling distribution of a sample mean ($\bar{x}$), which is the distribution of all possible sample means for a given sample size. You will master the ability to determine its shape, center, and spread by applying the Central Limit Theorem and correctly checking the necessary conditions for normality and standard deviation calculations. Ultimately, you will be able to calculate the probability of observing a specific sample mean from a given population.

Key Concepts

The core idea of a sampling distribution is a shift in thinking: instead of looking at one sample, we are looking at the distribution of a statistic (in this case, the sample mean, $\bar{x}$) that would be formed from every possible sample of a certain size.

[Image: Diagram showing a large population distribution. Multiple samples of size n are drawn from it. The mean ($\bar{x}$) is calculated for each sample. These sample means are then plotted, forming a new, less spread-out distribution called the sampling distribution of $\bar{x}$.]

1. Center: The Mean of the Sampling Distribution (${\mu }_{\bar{x}}$)

The mean of the sampling distribution of $\bar{x}$ is an unbiased estimator of the true population mean, $\mu$. This means that if you were to average all possible sample means, you would get the exact population mean.

• Formula:${\mu }_{\bar{x}}=\mu$

• In Words: The center of the distribution of sample means is the true mean of the population.

• Key Implication: The sample mean $\bar{x}$ is a good guess for the population mean $\mu$. It doesn't systematically over or underestimate the true value.

2. Spread: The Standard Deviation of the Sampling Distribution (${\sigma }_{\bar{x}}$)

This value, often called the standard error of the mean, measures the typical amount that a sample mean $\bar{x}$ varies from the population mean $\mu$.

• Formula:${\sigma }_{\bar{x}}=\frac{\sigma }{\sqrt{n}}$

  • $\sigma$ is the standard deviation of the population.

  • $n$ is the sample size.

• Key Implication: As the sample size ($n$) increases, the standard deviation of the sampling distribution decreases. This means larger samples give more precise estimates; their sample means will be more tightly clustered around the true population mean $\mu$.

• The 10% Condition (Independence): This formula is only valid if the observations in the sample are independent. When sampling without replacement, we must check that the sample size is no more than 10% of the population size ($n\le 0.10N$). If this condition is met, we can proceed as if the observations are independent.

3. Shape: The Condition for Normality

The shape of the sampling distribution of $\bar{x}$ tells us if we can use a Normal distribution to calculate probabilities. There are two ways the sampling distribution can be Normal or Approximately Normal.

• Condition 1: The Population Distribution is Normal.

  • If the original population from which the samples are drawn is stated to be Normally distributed, then the sampling distribution of $\bar{x}$ will also be Normal, regardless of the sample size n.

• Condition 2: The Central Limit Theorem (CLT).

  • This is one of the most important theorems in all of statistics.

  • The CLT states: If the sample size $n$ is sufficiently large, the sampling distribution of $\bar{x}$ will be approximately Normal, even if the original population distribution is not Normal (e.g., it could be skewed or bimodal).

  • Rule of Thumb: A sample size of $n\ge 30$ is generally considered "sufficiently large" for the CLT to apply.

[Image: A series of four graphs. The first shows a right-skewed population distribution. The next three show the sampling distributions of $\bar{x}$ for n=2, n=10, and n=30. The graphs clearly illustrate the distribution becoming more symmetric and bell-shaped as n increases, with the n=30 graph looking approximately Normal.]

Key Vocabulary

• Sampling Distribution of a Sample Mean ($\bar{x}$): The probability distribution of the statistic $\bar{x}$ based on all possible simple random samples of the same size from the same population.

• Central Limit Theorem (CLT): The theorem stating that the sampling distribution of the sample mean will be approximately normal if the sample size is sufficiently large ($n\ge 30$), regardless of the population's shape.

• Population Mean ($\mu$): A parameter representing the true arithmetic average of a quantitative variable for an entire population.

• Sample Mean ($\bar{x}$): A statistic representing the arithmetic average of a quantitative variable for a single sample. It is used to estimate the population mean.

• Standard Error of the Mean: The standard deviation of the sampling distribution of the sample mean (${\sigma }_{\bar{x}}=\frac{\sigma }{\sqrt{n}}$). It quantifies the precision of the sample mean as an estimate of the population mean.

