Subjects statistics

Sampling Distribution C26Ac4

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1. **Problem Statement:** We want to understand the mean and variance of the sampling distribution of the sample mean, including how to calculate them and their characteristics when the population variance is known or unknown. 2. **Definitions:** - The **sampling distribution of the sample mean** is the probability distribution of the means of all possible samples of size $n$ drawn from a population. - The **mean of the sampling distribution** (also called the expected value) is denoted as $\mu_{\bar{x}}$. - The **variance of the sampling distribution** is denoted as $\sigma^2_{\bar{x}}$. 3. **Formulas and Important Rules:** - The mean of the sampling distribution of the sample mean is equal to the population mean: $$\mu_{\bar{x}} = \mu$$ - The variance of the sampling distribution of the sample mean is the population variance divided by the sample size: $$\sigma^2_{\bar{x}} = \frac{\sigma^2}{n}$$ - The standard deviation of the sampling distribution (called the standard error) is: $$\text{SE} = \frac{\sigma}{\sqrt{n}}$$ 4. **When Population Variance is Known:** - The sampling distribution of the sample mean is normally distributed (or approximately normal for large $n$) with mean $\mu$ and standard error $\frac{\sigma}{\sqrt{n}}$. - This means sample means tend to cluster around the population mean with variability decreasing as sample size increases. 5. **When Population Variance is Unknown:** - We estimate the population variance using the sample variance $s^2$. - The standard error is estimated as: $$\text{SE} = \frac{s}{\sqrt{n}}$$ - The sampling distribution of the sample mean follows a $t$-distribution with $n-1$ degrees of freedom, which accounts for extra uncertainty. 6. **Example Calculation:** Suppose the population mean $\mu = 50$, population variance $\sigma^2 = 25$, and sample size $n = 25$. - Mean of sampling distribution: $$\mu_{\bar{x}} = 50$$ - Variance of sampling distribution: $$\sigma^2_{\bar{x}} = \frac{25}{25} = 1$$ - Standard error: $$\text{SE} = \frac{5}{\sqrt{25}} = \frac{5}{5} = 1$$ This means the sample means will be centered at 50 with a standard deviation of 1. 7. **Interpretation:** - A smaller standard error means sample means are more tightly clustered around the population mean, indicating more reliable estimates. - Cassielle’s observation of consistent average ratings with low variance suggests her sample means are stable and trustworthy.