Sampling Distribution and Central Limit Theorem

In statistical studies, researchers often sample individuals and compute numerical summaries.
The sample mean $\bar{x}$ is frequently computed.
If multiple samples are drawn, the values of $\bar{x}$ will likely vary.
Therefore, $\bar{x}$ is considered a random variable.

For each value of the random variable $\bar{x}$ , a probability can be computed.
The probability distribution of $\bar{x}$ is known as the sampling distribution of $\bar{x}$ .

Tetrahedral dice are pyramid-shaped with four faces, each numbered from 1 to 4.
Tossing a tetrahedral die simulates sampling from the population {1, 2, 3, 4}.
Population mean: $\mu = 2.5$
Population standard deviation: $\sigma = 1.118$
If a tetrahedral die is tossed three times, this is a sample of size 3 drawn with replacement.
Consider all possible samples of size 3 and their respective sample means $\bar{x}$ .
The mean of all $\bar{x}$ values is $\mu_{\bar{x}} = 2.5$ .
The standard deviation of all $\bar{x}$ values is $\sigma_{\bar{x}} = 0.6455$ .

The mean of the sampling distribution $\mu{\bar{x}} = 2.5$ is equal to the population mean $\mu = 2.5$ . This relationship always holds: $\mu{\bar{x}} = \mu$ .
The standard deviation of the sampling distribution $\sigma_{\bar{x}} = 0.6455$ is less than the population standard deviation $\sigma = 1.118$ .
The relationship between these quantities is: $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$ , where $n$ is the sample size.

Also known as the standard error, it is denoted by $\sigma{\bar{x}}$ and is calculated as: $\sigma{\bar{x}} = \frac{\sigma}{\sqrt{n}}$ .

The mean number of hours of television watched per week by college students is $\mu = 10.5$ with a standard deviation of $\sigma = 3.6$ .
A random sample of 16 students is chosen ( $n = 16$ ).
The mean of the sampling distribution is $\mu_{\bar{x}} = \mu = 10.5$ .
The standard deviation of the sampling distribution is $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{3.6}{\sqrt{16}} = 0.9$ .

In the tetrahedral die example, each number has a probability of 0.25.
The probability histogram for the population is uniform.
The probability histogram for the sampling distribution of $\bar{x}$ with $n = 3$ resembles a normal curve.

For any population, if the sample size is large enough, the sample mean $\bar{x}$ will be approximately normally distributed.
For symmetric populations, $\bar{x}$ is approximately normally distributed even for small sample sizes (e.g., $n = 3$ ).
For skewed populations, a larger sample size is needed for the sampling distribution of $\bar{x}$ to be approximately normal.
A sample size of n > 30 is generally considered large enough.
The CLT states that if $\bar{x}$ is the mean of a large (n > 30) simple random sample, then $\bar{x}$ has an approximately normal distribution with:
- Mean: $\mu_{\bar{x}} = \mu$
- Standard deviation: $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$
If the population is normal, the sample mean $\bar{x}$ will be normal for any sample size.

Example 1: A sample of size 45 is drawn from a population with $\mu = 15$ and $\sigma = 3.5$ . Since n > 30, $\bar{x}$ is approximately normally distributed.
Example 2: A sample of size 8 is drawn from a normal population with $\mu = -60$ and $\sigma = 5$ . Since the population is normal, $\bar{x}$ is approximately normally distributed.
Example 3: A sample of size 24 is drawn from a population with $\mu = 35$ and $\sigma = 1.2$ . Since $n$ is not greater than 30 and the population is not known to be normal, we cannot be certain that $\bar{x}$ has an approximately normal distribution.

The mean age of college students is $\mu = 25$ years with a standard deviation of $\sigma = 9.5$ years.
A simple random sample of 125 students is drawn ( $n = 125$ ).
The goal is to find the probability that the sample mean age is greater than 26.
Since n > 30, the CLT applies.
Mean of the sampling distribution: $\mu_{\bar{x}} = \mu = 25$
Standard error: $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{9.5}{\sqrt{125}} = 0.85$
Using the normal curve with $\mu = 25$ and $\sigma_{\bar{x}} = 0.85$ , we find the probability that \bar{x} > 26.
Using Excel: =1-NORM.DIST(26, 25, 0.85, TRUE) yields a result of 0.1197.

The mean weight of a one-year-old Hereford bull is $\mu = 1135$ pounds with a standard deviation of $\sigma = 97$ pounds.
The question is whether it would be unusual for the mean weight of 100 head of cattle to be less than 1100 pounds ( $n = 100$ ).
Since n > 30, the normal curve can be used.
Mean of the sampling distribution: $\mu_{\bar{x}} = \mu = 1135$
Standard error: $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{97}{\sqrt{100}} = 9.7$
Finding the probability that \bar{x} < 1100.
Using Excel: =NORM.DIST(1100, 1135, 9.7, TRUE) yields a probability of 0.00015.
Since this probability is much smaller than 0.05, it would be unusual.

The mean age of college students is $\mu = 25$ years with a standard deviation of $\sigma = 9.5$ years.
A simple random sample of 125 students is drawn ( $n = 125$ ).
The goal is to find the 30th percentile of the sample mean $\bar{x}$ .
Since n > 30, the sampling distribution of $\bar{x}$ is normally distributed.
Mean of the sampling distribution: $\mu_{\bar{x}} = \mu = 25$
Standard error: $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{9.5}{\sqrt{125}} = 0.8497$
The 30th percentile is the value with an area of 0.3 to its left.
Using Excel: =NORM.INV(0.3, 25, 0.8497) yields a value of approximately 24.55.
Using Table A-2, a z-score of -0.52 is found.
The value from the sampling distribution is: $x = \mu{\bar{x}} + z \times \sigma{\bar{x}} = 25 + (-0.52) \times 0.8497 = 24.56$ .