• Unbiased Estimator: A statistic whose sampling distribution has a mean equal to the true value of the parameter it is intended to estimate. $\bar{x}$ is an unbiased estimator of $\mu$.

• 10% Condition: A condition that must be checked when sampling without replacement to ensure that observations are reasonably independent. It requires that the sample size $n$ is no more than 10% of the population size $N$.

Calculator Tech (TI-84)

The primary calculator function for this topic is $normalcdf$, used to find the probability for a specific interval on a Normal distribution.

• Function:$normalcdf(lowerbound,upperbound,mean,standarddeviation)$

• How to Use for Sampling Distributions:

  • $lowerbound$: The lower value of the $\bar{x}$ you are interested in.

  • $upperbound‘:Theuppervalueofthe$\bar{x}you are interested in. - $mean: Use the mean of the sampling distribution, ${\mu }_{\bar{x}}=\mu$.

  • $standarddeviation‘:Usethestandarddeviationofthesamplingdistribution,$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}. ## How to Show Work on the FRQ Free Response Questions involving sampling distributions require you to do more than just calculate a number. You must clearly communicate your understanding by defining the sampling distribution's characteristics and checking the necessary conditions. ### Template for Describing the Sampling Distribution of\bar{x}$Whenaskedtodescribethesamplingdistribution,youmustaddressitsShape,Center,andSpread(SCS).1.\ast \ast Shape:\ast \ast JustifywhythedistributionisNormalorApproximatelyNormal.\ast "Becausethepopulationof[context]isstatedtobeNormallydistributed,thesamplingdistributionofthesamplemean$\bar{x}is also Normal." * **OR** "Because the sample size $n = [value] is $\ge 30$, the Central Limit Theorem (CLT) applies. Therefore, the sampling distribution of the sample mean $\bar{x}$ is approximately Normal."

2. Center: State the mean of the sampling distribution.

   • "The mean of the sampling distribution is ${\mu }_{\bar{x}}=\mu =$ [state value]."

3. Spread: State the standard deviation of the sampling distribution, making sure to check the 10% condition.

   • "The standard deviation of the sampling distribution is ${\sigma }_{\bar{x}}=\frac{\sigma }{\sqrt{n}}=\frac{[\text{value}]}{\sqrt{[\text{value}]}}=$ [calculated value]. We can use this formula because our sample size n = [value] is less than 10% of the total population of all [context] (assuming this population is at least [10 times n])."

Template for Calculating a Probability

Follow the four-step process: State, Plan, Do, Conclude.

1. State: Define the probability you are trying to find in terms of $\bar{x}$.

   • "We want to find the probability that the mean [context] in a random sample of size $n$ is [greater than/less than] [value]. P(\bar{x} > \text{value})`" 2. **Plan:** Describe the sampling distribution using the SCS template above. This step is where you check all conditions. 3. **Do:** Perform the calculation. Show your work for the z-score and state the calculator command used. A sketch is highly recommended. * $z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}} = \frac{[\text{value}] - [\text{value}]}{[\text{value}]} = [\text{z-score}]

   • P(\bar{x} > \text{value}) = P(Z > \text{z-score}) =` [probability]. * (Calculator input: `normalcdf(lower, upper, mean, std dev)`) 4. **Conclude:** Write a sentence that answers the question in the context of the problem. * "There is a [probability] probability of selecting a random sample of size `n` that has a mean [context] of at least [value]." ## Practice Problems **Problem 1:** The weights of individual bags of trail mix produced at a factory are known to follow a Normal distribution with a mean\mu$ of 16.1 ounces and a standard deviation $\sigma$ of 0.4 ounces. A quality control inspector randomly selects 9 bags. What is the probability that the mean weight of the 9 selected bags is less than 16.0 ounces?

Solution:

We want to find the probability that the mean weight of a random sample of 9 bags is less than 16.0 ounces, or P(\bar{x} < 16.0).

First, we must describe the sampling distribution of the sample mean weight, $\bar{x}$.

• Shape: Because the population of bag weights is stated to be Normally distributed, the sampling distribution of $\bar{x}$ is also Normal, even with a small sample size (n=9).

