Stats

Standard Deviation of the Sample Mean ( $\sigma_{\bar{x}}$ ): This is also referred to as the Standard Error or the standard deviation of the mean of the sample. * Notation: * $\mu_{\bar{x}}$ : Represents the mean of the sample mean. * $\sigma_{\bar{x}}$ : Represents the standard deviation of the sample mean. * Formula and Calculation Example: * If given a value of $9.6$ and a sample division is required, for example: $\frac{9.6}{8} = 1.2$ . Here, $1.2$ is the standard error. * Problem 1: If $\mu = 57$ , $\sigma = 42$ , and sample size $n = 49$ : * $\mu_{\bar{x}} = \mu = 57$ . * $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{42}{\sqrt{49}} = \frac{42}{7} = 6$ . * Problem 2: If $\sigma = 25$ and $n = 100$ , the standard error is $\frac{25}{\sqrt{100}} = \frac{25}{10} = 2.5$ .

Conceptual Differences: * Standard Normal Distribution ( $z$ ): Has an established mean of $0$ and a standard deviation of $1$ . * t-Distribution: The mean of the $t$ -distribution is not always fixed at zero in every situation in the same way as the standard normal; it depends on the specific situation or degrees of freedom. It is crucial not to mix up the mean of the standard normal with the mean of the $t$ -distribution.

Scenario: Heights of Men * Parameters: Mean $\mu = 68.7\,inches$ , standard deviation $\sigma = 3.5\,inches$ . * Sample Case: If one man is randomly selected, find the probability that their height is greater than $69.7\,inches$ . This uses the standard population deviation $\sigma = 3.5$ . * Sample Mean Case ( $n=100$ ): If a sample of $100$ men is selected, find the probability that their mean height $\bar{x}$ is greater than $69.7$ . * $\mu_{\bar{x}} = 68.7$ . * $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{3.5}{\sqrt{100}} = \frac{3.5}{10} = 0.35$ . * Calculator Steps (Normal CDF): * Lower Bound: $69.7$ . * Upper Bound: Infinity (entered as $E99$ ). * Mean ( $\mu_{\bar{x}}$ ): $68.7$ . * Standard Deviation ( $\sigma_{\bar{x}}$ ): $0.35$ . * Result: $0.0021$ .

Point Estimate for Population Proportion ( $p$ ): The best point estimate for the population proportion is the sample proportion, denoted as $\hat{p}$ . * Formula: $\hat{p} = \frac{x}{n}$ , where $x$ is the number of successes and $n$ is the sample size. * Example Case: A village received $628$ completed surveys out of $800$ . * $\hat{p} = \frac{628}{800} = 0.785$ .

Mean of the Sampling Distribution ( $\mu_{\hat{p}}$ ): $\mu_{\hat{p}} = p$ .
Standard Deviation of the Sampling Distribution ( $\sigma_{\hat{p}}$ , also called Standard Error of the Proportion): * Formula: $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$ . * Example Calculation: If $p = 37\%$ (or $0.37$ ) and $n = 144$ : * $1 - p = 0.63$ . * $\sigma_{\hat{p}} = \sqrt{\frac{0.37 \times 0.63}{144}}$ .

Large Sample Condition: The distribution of $\hat{p}$ is approximately normal if $n \times p \times (1 - p) \geq 10$ . * Example Check: For $n = 144, p = 0.37, (1-p) = 0.63$ : * $144 \times 0.37 \times 0.63 = 33.5$ . * Since $33.5 \geq 10$ , the distribution is approximately normal.
Finite Population Condition: If sampling from a finite population of size $N$ , the sample size $n$ must be less than or equal to $5\%$ of the population ( $n \leq 0.05N$ ). * Example Check: If population $N = 23,000$ and sample $n = 400$ : * $5\% \text{ of } 23,000 = 0.05 \times 23,000 = 1,150$ . * Since $400 \leq 1,150$ , the requirement is met.

Structure: Confidence intervals are expressed as $\hat{p} \pm E$ , where $E$ is the margin of error. The sample proportion $\hat{p}$ is always located exactly in the middle of the confidence interval.
Calculating Margin of Error ( $E$ ): * $E = z_{\alpha/2} \times \sigma_{\hat{p}}$ . * $E = z_{\alpha/2} \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ .
Manual Calculation vs. Calculator: * Calculator Tool: Use 1-PropZInt (One-proportion z-interval). * Example - University Financial Aid: $n = 200, x = 118$ , Confidence Level = $98\%$ . * $\hat{p} = \frac{118}{200} = 0.59$ . * Critical Value ( $z_{\alpha/2}$ ) for $98\%$ confidence is $2.33$ . * Interval result from calculator: $(0.5091, 0.671)$ . * Finding Margin of Error from the interval: Take the upper endpoint and subtract $\hat{p}$ : $0.671 - 0.59 = 0.081$ . * Final form: $0.59 \pm 0.081$ .

Question: What if we are finding the probability a single male has a height greater than a specific value instead of a sample mean? * Response: If you are dealing with an individual ( $x$ ) rather than a mean ( $\bar{x}$ ), you do not use the standard error ( $\sigma_{\bar{x}}$ ). Instead, you use the population standard deviation ( $\sigma$ ) in your calculations.
Question: How do I find the margin of error if I only have the interval from the calculator? * Response: You can determine the margin of error by taking the upper limit of the interval and subtracting the sample mean (or sample proportion) from it. Alternatively, calculate the total width of the interval and divide by two.
Interaction Note: The instructor emphasizes checking the top of the calculator screen to ensure the correct function (e.g., Normal CDF vs. Test) is being used before submitting an answer. Writing down the steps used on the calculator is encouraged to help earn partial credit even if a calculation error occurs.