• Center: The mean of the sampling distribution is ${\mu }_{\bar{x}}=\mu =16.1$ ounces.

• Spread: The standard deviation of the sampling distribution is ${\sigma }_{\bar{x}}=\frac{\sigma }{\sqrt{n}}=\frac{0.4}{\sqrt{9}}\approx 0.1333$ ounces. We can use this formula because a sample of 9 bags is surely less than 10% of all bags of trail mix produced at the factory.

Now we can calculate the probability. The z-score is $z=(16.0-16.1)/0.1333=-0.75$. The probability is $P(\bar{x}<16.0)=P(Z<-0.75)\approx 0.2266$.

(Using $normalcdf(lower:-1E99,upper:16.0,mean:16.1,stddev:0.1333)‘).Inconclusion,thereisapproximatelya0.2266probabilitythatarandomsampleof9bagswillhaveameanweightlessthan16.0ounces.\ast \ast Problem2:\ast \ast Alargeinternetserviceproviderclaimsthattheaveragedownloadspeedforitscustomersis100Mbps.However,theactualspeedsvary,withastandarddeviationof20Mbps.Thedistributionofdownloadspeedsisknowntoberight-skewed.Anindependentagencytakesasimplerandomsampleof100customerstotesttheclaim.Whatistheprobabilitythattheaveragedownloadspeedinthissampleisgreaterthan104Mbps?\ast \ast Solution:\ast \ast Wewanttofindtheprobabilitythatthemeandownloadspeedinarandomsampleof100customersisgreaterthan104Mbps,or‘P(\bar{x}>104)‘.First,wemustdescribethesamplingdistributionofthesamplemeandownloadspeed,$\bar{x}. * **Shape:** Although the population distribution of download speeds is right-skewed, our sample size is $n = 100, which is greater than 30. Therefore, by the Central Limit Theorem (CLT), the sampling distribution of $\bar{x}$ is approximately Normal.

• Center: The mean of the sampling distribution is ${\mu }_{\bar{x}}=\mu =100$ Mbps.

• Spread: The standard deviation of the sampling distribution is ${\sigma }_{\bar{x}}=\frac{\sigma }{\sqrt{n}}=\frac{20}{\sqrt{100}}=2$ Mbps. We can use this formula because the sample of 100 customers is less than 10% of all customers of a large internet service provider.

Now we can calculate the probability. The z-score is $z=(104-100)/2=2$. The probability is $P(\bar{x}>104)=P(Z>2)\approx 0.0228$.

(Using normalcdf(lower: 104, upper: 1E99, mean: 100, std dev: 2)).

In conclusion, there is approximately a 0.0228 probability of observing a sample mean download speed of 104 Mbps or higher in a random sample of 100 customers, if the true mean speed is 100 Mbps.

Common Mistakes to Avoid

• Confusing the Sample with the Sampling Distribution: A very common error is to say "the sample is approximately normal." This is incorrect. The sampling distribution of the mean is what becomes approximately normal due to the CLT. The distribution of data within one large sample will still reflect the shape of the population (e.g., it will still be skewed if the population is skewed).

• Forgetting to Divide by the Square Root of n: When calculating the standard deviation for the sampling distribution (${\sigma }_{\bar{x}}$), students often forget to divide the population standard deviation $\sigma$ by $\sqrt{n}$. This leads to a drastically incorrect spread and wrong probability calculations.

• Misstating the Central Limit Theorem: Do not say the CLT makes the population or a sample normal. The CLT applies only to the shape of the sampling distribution of a statistic (like $\bar{x}$ or $\hat{p}$). Also, only invoke the CLT when the sample size is large ($n\ge 30$) and the population is not already known to be normal. If the population is normal, the sampling distribution is normal because of that fact, not because of the CLT.

• Failing to Check the 10% Condition: On an FRQ, you must explicitly state and check the 10% condition ($n\le 0.10N$) to justify the calculation of the standard deviation. Simply writing the formula is not enough.

• Using Incorrect Notation: Be precise. $\mu$ is the population mean. $\bar{x}$ is the sample mean. ${\mu }_{\bar{x}}$ is the mean of the sampling distribution of the sample means. Using these interchangeably will result in a loss of credit.